Example 0.1.Vector equation of a line

[Pages:4]Example 0.1. Vector equation of a line.

Find a vector equation for the line through (4, 6, -3) and parallel to v = 5i - 10j + 2k.

Solution: With the identifications x0 = 4, y0 = 6, z0 = -3, a = 5, b = -10 and c = 2 we obtain from

< x, y, z >=< x(t), y(t), z(t) >=< x0 + at, y0 + bt, z0 + ct >=< x0, y0, z0 > +t < a, b, c >, a vector equation of the line:

< x, y, z >=< 4, 6-, 3 > +t < 5, -10, 2 > .

Example 0.2. Vector equation of a line.

Find a vector equation for the line though (2, -1, 8) and (5, 6, -3).

Solution: If we label the points as P0(2, -1, 8) and P1(5, 6, -3), then a direction vector for the line through P0 and P1 is

# ? #? #? v = P0P1 = OP1 - OP0 =< 5 - 2, 6 - (-1), -3 - 8 >=< 3, 7, -11 > .

From the vector equation of that line

r = r0 + t(r1 - r0),

# ?

# ?

# ?

(here r = OP , r0 = OP0, r1 = OP1) a vector equation of the line is

< x, y, z >=< 2, -1, 8 > +t < 3, 7, -11 > .

This is one of many possible equations of the line. For example, two alternative equations are

< x, y, z > =< 5, 6, -3 > +t < 3, 7, -11 >, < x, y, z > =< 5, 6, -3 > +t < -3, -7, 11 > .

Example 0.3. Parametric equation of a line.

Find a parametric equation for the line through (5, 2, 4) parallel to v = 4i + 7j - 9k. Solution:

Make identifications x0 = 5, y0 = 2, z0 = 4, a = 4, b = 7 and c = -9 so that from the parametric equation

x(t) = x = x0 + at, y(t) = y = y0 + bt, z(t) = z = z0 + ct, we get

x = 5 + 4t, y = 2 + 7t, z = 4 - 9t.

Definition 0.1. Two lines L1 and L2 with direction vectors v1 and v2, respectively, are i) perpendicular if v1 ? v2 = 0, and ii) parallel if v2 = kv1 for some non-zero scalar k.

Example 0.4. Perpendicular lines.

Determine whether the lines

(0.1)

L1 :

(0.2)

L2 :

are perpendicular.

x = -6 - t, y = 20 + 3t, x = 5 + 2s y = -9 - 4s,

1

z = 1 + 2t z = 1 + 7s

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Solution: Reading off the coefficients of the parameters t and s, we see that

v1 = -i + 3j + 2k and

v2 = 2i - 4j + 7k are the direction vectors for L1 and L2. Because v1 ? v2 = -2 - 12 + 14 = 0, we conclude that the lines are perpendicular.

Example 0.5. Parallel lines.

Determine whether the lines

(0.3) (0.4) are parallel.

L1 : x = 4 - 2t, y = 1 + 4t, z = 3 + 10t

1

L2 :

x = s y = 6 - 2s,

z = - 5s 2

Solution: Reading off the coefficients of the parameters t and s, we see that

v1 = -2i + 4j + 10k

and v2 = i - 2j - 5k

are the direction vectors for L1 and L2. Because v1 = -2v2, we conclude that the lines are parallel.

Example 0.6. Equation of a plane.

Find an equation of the plane that contains the point (4, -1, 3) and is perpendicular to the vector n = 2i + 8j - 5k.

Solution: It follows immediately from the equation of the plane containing P0(x0, y0, z0) and with normal vector n = ai + bj + ck, that is,

a(x - x0) + b(y - y0) + c(z - z0) = 0, that the identification x0 = 4, y0 = -1, z0 = 3, a = 2, b = 8, c = -5 yields

2(x - 4) + 8(y + 1) - 5(z - 3) = 0

Example 0.7. Normal vector a plane.

The graph of a linear equation ax + by + cz + d = 0, with a, b, c NOT all zero, is a plane with normal vector n = ai + bj + ck.

For instance by reading off the coefficients x, y, z in the linear equation 3x - 4y + 10z - 8 = 0, we obtain the normal vector

n = 3i - 4j + 10k

Of course a non-zero scalar multiple of a normal vector n is still perpendicular to the plane.

Three non collinear points P1, P2, P3 also determine a plane S. To obtain an equation of the plane, we need only form two vectors between two pairs of the points. The cross product is a vector normal to the plane containing these vectors. If P (x, y, z) represents any point on the plane and

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# ?

# ?

# ?

# ?

r = OP , r1 = OP1, r2 = OP2, r3 = OP3, then r - r1 (or, for that matter, r - r2 or r - r3) IS IN

THE PLANE. Hence

[(r2 - r1) ? (r3 - r1)] ? (r - r1) = 0

is a vector equation of the plane S.

Example 0.8. Normal vector a plane.

Find the equation of the plane that contains (1, 0, -1), (3, 1, 4), (2, -2, 0).

Solution: We need 3 vectors. Pairing the points as shown yields (the order in which we subtract is irrelevant)

(0.5)

(1, 0, -1)&(3, 1, 4) u = 2i + j + 5k,

next

(0.6)

(3, 1, 4)&(2, -2, 0) v = i + 3j + 4k,

and

(0.7)

(2, -2, 0)&(x, y, z) w = (x - 2)i + (y + 2)j + zk.

We also have that the cross product

u ? v = -11i - 3j + 5k

is a vector normal to the plane containing the given points. Hence from the vector equation of the plane

n ? (r - r0) = 0

we see that a vector equation of the plane is (u ? v) ? w = 0. This last equation implies

-11(x - 2) - 3(y + 2) + 5z = 0

or

-11x - 3y + 5z + 16 = 0.

Finally, we have the following definition. Two planes S1 and S2 with normal vectors n1 and n2, respectively, are i) perpendicular if n1 ? n2 = 0, ii) parallel if n2 = kn1, for some nonzero scalar k.

Example 0.9. Line of intersection.

Find parametric equations for the line of intersection of x - y + 2z = 1 and x + y + z = 3

Solution: In a system of two equations and three unknowns, we choose one variable arbitrarily, say z = t, and solve for x and y from

(0.8) (0.9)

x - y = 1 - 2t x + y = 3 - t.

Solving the system then gives

3

1

x = 2 - t, y = 1 + t, z = t.

2

2

These are the parametric equations for the line L of intersection of the given planes.

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Example 0.10. Point of intersection.

Find the point of intersection of the plane 3x-2y +z = -5 and the line x = 1+t, y = -2+2t, z = 4t.

Solution: If (x0, y0, z0) denotes the point of intersection, then we must have

3x0 - 2y0 + z0 = -5, x0 = 1 + t0, y0 = -2 + 2t, z = 4t0, for some number t0. Substituting the latter equations into the equation of the plane gives

3(1 + t0) - 2(-2 + 2t0) + 4t0 = -5 or

t0 = -4. From the parametric equations for the lines, we then obtain x0 = -3, y0 = -10 and z0 = -16. So the point is (-3, -10, -16).

Example 0.11. Distance D from a point to a plane.

Let P0(x0, y0, z0) be a point on the plane ax + by + cz + d = 0 and let n be a normal vector to the plane.

Solution: If P1(x1, y1, z1) is any point not on the plane, then the distance D from a point to a plane is given by

D = |ax1+ by1 + cz1 + d| a2 + b2 + c2

Homework: prove this formula.

Example 0.12. Distance D from a point to a plane.

Let P0(2, 1, 4) be a point on the plane x - 3y + z - 6 = 0. Find the distance from P0 to the plane. Solution: Do this as homework.

Example 0.13. Angle between planes.

The angle between two planes is defined to be the acute angle between their normal vectors.

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