Vector algebra problems and solutions pdf class 11 solutions english ...

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Vector algebra problems and solutions pdf class 11 solutions english answers

Show that each of the given three vectors is a unit vector, which are mutually perpendicular to each other. First one is a Scalar (or dot) product where the result is a scalar, whereas in the second one is a Vector (or cross) product of two vectors where the result is a vector.10.6.1 Scalar (or dot) Product of Two VectorsIn these two non zero vectors, a and b are multiplied with respect to cos which denotes the angle between both the vectors.The scalar product can be written as:a.b=|a||b| cosProperty 1: Distributivity of scalar product over addition.Let \[\overrightarrow{a}\], \[\overrightarrow{b}\] and \[\overrightarrow{c}\] be any three vectors, then:\[\overrightarrow{a}\] . A girl walks 4 km towards west, then she walks 3 km in a direction ${30^\circ }$ east of north and stops. Not only for the understanding part but it also provides important practice questions made on CBSE pattern. $0 < \theta < \dfrac{\pi }{2}$b. Justify your answer with an example.Ans: Let $\vec a = 2\hat i + 3\hat j + 4\hat k\;,\;\vec b = 4\hat i + 6\hat j + 8\hat k\;,\;\vec a \times \vec b = \vec 0$$\vec a \times \vec b = \left| {\begin{array}{*{20}{l}} {\hat i}&{\hat j}&{\hat k} \\ 2&3&4 \\ 4&6&8 \end{array}} \right| = \hat i(24 - 24) - \hat j(16 - 16) + \hat k(12 - 12) = 0\hat i + 0\hat j + 0\hat k$$|\vec a| = \sqrt {{2^2} + {3^2} + {4^2}} = \sqrt {29} $$\therefore \vec a e \vec 0$$|\vec b| = \sqrt {{4^2} + {6^2} + {8^2}} = \sqrt {116} $$\therefore \vec b e \vec 0$Hence, the converse of the statement need not be true.9. Find the area of the triangle with vertices A $(1,1,2)$, B $(2,3,5)$ and $C(1,5,5)$. Ans: $\overrightarrow {{\text{AB}}} = (2 - 1)\hat i + (3 - 1)\hat j + (5 - 2)\hat k = \hat i + 2\hat j + 3\hat k$$\overrightarrow {{\text{BC}}} = (1 2)\hat i + (5 - 3)\hat j + (5 - 5)\hat k = - \hat i + 2\hat j$Area $ = \dfrac{1}{2}|\overrightarrow {{\text{AB}}} \times \overrightarrow {{\text{BC}}} |$$\overrightarrow {{\text{AB}}} \times \overrightarrow {{\text{BC}}} = \left| {\begin{array}{*{20}{l}} {\hat i}&{\hat j}&{\hat k} \\ 1&2&3 \\ { - 1}&2&0 \end{array}} \right| = \hat i( - 6) - \hat j(3) + \hat k(2 + 2) = - 6\hat i - 3\hat j + 4\hat k$$|\overrightarrow {{\text{AB}}} \times \overrightarrow {{\text{BC}}} | = \sqrt {{{( - 6)}^2} + {{( - 3)}^2} + {4^2}} = \sqrt {36 + 9 + 16} = \sqrt {61} $So, the area of $\Delta {\text{ABC}}$ is $\dfrac{{\sqrt {61} }}{2}$ sq units.10. Find the area of the parallelogram whose adjacent sides are determined by the vector $\vec a = \hat i - \hat j + 3\hat k$ and $\overrightarrow {\text{b}} = 2\hat i - 7\hat j + \hat k$Ans: $\vec a \times \overrightarrow {\text{b}} = \left| {\begin{array}{*{20}{c}} {\hat i}&{\hat j}&{\hat k} \\ 1&{ - 1}&3 \\ 2&{ - 7}&1 \end{array}} \right| = \hat i( - 1 + 21) - \hat j(1 - 6) + \hat k( - 7 + 2) = 20\hat i + 5\hat j 5\hat k$$\therefore |\vec a \times \vec b| = \sqrt {{{20}^2} + {5^2} + {5^2}} = \sqrt {400 + 25 + 25} = 15\sqrt 2 $So, area of parallelogram is $15\sqrt 2 $ sq units11. Additive Identity: For any vector a,\[\overrightarrow{a}\] + \[\overrightarrow{0}\] = \[\overrightarrow{0}\] + \[\overrightarrow{a}\] = \[\overrightarrow{a}\]10.5 Multiplication of a Vector by a ScalarIncase of multiplication of vector \[\overrightarrow{(a)}\] by a scalar say (\[\gamma\] ):\[\gamma\] \[\overrightarrow{(a)}\] = | \[\overrightarrow{\gamma \;a}\] |The direction of the vector \[\overrightarrow{\gamma \;a}\] depends upon the value of " \[\gamma\] " .If " \[\gamma\] " is positive, the direction will be the same but if " \[\gamma\] " is negative, the direction will be opposite.This can be related to the additive inverse where,\[\overrightarrow{a}\] + \[\overrightarrow{(-a)}\] = \[\overrightarrow{(-a)}\] + \[\overrightarrow{a}\] = \[\overrightarrow{0}\]10.5.1 Components of a Vectorr = x ? + y + z k The above form of any vector is known as the component form of a vector in a 3D plane.Where, OA = x ?OB = yOC = z k Magnitude of such a vector can be calculated by using pythagoras theorem:r = x ? + y + z k |r| = ( x2 + y2 + z2)10.5.2 Vector Joining Two PointsA vector obtained by joining two points P1 and P2 in a plane will be P1P2.(Image to be added soon)In order to obtain magnitude (P1P2) you can apply the triangle law:\[\overrightarrow{OP_{1}}\] + \[\overrightarrow{P_{1} \; P_{2}}\] = \[\overrightarrow{OP_{2}}\](or)\[\overrightarrow{P_{1} \; P_{2}}\] = \[\overrightarrow{OP_{2}}\] - \[\overrightarrow{OP_{1}}\]That P1 P2 = (x2 ? + y2 + z2 k ) - (x1 ? + y1 + z1 k )= ( x2 ? x1) ? + ( y2 ? y1 ) + ( z2 ? z1 ) k Thus, the above equation represents the vector P1P2. Its magnitude can be given by:P1 P2 = [( x2-x1 )2 + ( y2-y1 )2 + ( z2-z1 )2]10.5.3 Section FormulaSection formula is a method to find the position of a line segment that divides a vector externally or internally in a given ratio.CASE 1: When PQ is divided internally by RYou can assume that PQ is divided by R in the ratio of m:n.(Image to be added soon)\[\overrightarrow{OR}\] = \[\frac{mb+na}{mn}\]CASE 2: When PQ is divided externally by RYou can assume that PQ is divided by R in the ratio of m:n.(Image to be added soon)\[\overrightarrow{OR}\] = \[\frac{mb-na}{mn}\]10.6 Product of Two VectorsIn this topic, you will understand the multiplication( product) of two vectors. Also, find its area. Ans: Diagonal of a parallelogram is $\vec a + \vec b$ $\vec a + \vec b = (2 + 1)\hat i + ( - 4 - 2)\hat j + (5 - 3)\hat k = 3\hat i - 6\hat j + 2\hat k$So, the unit vector parallel to diagonal is $\dfrac{{\vec a + \vec b}}{{|\vec a + \vec b|}} = \dfrac{{3\hat i - 6\hat j + 2\hat k}}{{\sqrt {{3^2} + {{( - 6)}^2} + {2^2}} }} = \dfrac{{3\hat i - 6\hat j + 2\hat k}}{{\sqrt {9 + 36 + 4} }} = \dfrac{{3\hat i - 6\hat j + 2\hat k}}{7} = \dfrac{3}{7}\hat i - \dfrac{6}{7}\hat j + \dfrac{2}{7}\hat k$$\vec a \times \vec b = \left| {\begin{array}{*{20}{r}} {\hat i}&{\hat j}&k \\ 2&{ - 4}&3 \\ 1&{ - 2}&{ - 3} \end{array}} \right|$$ = \hat i(12 + 10) - \hat j( 6 - 5) + \bar k( - 4 + 4)$$ = 22\hat i + 11\hat j$$ = 11(2\hat i + \hat j)$$\therefore |\vec a \times \vec b| = 11\sqrt {{2^2} + {1^2}} = 11\sqrt 5 $So, area of parallelogram is $11\sqrt 5 $ sq units11. 4Ans: $\overrightarrow {{\text{AB}}} = (1 + 1)\hat i + \left( {\dfrac{1}{2} - \dfrac{1}{2}} \right)\hat j + (4 - 4)\hat k = 2\hat i$$\overrightarrow {{\text{BC}}} = (1 - 1)\hat i + \left( { - \dfrac{1}{2} - \dfrac{1}{2}} \right)\hat j + (4 - 4)\hat k = - \hat j$$\overrightarrow {{\text{AB}}} \times \overrightarrow {{\text{BC}}} = \left| {\begin{array}{*{20}{c}} {\hat i}&{\hat j}&{\hat k} \\ 2&0&0 \\ 0&{ - 1}&0 \end{array}} \right| = \hat k( - 2) = - 2\hat k$$|\overrightarrow {{\text{AB}}} \times \overrightarrow {{\text{BC}}} | = \sqrt {{{( - 2)}^2}} = 2$So, the area of the required rectangle is 2 square units.Miscellaneous Exercise1. $\dfrac{\pi }{2}$d. The scalar product of the vector $\hat i + \hat j + \hat k$ with a unit vector along the sum of vectors $2\hat i + 4\hat j - 5\hat k$ ?nd $\lambda \hat i + 2\hat j + 3\hat k$ is equal to one. This has been done for better understanding and clarity of the topic for students. Find the value of $\lambda $. Ans: $(2\hat i + 4\hat j - 5\hat k) + (\lambda \hat i + 2\hat j + 3\hat k) = (2 + \lambda )\hat i + 6\hat j - 2\hat k$So, unit vector along $(2\hat i + 4\hat j - 5\hat k) + (\lambda \hat i + 2\hat j + 3\hat k)$ is $\left( {\dfrac{{(2 + \lambda )\hat i + 6\hat j - 2\hat k}}{{\sqrt {{\lambda ^2} + 4\lambda + 44} }}} \right)$$(\hat i + \hat j + \hat k) \cdot \left( {\dfrac{{(2 + \lambda )\hat i + 6\hat j - 2\hat k}}{{\sqrt {{\lambda ^2} + 4\lambda + 44} }}} \right) = 1$$ \Rightarrow \dfrac{{(2 + \lambda ) + 6 - 2}}{{\sqrt {{\lambda ^2} + 4\lambda + 44} }} = 1$$ \Rightarrow \sqrt {{\lambda ^2} + 4\lambda + 44} = \lambda + 6$$ \Rightarrow {\lambda ^2} + 4\lambda + 44 = {(\lambda + 6)^2}$$ \Rightarrow {\lambda ^2} + 4\lambda + 44 = {\lambda ^2} + 12\lambda + 36$$ \Rightarrow 8\lambda = 8$$ \Rightarrow \lambda = 1$14. ${40^\circ }$Ans: ${40^\circ }$ is a scalar quantity because it has only magnitude, not direction.iv. Write two different vectors having the same magnitude. Ans: $\vec a = (\hat i - 2\hat j + 3\hat k)$ and $\vec b = (2\hat i + \hat j - 3\hat k)$.$|\vec a| = \sqrt {{1^2} + {{( - 2)}^2} + {3^2}} = \sqrt {1 + 4 + 9} = \sqrt {14} $$|\vec b| = \sqrt {{2^2} + {1^2} + {{( - 3)}^2}} = \sqrt {4 + 1 + 9} = \sqrt {14} $But $\vec a e \vec b$3. }}0}&{{\text{ b. The two adjacent sides of a parallelogram are $2\hat i - 4\hat j + 5\hat k$ and $\hat i - 2\hat j - 3\hat k$. $\dfrac{\pi }{2}$Ans:$|\vec a \times \vec b| = 1$$ \Rightarrow ||a||\vec b|\sin \theta | = 1$$ \Rightarrow |\vec a||\overrightarrow {\text{b}} ||\sin \theta | = 1$$ \Rightarrow 3 \times \dfrac{{\sqrt 2 }}{3} \times \sin \theta = 1$$ \Rightarrow \sin \theta = \dfrac{1}{{\sqrt 2 }}$$ \Rightarrow \theta = \dfrac{\pi }{4}$12. the respective components of $\vec a$ and $\vec b$ are proportionald. $\overrightarrow {{\text{AB}}} + \overrightarrow {{\text{BC}}} - \overrightarrow {{\text{CA}}} = \vec 0$d. \[\overrightarrow{c}\]Property 2: Let a and b be any two vectors, and l be any scalar. $\dfrac{1}{2}$b. $0 < \theta < \pi $d. then $\lambda \vec a$ is unit vector ifa. $\overrightarrow {{\text{AB}}} + \overrightarrow {{\text{BC}}} - \overrightarrow {{\text{AC}}} = \vec 0$c. Show that $|\vec a|\vec b + |\vec b|\vec a$ is perpendicular to $|\vec a|\vec b - |\vec b|\vec a$ , For any two nonzero vectors $\vec a$ and $\vec b$.Ans: $(|\vec a|\vec b + |\vec b|\vec a) \cdot (|\vec a|\vec b - |\vec b|\vec a)$$ = |\vec a{|^2}\vec b\vec b - |\vec a||\vec b|\vec b\vec a + |\overrightarrow b ||a|\vec a\vec b - |\vec b{|^2}\vec a \cdot \vec a$$ = |\vec a{|^2}|\overrightarrow b {|^2} - |\overrightarrow b {|^2}|\overrightarrow a {|^2}\;\; = \;\;0$12. \; b)}\] = \[\overrightarrow{a}\] . If the vertices ${\text{A}},{\text{B}},{\text{C}}$ of a triangle ${\text{ABC}}$ are $(1,2,3),( - 1,0,0),(0,1,2)$, respectively, then find $\angle {\text{ABC}}$. Let the vectors $\vec a,\vec b,\vec c$ given as ${a_1}\hat i + {a_2}\hat j + {a_3}\hat k\;,\;{b_1}\hat i + {b_2}\hat j + {b_3}\hat k\;,\;{c_1}\hat i + {c_2}\hat j + {c_3}\hat k$.Then show that $ = \vec a \times (\vec b + \vec c) = \vec a \times \vec b + \vec a \times \vec c$Ans: $(\vec b + \vec c) = \left( {{b_1} + {c_1}} \right)\hat i + \left( {{b_2} + {c_2}} \right)\hat j + \left( {{b_3} + {c_3}} \right)\hat k$Now, $\vec a \times (\vec b + \vec c)\left| {\begin{array}{*{20}{c}} {\hat i}&{\hat j}&{\hat k} \\ {{a_1}}&{{a_2}}&{{a_3}} \\ {{b_1} + {c_1}}&{{b_2} + {c_2}}&{{b_3} + {c_3}} \end{array}} \right|$$ = i\left[ {{a_2}\left( {{b_3} + {c_3}} \right) - {a_3}\left( {{b_2} + {c_2}} \right)} \right] - \hat j\left[ {{a_1}\left( {{b_3} + {c_3}} \right) - {a_3}\left( {{b_1} + {c_i}} \right)} \right] + \hat k\left[ {{a_1}\left( {{b_2} + {c_2}} \right) - {a_2}\left( {{b_1} + {c_1}} \right)} \right]$$ = i\left[ {{a_2}{b_3} + {a_2}{c_3} - {a_3}{b_2} - {a_3}{c_2}} \right] + \hat j\left[ { - {a_1}{b_3} - {a_1}{c_3} + {a_3}{b_1} + {a_3}{c_1}} \right] + \hat k\left[ {{a_1}{b_2} + {a_1}{c_2} - {a_2}{b_1} - {a_2}{c_i}} \right]$$\vec a \times \vec b = \left| {\begin{array}{*{20}{c}} {\hat i}&{\hat j}&{\hat k} \\ {{a_1}}&{{a_2}}&{{a_3}} \\ {{b_1}}&{{b_2}}&{{b_3}} \end{array}} \right|$$ = i\left[ {{a_2}{b_3} - {a_3}{b_2}} \right] + \hat j\left[ {{b_1}{a_3} - a{b_3}} \right] + \hat k\left[ {{a_1}{b_2} - {a_2}h} \right]$$\vec a \times \vec c = \left| {\begin{array}{*{20}{l}} {\hat i}&{\hat j}&{\hat k} \\ {{a_1}}&{{a_2}}&{{a_3}} \\ {{c_1}}&{{c_2}}&{{c_3}} \end{array}} \right|$$ = \hat i\left[ {{a_2}{c_3} - {a_3}{c_2}} \right] + \hat j\left[ {{a_3}{c_1} - {a_1}{c_3}} \right] + \hat k\left[ {a{b_2} - {a_2}b} \right]$$\left. What can you conclude about $\vec a$ and $\vec b$ ? Ans: $\vec a\;.\;\vec b = 0$i. \[\overrightarrow{b}\] = \[\overrightarrow{(\gamma a)}\] . If either vector $\vec a = \vec 0$ or $\vec b = \vec 0$, then $\vec a \cdot \vec b = 0.$ But the converse need not be true. Find the position vector of a point ${\text{R}}$ which divides the line joining two points P and Q whose position vectors are $\hat i + 2\hat j - \hat k$ and $ - \hat i + \hat j + \hat k$ respectively, in the ratio 2 : 1 ,i. ( \[\overrightarrow{b}\] + \[\overrightarrow{c}\] ) = \ [\overrightarrow{a}\] . }}3} \end{array}$Ans:$\hat i \cdot (\hat j \times \hat k) + \hat j \cdot (\hat i \times \hat k) + \hat k \cdot (\hat i \times \hat j)$$ = \hat i\;.\;\hat i + \hat j \cdot ( - \hat j) + \hat k\;.\;\hat k$$ = 1 - 1 + 1$$ = 1$The correct answer is C.19. Thus, any vector $\vec b$ can satisfy $\vec a\;.\;\vec b = 0$.13. There are various quantities connected with vectors like displacement, velocity acceleration, force, weight, momentum, electric field intensity, etc. Find the direction cosines of the vector $\hat i + 2\hat j + 3\hat k$.Ans: $\vec a = \hat i + 2\hat j + 3\hat k$.$|\vec a| = \sqrt {{1^2} + {2^2} + {3^2}} = \sqrt {1 + 4 + 9} = \sqrt {14} $So, the Direction cosines of $\vec a$ are $\left( {\dfrac{1}{{\sqrt {14} }},\dfrac{2}{{\sqrt {14} }},\dfrac{3}{{\sqrt {14} }}} \right)$13. Find the direction cosines of the vector joining the points ${\text{A}}(1,2, - 3)$ and ${\text{B}}( - 1, - 2,1)$ directed from A to b.Ans: $\overrightarrow {{\text{AB}}} = ( - 1 - 1)\hat i + ( - 2 - 2)\hat j + \{ 1 - ( - 3)\} \hat k$$ \Rightarrow \overrightarrow {{\text{AB}}} = - 2\hat i - 4\hat j + 4\hat k$$|\overrightarrow {{\text{AB}}} | = \sqrt {{{( - 2)}^2} + {{( - 4)}^2} + {4^2}} = \sqrt {4 + 16 + 16} = \sqrt {36} = 6$So, the Direction cosines of $\overrightarrow {{\text{AB}}} $ are $\left( { - \dfrac{2}{6}, - \dfrac{4}{6},\dfrac{4}{6}} \right) = \left( { - \dfrac{1}{3}, - \dfrac{2}{3},\dfrac{2} {3}} \right)$14. Show that the vector $\hat i + \hat j + \hat k$ is equally inclined to the axes OX, OY and OZ .Ans: $\vec a = \hat i + \hat j + \hat k$$|\vec a| = \sqrt {{1^2} + {1^2} + {1^2}} = \sqrt 3 $So, the Direction Cosines of $\vec a$ are $\left( {\dfrac{1}{{\sqrt 3 }},\dfrac{1}{{\sqrt 3 }},\dfrac{1}{{\sqrt 3 }}} \right)$$\cos \alpha = \dfrac{1}{{\sqrt 3 }},\cos \beta = \dfrac{1}{{\sqrt 3 }},\cos \gamma = \dfrac{1}{{\sqrt 3 }}$So, the vector is equally inclined to OX, OY, and ${\text{OZ}}$.15. 0 b. \[\overrightarrow{b}\] + \[\overrightarrow{a}\] . Find the value of $x$ for which $x(\hat i + \hat j + \hat k)$ unit vector.Ans: $|x(\hat i + \hat j + \hat k)| = 1$$ \Rightarrow \sqrt {{x^2} + {x^2} + {x^2}} = 1$$ \Rightarrow \sqrt {3{x^2}} = 1$$ \Rightarrow \sqrt 3 x = 1$$ \Rightarrow x = \pm \dfrac{1}{{\sqrt 3 }}$6. Find a vector of magnitude 5 units, and parallel to the resultant of the vectors $\vec a = 2\hat i + 3\hat j - \hat k$ and $\vec b = \hat i - 2\hat j + \hat k$Ans: $\vec c = \vec a + \vec b = (2 + 1)\hat i + (3 2)\hat j + ( - 1 + 1)\hat k = 3\hat i + \hat j$$|\vec c| = \sqrt {{3^2} + {1^2}} = \sqrt {9 + 1} = \sqrt {10} $$\therefore \hat c = \dfrac{{\vec c}}{{|\vec c|}} = \dfrac{{(3\hat i + \hat j)}}{{\sqrt {10} }}$So, a vector of magnitude 5 and parallel to the resultant of $\vec a$ and $\vec b$ is $ \pm 5(\hat c) = \pm 5\left( {\dfrac{1}{{\sqrt {10} }}(3\hat i + \hat j)} \right) = \pm \dfrac{{3\sqrt {10} }}{2}\hat i \pm \dfrac{{\sqrt {10} }}{2}\hat j$7. If $\vec a = \hat i + \hat j + \hat k\;,\;\vec b = 2\hat i - \hat j + 3\hat k$ and $\vec c = \hat i - 2\hat j + \hat k$, find a unit vector parallel to the vector $2\vec a - \vec b + 3\vec c$.Ans: $2\vec a - \vec b + 3\vec c = 2(\hat i + \hat j + \hat k) - (2\hat i - \hat j + 3\hat k) + 3(\hat i - 2\hat j + \hat k)$$ = 2\hat i + 2\hat j + 2\hat k - 2\hat i + j - 3\vec k + 3\hat i - 6\hat j + 3\hat k$$ = 3\hat i - 3\hat j + 2\hat k$$|2\vec a - \vec b + 3c| = \sqrt {{3^2} + {{( - 3)}^2} + {2^2}} = \sqrt {9 + 9 + 4} = \sqrt {22} $Thus ,required unit vector is $\dfrac{{2\vec a - \vec b + 3\vec c}}{{|2\vec a - \vec b + 3\vec c|}} = \dfrac{{3\hat i - 3\hat j + 2\hat k}}{{\sqrt {22} }} = \dfrac{3}{{\sqrt {22} }}\hat i - \dfrac{3}{{\sqrt {22} }}\hat j + \dfrac{2}{{\sqrt {22} }}\hat k$8. $|\overrightarrow a |\; = \;0\;{\text{or}}\;|\vec b|\; = \;0$ or (if $|\vec a| e 0$ and $|\vec b| e 0)$But $\overrightarrow a $ and $\overrightarrow b $ cannot be parallel and perpendicular at same time.So, $|\vec a| = 0$ or $|\vec b| = 0$7. Find the angle between the vectors $i - 2\hat j + 3\hat k$ and $3\hat i - 2\hat j + \hat k$. Ans: $|\vec a| = \sqrt {{1^2} + {{( - 2)}^2} + {3^2}} = \sqrt {1 + 4 + 9} = \sqrt {14} $$|\vec b| = \sqrt {{3^2} + {{( - 2)}^2} + {1^2}} = \sqrt {9 + 4 + 1} = \sqrt {14} $$\vec a\;.\;\vec b = (i - 2\hat j + 3\hat k) (3\hat i - 2\hat j + \hat k)$$ = 1.3 + ( - 2)( - 2) + 3.1$$ = 3 + 4 + 3$$ = 10$$\therefore 10 = \sqrt {14} \sqrt {14} \cos \theta $$ \Rightarrow \cos \theta = \dfrac{{10}}{{14}}$$ \Rightarrow \theta = {\cos ^{ - 1}}\left( {\dfrac{5}{7}} \right)$3. Find $\lambda $ and $\mu $ if $(2\hat i + 6\hat j + 27\hat k) \times (\hat i + \lambda \hat j + \mu \hat k) = \vec 0$Ans: $(2\hat i + 6\hat j + 27\hat k) \times (\hat i + \lambda \hat j + \mu \hat k) = \vec 0$$ \Rightarrow \left| {\begin{array}{*{20}{c}} {\hat i}&{\hat j}&{\hat k} \\ 2&6&{27} \\ 1&\lambda &\mu \end{array}} \right| = 0\hat i + 0\hat j + 0\hat k$$ \Rightarrow \hat i(6\mu - 27\lambda ) - \hat j(2\mu - 27) + \hat k(2\lambda - 6) = 0\hat i + 0\hat j + 0\hat k$$6\mu - 27\lambda = 0$$2\mu - 27 = 0$$2\lambda - 6 = 0$$\lambda = 3$$\mu = \dfrac{{27}}{2}$6. Given that $\vec a\;.\;\vec b = 0$ and $\vec a \times \vec b = 0$. 1c. $\theta = \dfrac{\pi }{3}$c. Write two different vectors having same direction. Ans: $\vec p = (\hat i + \hat j + \hat k)$ and $\vec q = (2\hat i + 2\hat j + 2\hat k)$.The Direction Cosines of $\vec p$ are $a = \dfrac{1}{{\sqrt {{1^2} + {1^2} + {1^2}} }} = \dfrac{1}{{\sqrt 3 }},b = \dfrac{1}{{\sqrt {{1^2} + {1^2} + {1^2}} }} = \dfrac{1}{{\sqrt 3 }},c = \dfrac{1}{{\sqrt {{1^2} + {1^2} + {1^2}} }} = \dfrac{1}{{\sqrt 3 }}$The Direction Cosines of $\vec q$ are$a = \dfrac{2}{{\sqrt {{2^2} + {2^2} + {2^2}} }} = \dfrac{2}{{2\sqrt 3 }} = \dfrac{1}{{\sqrt 3 }},b = \dfrac{2}{{\sqrt {{2^2} + {2^2} + {2^2}} }} = \dfrac{2}{{2\sqrt 3 }} = \dfrac{1}{{\sqrt 3 }},c = \dfrac{2}{{\sqrt {{2^2} + {2^2} + {2^2}} }} = \dfrac{2}{{2\sqrt 3 }} = \dfrac{1}{{\sqrt 3 }}$The Direction Cosines of $\vec p$ and $\vec q$ are equal but $\vec p e \vec q$.4. Commutative Property: For any two vectors \[\overrightarrow{a}\] and \[\overrightarrow{b}\],\[\overrightarrow{a}\] + \[\overrightarrow{b}\] = \[\overrightarrow{b}\] + \[\overrightarrow{a}\]2. Show that the vectors $2\hat i - \hat j + \hat k\;,\;\hat i - 3\hat j - 5\hat k$ and $3\hat i - 4\hat j - 4\hat k$ form the vertices of a right angled triangle.Ans: $\overrightarrow {{\text{OA}}} = 2\hat i - \hat j + \hat k\;,\;\overrightarrow {{\text{OB}}} = \hat i - 3\hat j - 5\hat k\;,\;\overrightarrow {{\text{OC}}} = 3\hat i - 4\hat j - 4\hat k$$\overrightarrow {{\text{AB}}} = (1 - 2)\hat i + ( - 3 + 1)\hat j + ( - 5 - 1)\hat k = - \hat i - 2\hat j - 6 \hat k$$\overrightarrow {{\text{BC}}} = (3 - 1)\hat i + ( - 4 + 3)\hat j + ( - 4 + 5)\hat k = 2\hat i - \hat j + \hat k$$\overrightarrow {{\text{CA}}} = (2 - 3)\hat i + ( - 1 + 4)\hat j + (1 + 4)\hat k = - \hat i + 3\hat j + 5\hat k$$|\overrightarrow {{\text{AB}}} | = \sqrt {{{( - 1)}^2} + {{( - 2)}^2} + {{( - 6)}^2}} = \sqrt {1 + 4 + 36} = \sqrt {41} $$|\overrightarrow {BC} | = \sqrt {{2^2} + {{( 1)}^2} + {1^2}} = \sqrt {4 + 1 + 1} = \sqrt 6 $$|\overrightarrow {{\text{AC}}} | = \sqrt {{{( - 1)}^2} + {3^2} + {5^2}} = \sqrt {1 + 9 + 25} = \sqrt {35} $$\therefore |\overrightarrow {BC} {|^2} + |\overrightarrow {AC} {|^2} = 6 + 35 = 41 = |\overrightarrow {AB} {|^2}$Hence, $\Delta {\text{ABC}}$ is a right triangle.18. If $\vec a$ is a nonzero vector of magnitude 'a' and $\lambda $ a nonzero scalar. $\overrightarrow {{\text{AB}}} - \overrightarrow {{\text{CB}}} + \overrightarrow {{\text{CA}}} = \vec 0$Ans:(Image will be uploaded soon)$\overrightarrow {{\text{AB}}} + \overrightarrow {{\text{BC}}} = \overrightarrow {{\text{AC}}} $$ \Rightarrow \overrightarrow {{\text{AB}}} + \overrightarrow {{\text{BC}}} = - \overrightarrow {{\text{CA}}} $$ \Rightarrow \overrightarrow {{\text{AB}}} + \overrightarrow {{\text{BC}}} + \overrightarrow {{\text{CA}}} = \vec 0$$\overrightarrow {{\text{AB}}} + \overrightarrow {{\text{BC}}} = \overrightarrow {{\text{AC}}} $$ \Rightarrow \overrightarrow {{\text{AB}}} + \overrightarrow {{\text{BC}}} - \overrightarrow {{\text{AC}}} = \vec 0$$\overrightarrow {{\text{AB}}} + \overrightarrow {{\text{BC}}} + \overrightarrow {{\text{CA}}} = \vec 0$$\overrightarrow {{\text{AB}}} - \overrightarrow {{\text{CB}}} + \overrightarrow {{\text{CA}}} = \vec 0$If $\overrightarrow {{\text{AB}}} + \overrightarrow {{\text{BC}}} - \overrightarrow {{\text{CA}}} = \vec 0$$\overrightarrow {{\text{AC}}} = \overrightarrow {{\text{CA}}} $$ \Rightarrow \overrightarrow {{\text{AC}}} = - \overrightarrow {{\text{AC}}} $$ \Rightarrow \overrightarrow {{\text{AC}}} + \overrightarrow {{\text{AC}}} = \vec 0$$ \Rightarrow 2\overrightarrow {{\text{AC}}} = \vec 0$$ \Rightarrow \overrightarrow {{\text{AC}}} = \vec 0$, which is not true.So, the equation given in option ${\text{C}}$ is False.Hence, the correct answer is ${\text{C}}$.19. Determine the girl's displacement from her initial point of departure. Ans: (Image will be uploaded soon)$\overrightarrow {{\text{OA}}} = - 4\hat i$$\overrightarrow {{\text{AB}}} = \hat i|\overrightarrow {{\text{AB}}} |\cos {60^\circ } + \hat j|\overrightarrow {{\text{AB}}} |\sin {60^\circ }$$ = \hat i3 \times \dfrac{1}{2} + \hat j3 \times \dfrac{{\sqrt 3 }}{2}$$ = \dfrac{3}{2}\hat i + \dfrac{{3\sqrt 3 }}{2}\hat j$$\overrightarrow {{\text{OB}}} = \overrightarrow {{\text{OA}}} + \overrightarrow {{\text{AB}}} $$ = ( - 4\hat i) + \left( {\dfrac{3}{2}\hat i + \dfrac{{3\sqrt 3 }}{2}j} \right)$$ = \left( { - 4 + \dfrac{3}{2}} \right)\hat i + \dfrac{{3\sqrt 3 }}{2}\hat j$$ = \left( {\dfrac{{ - 8 + 3}}{9}} \right)\hat i + \dfrac{{3\sqrt 3 }}{9}\hat j$$ = \dfrac{{ - 5}}{2}\vec i + \dfrac{{3\sqrt 3 }}{2}\hat j$4. Two vectors can be multiplied by two methods. You can take an example of a directed line segment AB, where the initial point is A from where the vector starts and the terminating point is B where it ends. distanceAns: Distance is a scalar quantity because it has only magnitude.iii. The distance between the initial point and the terminating point is known as the magnitude of a vector. $|\vec a| = 0$ or $|\vec b| = 0$ or $\vec a \bot \vec b\quad ($ if $|\vec a| e 0$ and $|\vec b| e 0)$$\vec a \times \vec b = 0$ii. Let $\vec a$ and $\vec b$ be two unit vectors and $\theta $ is the angle between them. If $\theta $ is the angle between two vectors $\vec a$ and $\vec b$, then $\vec a\vec b \geqslant 0$ only whena. Then $\vec a + \vec b$ is a unit vector ifa. Associative Property: For any three vector \[\overrightarrow{a}\], \[\overrightarrow{b}\] and \[\overrightarrow{c}\],(\[\overrightarrow{a}\] + \[\overrightarrow{b}\]) + \[\overrightarrow{c}\] = \[\overrightarrow{a}\] + (\[\overrightarrow{b}\] + \ [\overrightarrow{c}\] ) 3. If $\vec a,\vec b,\vec c$ are mutually perpendicular vectors of equal magnitudes, show that the vector $\vec a + \vec b + \vec c$ is equally inclined to $\vec a\;,\;\vec b$ and $\vec c$. Ans: $\vec a\;.\;\vec b = \vec b\;.\;\vec c = \vec c\;.\;\vec a = 0$$|\vec a| = |\vec b| = |\vec c|$Let $\vec a + \vec b + \vec c$ be inclined to $\vec a,\vec b,\vec c$ at angles ${\theta _1},{\theta _2},{\theta _3}$ respectively. Show that the vectors $2\hat i - 3\hat j + 4\hat k$ and $ - 4\hat i + 6\hat j - 8\hat k$ are collinear.Ans: $\vec a = 2\hat i - 3\hat j + 4\hat k$ and $\vec b = - 4\hat i + 6\hat j - 8\hat k$$\vec b = - 4\hat i + 6\hat j - 8\hat k = - 2(2\hat i - 3\hat j + 4\hat k) = - 2\vec a$$\therefore \vec b = \lambda \vec a,\;\;\lambda = - 2$So, the given vectors are collinear.12. Prove that, $(\vec a + \vec b) \cdot (\vec a + \vec b) = \;|\vec a{|^2} + |\vec b{|^2}$ if and only if $\vec a,\vec b$ are perpendicular, given $\vec a e \vec 0,\vec b e 0$Ans: $(\vec a + \vec b) \cdot (\vec a + \vec b) = |\vec a{|^2} + |\vec b{|^2}$$ \Rightarrow \vec a \cdot \vec a + \vec a\;.\;\vec b + \vec b \cdot \vec a + \vec b\;.\;\vec b = |\vec a{|^2} + |\vec b{|^2}$$ \Rightarrow |\vec a{|^2} + 2\vec a\;.\;\vec b + |\vec b{|^2} = |a{|^2} + |\vec b{|^2}$$ \Rightarrow 2\vec a\;.\;\vec b = 0$$ \Rightarrow \vec a\;.\;\vec b = 0$So $\vec a$ and $\vec b$ are perpendicular.16. If $\vec a \cdot \vec a = 0$ and $\vec a\vec b = 0$, then what can be concluded above the vector $\vec b$ ? Ans: $\vec a \cdot \vec a = 0 \Rightarrow |\vec a{|^2} = 0 \Rightarrow |\vec a| = 0$$\therefore \vec a$ is the zero vector. $\vec r = \cos {30^\circ }\hat i + \sin {30^\circ }\hat j = \dfrac{{\sqrt 3 }}{2}\hat i + \dfrac{1}{2}\hat j$2. $\vec a = \pm \vec b$c. force Ans: Force is a vector quantity because it has both magnitude as well as direction.iv. Let $\vec a = \hat i + 4\hat j + 2\hat k$ and $\vec b = 3\hat i - 2\hat j + 7\hat k$ and $\vec c = 2\hat i - \hat j + 4\hat k$. In triangle ABC, which of the following is not true?(Image will be uploaded soon)a. Find the angle between two vectors $\vec a$ and $\vec b$ with magnitudes $\sqrt 3 $ and 2, respectively having $\vec a \cdot \vec b = \sqrt 6 $Ans: $|\vec a| = \sqrt 3 \;,\;|\vec b| = 2\;,\;\vec a\;.\;\vec b = \sqrt 6 $$\therefore \sqrt 6 = \sqrt 3 \times 2 \times \cos \theta $$ \Rightarrow \cos \theta = \dfrac{{\sqrt 6 }}{{\sqrt 3 \times 2}}$$ \Rightarrow \cos \theta = \dfrac{1}{{\sqrt 2 }}$$ \Rightarrow \theta = \dfrac{\pi }{4}$2. $\dfrac{\pi }{4}$c. Internally Ans: $\overrightarrow {{\text{OP}}} = \hat i + 2\hat j - \hat k$ and $\overrightarrow {{\text{OQ}}} = - \hat i + \hat j + \hat k$ The position vector of ${\text{R}}$ is $\overrightarrow {{\text{OR}}} = \dfrac{{2( - \hat i + \hat j + \hat k) + 1(\hat i + 2\hat j - \hat k)}}{{2 + 1}} = \dfrac{{( - 2\hat i + 2\hat j + 2\hat k) + (\hat i + 2\hat j - \hat k)}}{3}$$ = \dfrac{{ - \;\hat i + 4\hat j + \hat k}}{3} = - \dfrac{1}{3}\hat i + \dfrac{4}{3}\hat j + \dfrac{1}{3}\hat k$ii. $\theta = \dfrac{\pi }{2}$d. Equal Ans: Vectors $\vec b$ and $\vec d$ are equal.iii. $a = |\lambda |$d. $\dfrac{1}{7}(2\hat i + 3\hat j + 6\hat k)\;\;,\;\;\dfrac{1}{7}(3\hat i - 6\hat j + 2\hat k)\;\;,\;\;\dfrac{1}{7}(6\hat i + 2\hat j - 3\hat k)$Ans: $\vec a = \dfrac{1}{7}(2\hat i + 3\hat j + 6\hat k) = \dfrac{2}{7}\hat i + \dfrac{3}{7}\hat j + \dfrac{6}{7}\hat k$,$\vec b = \dfrac{1}{7}(3\bar i - 6\hat j + 2\hat k) = \dfrac{3}{7}\hat i - \dfrac{6}{7}\bar j + \dfrac{2}{7}\hat k$$\vec c = \dfrac{1}{7}(6\hat i + 2\hat j - 3\hat k) = \dfrac{6} {7}\hat i + \dfrac{2}{7}\hat j - \dfrac{3}{7}\hat k$$|\vec a| = \sqrt {{{\left( {\dfrac{2}{7}} \right)}^2} + {{\left( {\dfrac{3}{7}} \right)}^2} + {{\left( {\dfrac{6}{7}} \right)}^2}} = \sqrt {\dfrac{4}{{49}} + \dfrac{9}{{49}} + \dfrac{{36}}{{49}}} = 1$$|\vec b| = \sqrt {{{\left( {\dfrac{3}{7}} \right)}^2} + {{\left( { - \dfrac{6}{7}} \right)}^2} + {{\left( {\dfrac{2}{7}} \right)}^2}} = \sqrt {\dfrac{9}{{49}} + \dfrac{{36}}{{49}} + \dfrac{4}{{49}}} = 1$$|\vec c| = \sqrt {{{\left( {\dfrac{6}{7}} \right)}^2} + {{\left( {\dfrac{2}{7}} \right)}^2} + {{\left( { - \dfrac{3}{7}} \right)}^2}} = \sqrt {\dfrac{{36}}{{49}} + \dfrac{4}{{49}} + \dfrac{9}{{49}}} = 1$So, each of the vectors is a unit vector.$\vec a\;.\;\vec b = \dfrac{2}{7} \times \dfrac{3}{7} + \dfrac{3}{7} \times \left( {\dfrac{{ - 6}}{7}} \right) + \dfrac{6}{7} \times \dfrac{2}{7} = \dfrac{6}{{49}} - \dfrac{{18}}{{49}} + \dfrac{{12}}{{49}} = 0$$\vec b \cdot \vec c = \dfrac{3}{7} \times \dfrac{6}{7} + \left( {\dfrac{{ - 6}}{7}} \right) \times \dfrac{2}{7} + \dfrac{2}{7} \times \left( {\dfrac{{ - 3}}{7}} \right) = \dfrac{{18}}{{49}} - \dfrac{{12}}{{49}} - \dfrac{6}{{49}} = 0$$\vec c\;.\;\vec a = \dfrac{6}{7} \times \dfrac{2}{7} + \dfrac{2}{7} \times \dfrac{3}{7} + \left( {\dfrac{{ - 3}}{7}} \right) \times \dfrac{6}{7} = \dfrac{{12}}{{49}} + \dfrac{6}{{49}} - \dfrac{{18}} {{49}} = 0$So, given vectors are mutually perpendicular to each other.6. Find $|\vec a|$ and $|\vec b|$, if $(\vec a + \vec b) \cdot (\vec a - \vec b) = 8$ and $|\overrightarrow a |\; = \;8|\vec b|$.Ans: $(\vec a + \vec b) \cdot (\vec a - \vec b) = 8$$ \Rightarrow \vec a\vec a - \vec a\;.\;\vec b + \vec b \cdot \vec a - \vec b\vec b = 8$$ \Rightarrow |\vec a{|^2} - |\vec b{|^2} = 8$$ \Rightarrow {(8|\vec b|)^2} - |\vec b{|^2} = 8$$ \Rightarrow 64|\vec b{|^2} - |\vec b{|^2} = 8$$ \Rightarrow 63|\vec b{|^2} = 8$$ \Rightarrow |\vec b{|^2} = \dfrac{8}{{63}}$$ \Rightarrow |\vec b| = \sqrt {\dfrac{8}{{63}}} $$ \Rightarrow |\vec b| = \dfrac{{2\sqrt 2 }}{{3\sqrt 7 }}$$|\vec a| = 8|\vec b| = \dfrac{{8 \times 2\sqrt 2 }}{{3\sqrt 7 }} = \dfrac{{16\sqrt 2 }}{{3\sqrt 7 }}$7. $0 \leqslant \theta \leqslant \dfrac{\pi }{2}$c. Evaluate the product $(3\vec a - 5\vec b) \cdot (2\vec a + 7\vec b)$Ans: $(3\vec a - 5\vec b) \cdot (2\vec a + 7\vec b)$$ = 3\vec a \cdot 2\vec a + 3\vec a.7\vec b - 5\vec b \cdot 2\vec a - 5\vec b \cdot 7\vec b$$ = 6\vec a\vec a + 21\vec a\vec b - 10\vec a\vec b - 35\vec b\vec b$$ = 6|\vec a{|^2} + 11\vec a\vec b - 35|\vec b{|^2}$8. Find the magnitude of two vectors $\vec a$ and $\vec b$, having the same magnitude and such that the angle between them is ${60^\circ }$ and their scalar product is $\dfrac{1}{2}$ Ans: Let $\theta $ be angle between $\vec a$ and $\vec b$.$|\vec a| = |\vec b|,\vec a\;.\;\vec b = \dfrac{1}{2}$, and $\theta = {60^\circ }$$\therefore \dfrac{1}{2} = |\vec a|\vec b\mid \cos {60^\circ }$$ \Rightarrow \dfrac{1}{2} = |\vec a{|^2} \times \dfrac{1}{2}$$ \Rightarrow |\vec a{|^2} = 1$$ \Rightarrow |\vec a| = |\vec b| = 1$9. $\dfrac{\pi }{6}$b. Parallelogram Law of Vector Addition: According to this law, two vectors \[\overrightarrow{a}\] and \[\overrightarrow{b}\] representing adjacent sides of a parallelogram, when added, results into the diagonal of the parallelogram.(Image to be added soon)\[\overrightarrow{OA}\] + \[\overrightarrow{OB}\]= \[\overrightarrow{OC}\]Properties of Vector Addition1. The key features to consider are:You get a better mode to clarify your doubts as it offers in-depth knowledge in easy language.It makes the concept easier by simplifying and classifying each and every concept of the topic.Since it is made on the CBSE pattern, it gives you the idea of how to attempt the question paper.After a lot of research work, these solutions are made to build foundations for your higher studies. Then:\[\overrightarrow{(\gamma a)}\] . If $\vec a = 2\hat i + 2\hat j + 3\hat k,\vec b = - \hat i + 2\hat j + \hat k$ and $\vec c = 3\hat i + \hat j$ are such that $\vec a + \lambda \vec b$ is perpendicular to $\vec c$, then find the value of $\lambda $.Ans: $\vec a + \lambda \vec b = (2\hat i + 2\hat j + 3\hat k) + \lambda ( - \hat i + 2\hat j + \dot k) = (2 - \lambda )\hat i + (2 + 2\lambda )\hat j + (3 + \lambda )\hat k$$(\vec a + \lambda \vec b) \cdot \vec c = 0$$ \Rightarrow [(2 + - )\hat i + (2 + 2\lambda )\hat j + (3 + \lambda )\hat k] \cdot (3\hat i + j) = 0$$ \Rightarrow 3(2 - \lambda ) + (2 + 2\lambda ) + 0(3 + \lambda ) = 0$$ \Rightarrow 6 - 3\lambda + 2 + 2\lambda = 0$$ \Rightarrow - \lambda + 8 = 0$$ \Rightarrow \lambda = 8$11. $\theta = \dfrac{{2\pi }}{3}$Ans: $|\vec a| = |\vec b| = 1$$|\vec a + \vec b| = 1$$ \Rightarrow (\vec a + \overrightarrow b )(\vec a + \overrightarrow b ) = 1$$ \Rightarrow \vec a \cdot \vec a + \vec a\;.\;\vec b + \vec b\;.\;\vec a + \vec b\;.\;\vec b = 1$$ \Rightarrow |\vec a{|^2} + 2\vec a\;.\;\vec b + |\overrightarrow b {|^2} = 1$$ \Rightarrow {1^2} + 2|\vec a||\vec b|\cos \theta + {1^2} = 1$$ \Rightarrow 1 + 2.1.1\cos \theta + 1 = 1$$ \Rightarrow \cos \theta = - \dfrac{1}{2}$$ \Rightarrow \theta = \dfrac{{2\pi }}{3}$So, $\vec a + \vec b$ is unit vector if $\theta = \dfrac{{2\pi }}{3}$ The correct answer is D18. velocityAns: Velocity is a vector quantity because it has both magnitude as well as direction.v. work doneAns: Work done is a scalar quantity because it has only magnitude.4. In Figure, identify the following vectors.(Image will be uploaded soon)i. If $\theta $ is the angle between any two vectors $\vec a$ and $\vec b$, then $|\vec a\vec b| = |\vec a \times \vec b|$ when $\theta $ is equal toa. Show that the direction cosines of a vector equally inclined to the axes OX, OY and OZ are $\dfrac{1}{{\sqrt 3 }}$ $\dfrac{1}{{\sqrt 3 }},\dfrac{1}{{\sqrt 3 }}$.Ans: Let a vector be equally inclined to ${\text{OX}},{\text{OY}}$, and ${\text{OZ}}$ at an angle $\alpha $.So, the Direction Cosines of the vector are $\cos \alpha ,\cos \alpha $and $\cos \alpha $.${\cos ^2}\alpha + {\cos ^2}\alpha + {\cos ^2}\alpha = 1$$ \Rightarrow 3{\cos ^2}\alpha = 1$$ \Rightarrow \cos \alpha = \dfrac{1}{{\sqrt 3 }}$So, the DCs of the vector are $\dfrac{1}{{\sqrt 3 }},\dfrac{1}{{\sqrt 3 }},\dfrac{1}{{\sqrt 3 }}$.12. $0 \leqslant \theta \leqslant \pi $Ans: $\therefore \vec a\vec b \geqslant 0$$ \Rightarrow |\vec a||\vec b|\cos \theta \geqslant 0$$ \Rightarrow \cos \theta \geqslant 0\quad \because [|\vec a| \geqslant 0$ and $|\vec b| \geqslant 0]$$ \Rightarrow 0 \leqslant \theta \leqslant \dfrac{\pi }{2}$$\vec a \cdot \vec b \geqslant 0$ if $0 \leqslant \theta \leqslant \dfrac{\pi }{2}$So the right answer is B17. Two collinear vectors are always equal in magnitude.Ans: Falseiii. $a = \dfrac{1}{{|\lambda |}}$Ans: $|\lambda \vec a| = 1$$ \Rightarrow |\vec a| = \dfrac{1}{{|\lambda |}}$$ \Rightarrow a = \dfrac{1}{{|\lambda |}}$Exercise 10.41. Find $|\vec a \times \vec b|$, if $\vec a = \hat i - 7\hat j + 7\hat k$ and $\vec b = 3\hat i - 2\hat j + 2\hat k$Ans: We have, $\vec a = \hat i - 7\hat j + 7\hat k$ and $\vec b = 3\hat i - 2\hat j + 2\hat k$$\vec a \times \vec b = \left| {\begin{array}{*{20}{c}} {\hat i}&{\hat j}&{\hat k} \\ 1&{ - 7}&7 \\ 3&{ - 2}&2 \end{array}} \right|$$ = \hat i( - 14 + 14) - \hat j(2 - 21) + \hat k( - 2 + 21) = 19\hat j + 19\hat k$$\therefore |\vec a \times \vec b| = \sqrt {{{(19)}^2} + {{(19)}^2}} = \sqrt {2 \times {{(19)}^2}} = 19\sqrt 2 $2. Find a unit vector perpendicular to each of the vector $\vec a + \vec b$ and $\vec a - \vec b$, where $\vec a = 3\hat i + 2\hat j + 2\hat k$ and $\vec b = \hat i + 2\hat j - 2\hat k$.Ans: $\vec a = 3\hat i + 2\hat j + 2\hat k$ and $\vec b = \hat i + 2\hat j - 2\hat k$$\vec a + \vec b = 4\hat i + 4\hat j\;,\;\vec a - \vec b = 2\hat i + 4\hat k$$(\vec a + \vec b) \times (\vec a - \vec b) = \left| {\begin{array}{*{20}{c}} {\hat i}&{\hat j}&{\hat k} \\ 4&4&0 \\ 2&0&4 \end{array}} \right| = \hat i(16) - \hat j(16) + \hat k( - 8) = 16\hat i - 16\hat j - 8\hat k$$|(\vec a + \vec b) \times (\vec a - \vec b)| = \sqrt {{{16}^2} + {{( - 16)}^2} + {{( - 8)}^2}} $$ = \sqrt {{2^2} \times {8^2} + {2^2} \times {8^2} + {8^2}} $$ = 8\sqrt {{2^2} + {2^2} + 1} = 8\sqrt 9 = 8 \times 3 = 24$So, the unit vector is $ = \pm \dfrac{{(\vec a + \vec b) \times (\vec a - \vec b)}}{{|(\vec a + \vec b) \times (\vec a - \vec b)\mid }} = \pm \dfrac{{16\hat i - 16\hat j - 8\hat k}}{{24}}$$ = \pm \dfrac{{2\hat i - 2\hat j - \hat k}}{3} = \pm \dfrac{2}{3}\hat i \mp \dfrac{2}{3}\hat j \mp \dfrac{1}{3}\hat k$3. If a unit vector $\vec a$ makes an angle $\dfrac{\pi }{3}$ with $\hat i,\dfrac{\pi }{4}$ with $\hat j$ and an acute angle $\theta $ with $\hat k$, then find $\theta $ and hence, the components of $\vec a$ Ans: $\vec a = {a_1}\hat i + {a_2}\hat j + {a_3}\hat k$$|\vec a|\; = 1$. Write down a unit vector in XY-plane, making an angle of ${30^\circ }$ with the positive direction of x axis. Ans: Unit vector is $\vec r = \cos \theta \hat i + \sin \theta \hat j$, where $\theta $ is angle with positive ${\text{X}}$ axis. Find the projection of the vector $\hat i - \hat j$ on the vector $i + \hat j$. Ans: $\vec a = \hat i - \hat j$ and $\vec b = \hat i + \hat j$Projection of $\vec a$ on $\vec b$ is $\dfrac{1}{{\mid \vec b|}}(\vec a \cdot \vec b) = \dfrac{1}{{\sqrt {1 + 1} }}\{ 1.1 + ( - 1)(1)\} = \dfrac{1}{{\sqrt 2 }}(1 - 1) = 0$4. $\overrightarrow {{\text{OB}}} = \dfrac{{\lambda \overrightarrow {{\text{OC}}} + \overrightarrow {{\text{OA}}} }}{{(\lambda + 1)}}$$ \Rightarrow 5\hat i - 2\hat k = \dfrac{{\lambda (11\hat i + 3\hat j + 7\hat k) + (\hat i - 2\hat j - 8\hat k)}}{{\lambda + 1}}$$ \Rightarrow (\lambda + 1)(5\hat i - 2\hat k) = 11\lambda \hat i + 3\lambda \hat j + 7\lambda \hat k + \hat i - 2\hat j - 8\hat k$$ \Rightarrow 5(\lambda + 1)\hat i - 2(\lambda + 1)\hat k = (11\lambda + 1)\hat i + (3\lambda 2)\hat j + (7\lambda - 8)\hat k$$ \Rightarrow \lambda = \dfrac{2}{3}$So, the required ratio is 2:39. You have the real-life application of the vector. If $\vec a$ and $\vec b$ are two collinear vectors, then which of the following are incorrect?a. {(\vec a \times \vec b) + (\vec a \times \vec c) = \hat i\left[ {{a_2}{b_3} + {a_2}{c_3} - {a_3}{b_2} - {a_3} {c_2}} \right] + \hat j{{[{b_1}{a_3} + {a_3}{c_1} - {a_1}{b_3} - {a_i}\} }_3}} \right]$$ + \hat k\left[ {a,b, + {a_1}c, - a,{b_1} - a,{c_1}} \right]$$\vec a \times (\vec b + \vec c) = \vec a \times \vec b + \vec a \times \vec c$Hence, proved.8. If either $\vec a = 0$ or $b = 0$, then $\vec a \times b = 0$.Is the converse true? Find the projection of the vector $\hat i + 3\hat j + 7\hat k$ on the vector $7\hat i - \hat j + 8\hat k$. Ans: $\vec a = \hat i + 3\hat j + 7\hat k$ and $\vec b = 7\hat i - \hat j + 8\hat k$Projection of $\vec a$ on $\vec b$ is $\dfrac{1}{{|\vec b|}}(\vec a \cdot \vec b) = \dfrac{1}{{\sqrt {{7^2} + {{( - 1)}^2} + {8^2}} }}\{ 1(7) + 3( - 1) + 7(8)\} = \dfrac{{7 - 3 + 56}}{{\sqrt {49 + 1 + 64} }} = \dfrac{{60}}{{\sqrt {114} }}$5. }} - 1}&{{\text{ c. Complete data and explanation of the topic will be available. $\cos {\theta _1} = \dfrac{{(\vec a + \vec b + \vec c) \cdot \vec a}}{{|\vec a + \vec b + \vec c||\vec a\mid }} = \dfrac{{\vec a \cdot \vec a + \vec b \cdot \vec a + \vec c \cdot \vec a}}{{|\vec a + \vec b + \vec c||\vec a\mid }} = \dfrac{{|\vec a{|^2}}}{{|\vec a + \vec b + \vec c||\vec a|}} = \dfrac{{|\vec a|}}{{|\vec a + \vec b + \vec c|}}$$\cos {\theta _2} = \dfrac{{(\vec a + \vec b + \vec c)\vec b}}{{|\vec a + \vec b + \vec c||\vec b\mid }} = \dfrac{{\vec a\vec b + \vec b\vec b + \vec c\vec b}}{{|\vec a + \vec b + \vec c||\vec b\mid }} = \dfrac{{|\vec b{|^2}}}{{|\vec a + \vec b + \vec c||\vec b|}} = \dfrac{{|\vec b|}}{{|\vec a + \vec b + \vec c|}}$$\operatorname{os} {\theta _3} = \dfrac{{(\vec a + \vec b + \vec c) \cdot \vec c}}{{|\vec a + \vec b + \vec c||\bar c|}} = \dfrac{{\vec a\vec c + \vec b\vec c + \vec c\vec c}}{{|\vec a + \vec b + \vec c||\vec c\mid }} = \dfrac{{|\vec c{|^2}}}{{|\vec a + \vec b + \vec c||\vec c|}} = \dfrac{{|\vec c|}}{{|\vec a + \vec b + \vec c|}}$Since, $|\vec a| = |\vec b| = |\vec c|\; \Rightarrow \;\cos {\theta _1} = \cos {\theta _2} = \cos {\theta _3}$, So, ${\theta _1} = {\theta _2} = {\theta _3}$15. Find the unit vector parallel to its diagonal. Justify your answer. Ans: $\operatorname{In} \Delta {\text{ABC}},\overrightarrow {\;CB} = \vec a,\;\overrightarrow {CA} = b,\;\overrightarrow {AB} = \vec c$(Image will be uploaded soon)$\vec a = \vec b + \vec c$, by triangle law of addition for vectors.$|\vec a|\; < \;|\vec b| + |\vec c|$, by triangle inequality law of lengths. Hence, it's not true that $|\vec a| = |\vec b| + |\vec c|$5. $\cos \dfrac{\pi }{3} = \dfrac{{{a_1}}}{{|\vec a|}}$$ \Rightarrow \dfrac{1}{2} = {a_1}$$\cos \dfrac{\pi }{4} = \dfrac{{{a_2}}}{{|\vec a|}}$$ \Rightarrow \dfrac{1}{{\sqrt 2 }} = {a_2}$$\cos \theta = \dfrac{{{a_3}}}{{|\vec a|}}$$ \Rightarrow {a_3} = \cos \theta $$ \Rightarrow \sqrt {a_1^2 + a_2^2 + a_3^2} = 1$$ \Rightarrow {\left( {\dfrac{1}{2}} \right)^2} + {\left( {\dfrac{1}{{\sqrt 2 }}} \right)^2} + {\cos ^2}\theta = 1$$ \Rightarrow \dfrac{1}{4} + \dfrac{1}{2} + {\cos ^2}\theta = 1$$ \Rightarrow \dfrac{3}{4} + {\cos ^2}\theta = 1$$ \Rightarrow {\cos ^2}\theta = 1 - \dfrac{3}{4} = \dfrac{1}{4}$$ \Rightarrow \cos \theta = \dfrac{1}{2} \Rightarrow \theta = \dfrac{\pi }{3}$$\therefore {a_3} = \cos \dfrac{\pi }{3} = \dfrac{1}{2}$So, $\theta = \dfrac{\pi }{3}$ and components of $\vec a$ are $\left( {\dfrac{1}{2},\dfrac{1}{{\sqrt 2 }},\dfrac{1}{2}} \right)$4. Show that $(\vec a - \vec b) \times (\vec a + \vec b) = 2(\vec a \times \vec b)$Ans: $(\vec a - \vec b) \times (\vec a + \vec b)$$ = (a - b) \times \vec a + (a - b) \times b$$ = a \times \vec a - \vec b \times \vec d + a \times \vec b - \vec b \times \vec b$$ = 0 + \vec a \times \vec b + \vec a \times \vec b - 0$$ = 2\;(\vec a \times \vec b)$5. nAns: $|\vec a\vec b| = |\vec a \times \vec b|$$ \Rightarrow |\vec a|\vec b|\cos \theta = |\vec a||\vec b\mid \sin \theta $$ \Rightarrow \cos \theta = \sin \theta $$ \Rightarrow \tan \theta = 1$$ \Rightarrow \theta = \dfrac{\pi } {4}$The correct answer is BNCERT Solutions for Class 12 Maths PDF DownloadNCERT Solution Class 12 Maths of Chapter 10 All ExercisesChapter 10 - Vector Algebra Exercises in PDF FormatExercise 10.15 Questions & Solutions (5 Short Answers)Exercise 10.219 Questions & Solutions (5 Short Answers, 14 Long Answers)Exercise 10.318 Questions & Solutions (5 Short Answers, 13 Long Answers)Exercise 10.412 Questions & Solutions (4 Short Answers, 8 Long Answers)NCERT Solutions for Class 12 Maths Chapter 10 - Vector AlgebraNCERT Solutions for Class 12 Maths ? Free PDF DownloadStudents who are in search of Vector Algebra class 12 CBSE NCERT Solutions need to open the link available at Vedantu website. Students will be able to understand the topic in a much easier and quicker way.Important PointsA vector is defined as a quantity that has a magnitude as well as direction.According to the triangle law of vector addition, "If two vectors are represented by two sides of a triangle taken in order, then their sum or resultant is given by the third side taken in opposite order".The dot product or the scalar product of any two vectors x and y having an angle between them is defined as x.y = |x||y|cos .The vector or cross product of two given vectors p and q having an angle between them is defined as p x q = |p||q|sin .Chapter 10 ? Vector Algebra10.1 IntroductionIn Vector Algebra Class 12, you will be able to learn about the concept, operations and algebraic and geometric properties of vectors. Find the scalar components and magnitude of the vector joining the points ${\text{P}}\left( {{x_1},{y_1},{z_1}} \right)$ and ${\text{Q}}\left( {{x_2},{y_2},{z_2}} \right)$Ans: $\overrightarrow {{\text{PQ}}} = \left( {{x_2} - {x_1}} \right)\hat i + \left( {{y_2} - {y_1}} \right)\hat j + \left( {{z_2} - {z_1}} \right)\hat k$$|\overrightarrow {{\text{PQ}}} |\; = \;\sqrt {{{\left( {{x_2} - {x_1}} \right)}^2} + {{\left( {{y_2} - {y_1}} \right)}^2} + {{\left( {{z_2} - {z_1}} \right)}^2}} $3. Find a vector in the direction of vector $5\hat i - \hat j + 2\hat k$ which has magnitude 8 units.Ans: $\vec a = 5\hat i - \hat j + 2\hat k$$|\vec a| = \sqrt {{5^2} + {{( - 1)}^2} + {2^2}} = \sqrt {25 + 1 + 4} = \sqrt {30} $$\therefore \hat a = \dfrac{{\vec a}}{{|\vec a|}} = \dfrac{{5\hat i - \hat j + 2\hat k}}{{\sqrt {30} }}$So, a vector in direction of $5\hat i - \hat j + 2\hat k$ with magnitude 8 units is:$8\hat a = 8\left( {\dfrac{{5\hat i - \hat j + 2\hat k}}{{\sqrt {30} }}} \right) = \dfrac{{40}}{{\sqrt {30} }}\hat i - \dfrac{8}{{\sqrt {30} }}\hat j + \dfrac{{16}}{{\sqrt {30} }}\hat k$11. You can understand the NCERT concept enriched with expert's knowledge who writes these solutions. For example, it is used to store positions, directions and velocities in games. Two vectors having same magnitude are collinear.Ans: Falseiv. For given vectors, $\vec a = 2\hat i - \hat j + 2\hat k$ and $\vec b = - \hat i + \hat j - \hat k$, find the unit vector in the direction of the vector $\vec a + \vec b$. Ans: $\vec a + \vec b = (2 - 1)\hat i + ( - 1 + 1)\hat j + (2 - 1)\hat k = 1\hat i + 0\hat j + 1\hat k = \hat i + \hat k$$|\vec a + \vec b| = \sqrt {{1^2} + {1^2}} = \sqrt 2 $So, unit vector $ = \dfrac{{(\vec a + \vec b)}}{{|\vec a + \vec b|}} = \dfrac{{\hat i + \hat k}}{{\sqrt 2 }} = \dfrac{1}{{\sqrt 2 }}\hat i + \dfrac{1}{{\sqrt 2 }}\hat k$10. By combining algebraic and geometric properties, you will be able to learn vital applicability of vectors in different areas.10.2 Some Basic ConceptsAny quantity or object that has both magnitude and direction is known as a vector. Collinear but not equal Ans: Vectors $\vec a$ and $\vec c$ are collinear but not equal.5. Answer the following as true or false.i.$\vec a$ and $ - \vec a$ and are collinear.Ans: Trueii. Let the vectors $\vec a$ and $\overrightarrow {\text{b}} $ be such that $|\vec a| = 3$ and $|\vec b| = \dfrac{{\sqrt 2 }}{3}$, then $\vec a \times \overrightarrow {\text{b}} $ is a unit vector, if the angle between $\vec a$ and $\vec b$is:a. Justify your answer with an example.Ans: $\overrightarrow {\text{a}} = 2\widehat {\text{i}} + 4\widehat {\text{j}} + 3\widehat {\text{k}}$ and $\overrightarrow b = 3\widehat {\text{i}} + 3\widehat {\text{j}} - 6\widehat {\text{k}}$$\vec a \cdot \vec b = 2.3 + 4.3 + 3( - 6) = 6 + 12 - 18 = 0$$|{\text{a}}| = \sqrt {{2^2} + {4^2} + {3^2}} = \sqrt {29} $$\therefore \vec a e \vec 0$$|\vec b| = \sqrt {{3^2} + {3^2} + {{( - 6)}^2}} = 54$$\therefore \vec b e \bar 0$So, it is clear from the above example that the converse of the given statement need not be true.15. The correct answer is ${\text{D}}$.Exercise 10.31. $\lambda = - 1$c. Find the unit vector in the direction of vector $\overrightarrow {PQ} $, where ${\text{P}}$ and ${\text{Q}}$ are the points $(1,2,3)$ and $(4,5,6)$, respectively.Ans: $\overrightarrow {PQ} = (4 1)\hat i + (5 - 2)\hat j + (6 - 3)k = 3\hat i + 3\hat j + 3k$$|\overrightarrow {PQ} | = \sqrt {{3^2} + {3^2} + {3^2}} = \sqrt {9 + 9 + 9} = \sqrt {27} = 3\sqrt 3 $So, unit vector $ = \dfrac{{\overrightarrow {PQ} }}{{|\overrightarrow {PQ} |}} = \dfrac{{3\hat i + 3\hat j + 3\hat k}}{{3\sqrt 3 }} = \dfrac{1}{{\sqrt 3 }}\hat i + \dfrac{1}{{\sqrt 3 }}\hat j + \dfrac{1}{{\sqrt 3 }}\hat k$9. The value of $\hat i.(\hat j \times \hat k) + \hat j \cdot (\hat i \times \hat k) + \hat k \cdot (\hat i \times \hat j)$ is$\begin{array}{*{20}{l}} {{\text{ a. Two collinear vectors having the same magnitude are equal.Ans: FalseExercise 10.21. Compute the magnitude of the following vectors:-$\vec a = \hat i + \hat j + \hat k\;;\;\vec b = 2\hat i - 7\hat j - 3\hat k\;;\;\vec c = \dfrac{1}{{\sqrt 3 }}\hat i + \dfrac{1}{{\sqrt 3 }}\hat j - \dfrac{1}{{\sqrt 3 }}\hat k$Ans: $|\vec a| = \sqrt {{{(1)}^2} + {{(1)}^2} + {{(1)}^2}} = \sqrt 3 $$|\vec b| = \sqrt {{{(2)}^2} + {{( - 7)}^2} + {{( - 3)}^2}} $$ = \sqrt {4 + 49 + 9} $$ = \sqrt {62} $$|\vec c| = \sqrt {{{\left( {\dfrac{1}{{\sqrt 3 }}} \right)}^2} + {{\left( {\dfrac{1}{{\sqrt 3 }}} \right)}^2} - {{\left( {\dfrac{1}{{\sqrt 3 }}} \right)}^2}} $$ = \sqrt {\dfrac{1}{3} + \dfrac{1}{3} + \dfrac{1}{3}} = 1$2. $\vec b = \lambda \vec a($ scalar $\lambda )$If $\lambda = \pm 1$, then $\vec a = \pm \vec b$If $\vec a = {a_1}\hat i + {a_2}\hat j + {a_3}\hat k$ and $\vec b = {b_1}\hat i + {b_2}\hat j + {b_3}\hat k$,$\vec b = \lambda \vec a \Rightarrow {b_1}\hat i + {b_2}\hat j + {b_3}\hat k = \lambda \left( {a\hat i + {a_2}\hat j + {a_3}\hat k} \right)$$ \Rightarrow {b_1}\hat i + {b_2}\hat j + {b_3}\hat k = \left( {\lambda {a_1}} \right)\hat i + \left( {\lambda {a_2}} \right)\hat j + \left( {\lambda {a_3}} \right)\hat k$$ \Rightarrow {b_1} = \lambda {a_1},{b_2} = \lambda {a_2},{b_3} = \lambda {a_3}$$ \Rightarrow \dfrac{{{b_1}}}{{{a_1}}} = \dfrac{{{b_2}}}{{{a_2}}} = \dfrac{{{b_3}}}{{{a_3}}} = \lambda $Thus, respective components of $\vec a$ and $\vec b$ are proportional. But, $\vec a$ and $\vec b$ may have different directions. $\dfrac{\pi }{3}$d. Find $|\vec x|$, if for a unit vector $\vec a\;,\;(\vec x - \vec a) \cdot (\vec x + \vec a) = 12$Ans: $(\vec x - \vec a) \cdot (\vec x + \vec a) = 12$$ \Rightarrow \vec x \cdot \vec x + \vec x\vec a - \vec a\vec x - \vec a \cdot \vec a = 12$$ \Rightarrow |\vec x{|^2} - |\vec a{|^2} = 12$$ \Rightarrow |\vec x{|^2} - 1 = 12$$ \Rightarrow |\vec x{|^2} = 13$$\therefore |\vec x| = \sqrt {13} $10. Show that the points ${\text{A}},{\text{B}}$ and ${\text{C}}$ with position vectors, $\vec a = 3\hat i - 4\hat j - 4\hat k\;,\;\vec b = 2\hat i - \hat j + \hat k$ and $\vec c = \hat i - 3\hat j - 5\hat k$, respectively form the vertices of a right angled triangle. Ans: $\overrightarrow {{\text{AB}}} = \vec b \vec a = (2 - 3)\hat i + ( - 1 + 4)\hat j + (1 + 4)\hat k = - \hat i + 3\hat j + 5\hat k$$\overrightarrow {{\text{BC}}} = \vec c - \vec b = (1 - 2)\hat i + ( - 3 + 1)\hat j + ( - 5 - 1)\hat k = - \hat i + 2\hat j - 6\hat k$$\overrightarrow {{\text{CA}}} = \vec a - \vec c = (3 - 1)\hat i + ( - 4 + 3)\hat j + ( - 4 + 5)\hat k = 2\hat i - \hat j + \hat k$$|\overrightarrow {{\text{AB}}} {|^2} = {( - 1)^2} + {3^2} + {5^2} = 1 + 9 + 25 = 35$$|\overrightarrow {{\text{BC}}} {|^2} = {( - 1)^2} + {( - 2)^2} + {( - 6)^2} = 1 + 4 + 36 = 41$$|\overrightarrow {CA} {|^2} = {2^2} + {( - 1)^2} + {1^2} = 4 + 1 + 1 = 6$$|\overrightarrow {{\text{AB}}} {|^2} + |\overrightarrow {{\text{CA}}} {|^2} = 35 + 6 = 41 = |\overrightarrow {{\text{BC}}} {|^2}$So, ABC is a right angled triangle.18. Find the position vector of a point $R$ which divides the line joining two points $P$ and $Q$ whose position vectors are $(2\vec a + \vec b)$ and $(\vec a - 3\vec b)$ externally in the ratio 1: 2. Show that the points ${\text{A}}(1,2,7),\;{\text{B}}(2,6,3)$ and ${\text{C}} (3,10, - 1)$ are collinear. Ans: $\overrightarrow {{\text{AB}}} = (2 - 1)\hat i + (6 - 2)\hat j + (3 - 7)\hat k = \hat i + 4\hat j - 4\hat k$$\overrightarrow {{\text{BC}}} = (3 - 2)\hat i + (10 - 6)\hat \jmath + ( - 1 - 3)\hat k = \hat i + 4\hat j - 4\hat k$$\overrightarrow {{\text{AC}}} = (3 - 1)\hat i + (10 - 2)\hat j + ( - 1 - 7)\hat k = 2\hat i + 8\hat j - 8\hat k$$|\overrightarrow {AB} | = \sqrt {{1^2} + {4^2} + {{( - 4)}^2}} = \sqrt {1 + 16 + 16} = \sqrt {33} $$|\overrightarrow {BC} | = \sqrt {{1^2} + {4^2} + {{( - 4)}^2}} = \sqrt {1 + 16 + 16} = \sqrt {33} $$|\overrightarrow {AC} | = \sqrt {{2^2} + {8^2} + {8^2}} = \sqrt {4 + 64 + 64} = 2\sqrt {33} $$|\overrightarrow {{\text{AC}}} | = |\overrightarrow {{\text{AB}}} | + |\overrightarrow {{\text{BC}}} |$Hence, the given points are collinear.17. $20\;{\text{m/}}{{\text{s}}^2}$Ans: $20\;{\text{m/}}{{\text{s}}^2}$ is a vector quantity because it has both magnitude as well as direction.3. Classify the following as scalar and vector quantities:-i. $\overrightarrow {{\text{AB}}} + \overrightarrow {{\text{BC}}} + \overrightarrow {{\text{CA}}} = \vec 0$b. $\vec b = \lambda \vec a$, for some scalar $\lambda $b. Here the initial and terminating point of the vector co-occurs together and thus gives no direction.Unit Vector: A vector whose magnitude is equivalent to its unit length i.e. 1 is known as a unit vector.Coinitial Vectors: Two or more vectors initiating from the same point are known as coinitial vectors.Collinear Vectors: Two or more vectors are referred to as collinear if they are parallel to the same line, irrespective of their magnitudes and directions.Equal Vectors: Irrespective of the initial points, if two or more vectors have equal magnitude and directions, they are known as equal vectors.Negative of a Vector: A vector with the same magnitude but a different direction that of a given vector is known as a negative vector of the given vector.10.4 Addition of VectorsFor the addition of vectors, you have two laws:1. Also, show that $P$ is the mid point of the line segment RQ. Ans: $\overrightarrow {{\text{OP}}} = 2\vec a + \vec b,\overrightarrow {{\text{OQ}}} = \overrightarrow {\text{a}} - 3\overrightarrow {\text{b}} $$\overrightarrow {OR} = \dfrac{{2(2\vec a + \vec b) - (\vec a - 3\vec b)}}{{2 - 1}} = \dfrac{{4\vec a + 2\vec b - \vec a - 3\vec b}}{1} = 3\vec a + 5\vec b$So, the position vector of ${\text{R}}$ is $3\overrightarrow {\text{a}} + 5\overrightarrow {\text{b}} $Position vector of midpoint of ${\text{RQ}} = \dfrac{{\overrightarrow {{\text{OQ}}} + \overrightarrow {{\text{OR}}} }}{2}$$ = \dfrac{{(a\sqrt 6 ) + (3\vec a + 5\bar b)}}{2}$$ = 2\vec a + \vec b$$ = \overrightarrow {OP} $Thus, ${\text{P}}$ is midpoint of line segment ${\text{RQ}}$10. }}1}& {{\text{ d. Co-initialAns: Vectors $\vec a$ and $\vec d$ are co-initial.ii. $[\angle {\text{ABC}}$ is the angle between the vectors $\overrightarrow {{\text{BA}}} $ and $\overrightarrow {{\text{BC}}} $ ] Ans: $\overrightarrow {{\text{BA}}} = \{ 1 - ( - 1)\hat i + (2 - 0)\hat j + (3 - 0)\hat k = 2\hat i + 2\hat j + 3\hat k$$\overrightarrow {{\text{BC}}} = \{ 0 - ( - 1)\} \hat i + (1 - 0)\hat j + (2 - 0)\hat k = \hat i + \hat j + 2\hat k$$\overrightarrow {{\text{BA}}} \cdot \overrightarrow {{\text{BC}}} = (2\hat i + 2\hat \jmath + 3\hat k) \cdot (\hat i + \hat \jmath + 2\hat k) = 2 \times 1 + 2 \times 1 + 3 \times 2 = 2 + 2 + 6 = 10$$|\overrightarrow {{\text{BA}}} | = \sqrt {{2^2} + {2^2} + {3^2}} = \sqrt {4 + 4 + 9} = \sqrt {17} $$|\overrightarrow {BC} | = \sqrt {1 + 1 + {2^2}} = \sqrt 6 $$\therefore 10 = \sqrt {17} \times \sqrt 6 \cos (\angle {\text{ABC}})$$ \Rightarrow \cos (\angle {\text{ABC}}) = \dfrac{{10}}{{\sqrt {17} \times \sqrt 6 }}$$ \Rightarrow (\angle {\text{ABC}}) = {\cos ^{ - 1}}\left( {\dfrac{{10}} {{\sqrt {102} }}} \right)$16. both the vectors $\vec a$ and $\vec b$ have same direction, but different magnitudes Ans: If $\vec a$ and $\vec b$ are collinear vectors, they are parallel. Triangle Law of Vector Addition: According to this law, the initial point of one vector should coincide with the terminating point of the other vector.(Image to be added soon)So it can be concluded that: \[\overrightarrow{AC}\]= \[\overrightarrow{AB}\]+ \[\overrightarrow{BC}\]2. If $\vec a,\vec b,\vec c$ are unit vectors such that $\vec a + \vec b + \vec c = 0$, find the value of $\vec a \cdot \vec b + \vec b \cdot \vec c + \vec c\;.\;\vec a$.Ans: $|\vec a + \vec b + \vec c{|^2} = (\vec a + \vec b + \vec c) \cdot (\vec a + \vec b + \vec c) = |\vec a{|^2} + |\vec b{|^2} + |\vec c{|^2} + 2(\vec a\vec b + \vec b\vec c + \vec c \cdot \vec a)$$ \Rightarrow 0 = 1 + 1 + 1 + 2(\vec a\vec b + \vec b \cdot \vec c + \vec c \cdot \vec a)$$ \Rightarrow (\vec a\;.\;\vec b + \vec b \cdot \vec c + \vec c\;.\;\vec a) = \dfrac{{ - 3}}{2}$14. \[\overrightarrow{(\gamma b)}\]10.6.2 Projection of a Vector on a Line(Image to be added soon)When a vector \[\overrightarrow{AB}\] with a given directed line "l" makes an angle in an anticlockwise direction, the vector \[\overrightarrow{p}\] formed will be termed as the projection vector and its magnitude will be termed as the projection of the vector \[\overrightarrow{AB}\] on the directed line "l".10.6.3 Vector (or Cross) Product of Two VectorsIn this two non zero vectors \[\overrightarrow{a}\] and \[\overrightarrow{b}\] are multiplied with respect to sin which denotes the angle between both the vectors and is a unit vector perpendicular to both \[\overrightarrow{a}\] and \[\overrightarrow{b}\], in order to form a right-handed system wherein the right-handed system rotated from \[\overrightarrow{a}\] to \[\overrightarrow{b}\] moves in the direction of .The vector product can be written as:\[\overrightarrow{a \times b}\] = |\[\overrightarrow{a}\] | |\[\overrightarrow{b}\] | \[\overrightarrow{sin \theta}\](Image to be added soon)Key Features of NCERT Solutions for Class 12 Maths Chapter 10NCERT Solutions for Class 10 Maths offered by Vedantu are the one which can guide you the best as they provide complete and comprehensive details of the topic. So, option D is incorrect. \[\overrightarrow{b}\] = \[\overrightarrow{\gamma}\] \[\overrightarrow{(a \; . Externally.Ans: The position vector of $R$ is $\overrightarrow {{\text{OR}}} = \dfrac{{2( - \hat i + \hat j + \hat k) - 1(\hat i + 2\hat j - \hat k)}}{{2 - 1}} = ( - 2\hat i + 2\hat j + 2\hat k) - (\hat i + 2\hat j - \hat k)$$ = - 3\hat i + 3\hat k$16. Find the position vector of the mid point of the vector joining the points P(2,3,4) and Q(4,1,-2).Ans: The position vector of $R$ is$\overrightarrow {{\text{OR}}} = \dfrac{{(2\hat i + 3\hat j + 4\hat k) + (4\hat i + \hat j - 2\hat k)}}{2} = \dfrac{{(2 + 4)\hat i + (3 + 1)\hat j + (4 - 2)\hat k}}{2}$$ = \dfrac{{6\hat i + 4\hat j + 2\hat k}}{2} = 3\hat i + 2\hat j + \hat k$17. Show that the points ${\text{A}}(1, - 2, - 8),{\text{B}}(5,0, - 2)$ and ${\text{C}}(11,3,7)$ are collinear, and find the ratio in which B divides Ac. Ans: $\overrightarrow {{\text{AB}}} = (5 - 1)\hat i + (0 + 2)\hat j + ( - 2 + 8)\hat k = 4\hat i + 2\hat j + 6\hat k$$\overrightarrow {{\text{BC}}} = (11 - 5)\hat i + (3 - 0)\hat j + (7 + 2)\hat k = 6\hat i + 3\hat j + 9\hat k$$\overrightarrow {AC} = (11 - 1)\hat i + (3 + 2)\hat j + (7 + 8)\hat k = 10\hat i + 5\hat j + 15\hat k$$|\overrightarrow {{\text{AB}}} | = \sqrt {{4^2} + {2^2} + {6^2}} = \sqrt {16 + 4 + 36} = \sqrt {56} = 2\sqrt {14} $$|\overrightarrow {{\text{BC}}} | = \sqrt {{6^2} + {3^2} + {9^2}} = \sqrt {36 + 9 + 81} = \sqrt {126} = 3\sqrt {14} $$|\overrightarrow {AC} | = \sqrt {{{10}^2} + {5^2} + {{15}^2}} = \sqrt {100 + 25 + 225} = \sqrt {350} = 5\sqrt {14} $$\therefore \;\;|\overrightarrow {{\text{AC}}} | = |\overrightarrow {{\text{AB}}} | + |\overrightarrow {{\text{BC}}} |$So, the points are collinear.Let B divide AC in ratio $\lambda :1$. Time periodAns: Time period is a scalar quantity because it has only magnitude.ii. $\lambda = 1$b. $\theta = \dfrac{\pi }{4}$b. Find the values of x and y so that the vectors $2\hat i + 3\hat j$ and $x\hat i + y\hat j$ are equal. Ans: $2\hat i + 3\hat j = x\hat i + y\hat j \Rightarrow x = 2,y = 3$5. Find the scalar and vector components of the vector with initial point $(2,1)$ and terminal point $( - 5,7)$.Ans: Let the points be ${\text{P}}(2,1)$ and ${\text{Q}}( - 5,7)$ $\overrightarrow {PQ} = ( - 5 2)\hat i + (7 - 1)\hat j$$ \Rightarrow \overrightarrow {PQ} = - 7\hat i + 6\hat j$So, scalar components of required vector are $ - 7$ and 6 and the vector components are $ - 7\hat i$ and $6\hat j$.6. Find the sum of the vectors $\vec a = \hat i - 2\hat j + \hat k\;,\;\vec b = - 2\hat i + 4\hat j + 5\hat k$ and $\vec c = \hat i - 6\hat j - 7\hat k$.Ans: $\vec a + \vec b + \vec c = (1 - 2 + 1)\hat i + ( - 2 + 4 - 6)\hat j + (1 + 5 - 7)\hat k{\text{ }} = 0\hat i - 4\hat j - 1\hat k{\text{ }} = - \;4\hat j - \hat k{\text{ }}$7. Find the unit vector in the direction of the vector $\vec a = \hat i + \hat j + 2\hat k$. Ans: $|\vec a| = \sqrt {{1^2} + {1^2} + {2^2}} = \sqrt {1 + 1 + 4} = \sqrt 6 $$\therefore \hat a = \dfrac{{\vec a}}{{|\vec a|}} = \dfrac{{\hat i + \hat j + 2\hat k}}{{\sqrt 6 }} = \dfrac{1}{{\sqrt 6 }}\hat i + \dfrac{1}{{\sqrt 6 }}\hat j + \dfrac{2}{{\sqrt 6 }}\hat k$8. Exercise 10.11. Find a vector $\vec d$ which isperpendicular to both $\vec a$ and $\vec b$ and $\vec c \cdot \vec d = 15$ Ans: $\vec d = {d_1}\hat i + {d_2}\hat \jmath + {d_3}\hat k$$\vec d\;.\;\vec a = 0 \Rightarrow {d_1} + 4{d_2} + 2{d_3} = 0$$\vec d\;.\;\vec b = 0 \Rightarrow 3{d_1} - 2{d_2} + 7{d_3} = 0$$\vec c \cdot \vec d = 15 \Rightarrow 2{d_1} - {d_2} + 4{d_3} = 15$Solving these equations, we get ${d_1} = \dfrac{{160}}{3},{d_2} = - \dfrac{5}{3},{d_3} = - \dfrac{{70}}{3}$$\therefore \vec d = \dfrac{{160}}{3}\hat i + \dfrac{5}{3}\widehat j + \dfrac{{70}}{3}\hat k = \dfrac{1}{3}(160\hat i + 5\hat j + 70\hat k)$13. 2d. Area of a rectangle having vertices ${\text{A}},{\text{B}},{\text{C}}$, and ${\text{D}}$ with position vectors $ - \hat i + \dfrac{1}{2}\hat j + 4\hat k,\hat i + \dfrac{1}{2}\hat j + 4\hat k,\hat i - \dfrac{1}{2}\hat j + 4\hat k$ and $ - \hat i - \dfrac{1}{2}\hat j + 4\hat k$ respectively isa. Represent graphically a displacement of $40\;{\text{km}},{30^\circ }$ east of north.Ans:(Image will be uploaded soon)Here, the vector OP is representing the displacement of $40\;{\text{km}},{30^\circ }$ in East of North direction.2. Classify the following measures as scalars and vectors.i. $10\;{\text{kg}}$Ans: $10\;{\text{kg}}$is a scalar quantity because it has only magnitude not direction.ii. 2 meters north-westAns: 2 meters north-west is a vector quantity because it has both magnitude as well as direction.iii. $40$ wattAns: 40 watts is a scalar quantity because it has only magnitude, not direction.v. ${10^{ 19}}$coulombAns: ${10^{ - 19}}$Coulomb is a scalar quantity because it has only magnitude not direction.vi. If $\overrightarrow a = \vec b + \vec c$, then is it true that $|\overrightarrow a | = |\vec b| + |\vec c|$ ? The direction of the arrow is represented by the arrow.Positional Vector: In a set of three-dimensional coordinates, the distance between two points is called a position vector.Direction Cosines: The cosine values of the angles made by the vector in a three-dimensional plane is called direction cosines.10.3 Types of VectorsZero Vector: A vector with magnitude 0 is called a zero vector.

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