1. Find the distance from the point (1, 4, 1) to the plane ...
[Pages:5]1. Find the distance from the point (1, 4, 1) to the plane 2x + 3y - z = -1.
We need to find some vector to the plane, for which we need a point in the
plane. If we pick x = 0 and y = 0, we get -z = -1, so z = 1. Thus, (0, 0, 1)
is a point in the plane. Then we can compute a vector from the point to the
plane as
1 0 1
4 - 0 = 4
1
1
0
The shortest distance from a point to a plane is along a line orthogonal to
the plane. That is, it is in the direction of the normal vector. We can project
the vector we found earlier onto the normal vector to find the shortest vector
1
2
from the point to the plane. If we let v = 4 and n = 3 , we can
0
-1
compute
v ? n = (1)(2) + (4)(3) + (0)(-1) = 14
||n||2 = 22 + 32 + (-1)2 = 14 v?n
projnv = ||n||2 n
14
2
2
= 3 = 3
14 -1
-1
Then the distance from the point to the plane is the length of this vector, 14.
2. Solve the following system of equations:
w + 3x - z = -1 w + 4x - 2y + 2z = 4 3w + 11x - 3y + 5z = 4
-2w - 7x + y = 0
We set up the augmented matrix, and then put the left side in reduced row echelon form.
1 3 0 -1
1 4 -2 2
3
11 -3
5
-2 -7 1 0
-1 .
4 -R1
4
-3R1
0 +2R1
1 3 0 -1
0 1 -2 3
0
2
-3
8
0 -1 1 -2
-1 -3R2
5 .
7
-2R2
-2 +R2
1 0 6 -10
0 1 -2 3
0
0
1
2
0 0 -1 1
-16 -6R3
5 +2R3
-3
.
3 +R3
1 0 0 -22 2 .
0 1 0 7
0
0
1
2
000 3
-1 .
-3
.
0
?
1 3
1 0 0 -22 2 +22R4
0 1 0 7
0
0
1
2
000 1
-1 -7R4
-3
-2R4
0.
1 0 0 0
0 1 0 0
0
0
1
0
0001
2
-1
-3
0
w 2
The
unique
solution
is
thus
x y
=
-1 -3
.
z
0
3. Let
2 1
3 2
-1
X
-1 5
0 4
-1
=
-1 3
-2 0
. Solve for the matrix X.
We can multiply by
23 12
on the left and
-1 0 54
on the right to get
23 12
2 1
3 2
-1
X
-1 0 -1 54
-1 0 54
=
IXI =
X=
23 12
-1 -2 30
7 -4 5 -2
-1 0 54
-27 -16 -15 -8
-1 0 54
4. Determine whether the following matrix is invertible, and if so, find its inverse:
1 2 0 0
-1 -1 -1 0
1
4
-1
3
2 1 1 -5
We put the original matrix on the left side of an augmented matrix, and put the identity matrix on the right side. We then put the left side into reduced row echelon form.
1 2 0 0
-1 -1 -1 0
1
4 -1 3
2 1 1 -5
1 0 0 0 .
0 1 0 0 +R1
0
0
1
0
-R1
0 0 0 1 -2R1
1 2 0 0
0 1 -1 0
0
2
-1
3
0 -3 1 -5
1 0 0 0 -2R2
1 1 0 0 .
-1
0
1
0
-2R2
-2 0 0 1 +3R2
1 0 2 0
0 1 -1 0
0
0
1
3
0 0 -2 -5
-1 -2 0 0 -2R3
1 1 0 0 +R3
-3
-2
1
0
.
1 3 0 1 +2R3
1 0 0 -6
0 1 0 3
0
0
1
3
000 1
5 2 -2 0 +6R4
-2 -1 1 0 -3R4
-3 -2
1
0
-3R4
-5 -1 2 1 .
1 0 0 0
0 1 0 0
0
0
1
0
0001
-25 -4 10 6
13 2 -5 -3
12
1
-5
-3
-5 -1 2 1
Thus, our matrix is invertible, and its inverse is
-25 -4 10 6
13 2 -5 -3
12
1
-5
-3
-5 -1 2 1
5. Three of these systems of linear equations have the same set of solutions. Determine the one that differs and justify your answer.
2 0
0
4 3 0
1 -5 0
-3 7 -4
x1 x2 x3 x4
=
0 0 0
4 0
0
5 3 0
7 -5 3
1 7 2
x1 x2 x3 x4
=
0 0 0
2 0
0
1 3 0
6 -5 0
-10 -1 2
x1 x2 x3 x4
=
0 0 0
4 0
0
5 6 0
7 -10
0
1 6 1
x1 x2 x3 x4
=
0 0 0
In the bottom left system, x4 is a free variable. In the rest, the bottom line of the matrix gives us that x4 = 0. Thus, the bottom left system is the one that differs from the others.
2
0
2
6. Let u = 5 , v = 2 , and w = 3 . Compute
-1
6
1
u - (u ? w)v ? w ? u - (u ? w)v .
Let x = u - (u ? w)v. We are asked to compute x ? (w ? x). The cross product of two vectors is orthogonal to both vectors, so in particular, w ? x is orthogonal to x. Therefore, x ? (w ? x) = 0.
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