VECTORS IN SPACE. LINES AND PLANES
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40.109
Show that the line X of intersection of the planes x + y - z = 0 and x- y -5z + 7 = 0 is parallel to the line
M determined by
The line Z£ must be perpendicular to the normal vectors of the planes, (1,1, -1) and (1, -1, -5). Note that
M is parallel to the vector (3, -2,1), which is perpendicular to both (1,1, -1) and (1, -1, -5). Hence, M is
parallel to £.
Show that the second-degree equation (2x + y - z - 3) + (x + 2y - 3r + 5) =0 represents a straight line in
space.
a
2 + b
2 = 0 if and only if a = 0 and b = 0. So, the given equation is equivalent to 2x + y — z — 3 = 0
and x + 2y — 3z + 5 = 0, a pair of intersecting planes that determine a straight line.
Find cos 0, where 0 is the angle between the planes 2x — y + 1z = 3 and 3* + 2y — 6z = 7.
)is equal to the angle between the normal vectors to the planes: (2, —1,2) and (3,2, -6). So,
Find an equation of the line through P(4,2, —1) and perpendicular to the plane 6x - 3y + z = 5.
The line is parallel to the normal vector to the plane, (6, -3, 2). Hence, the line has the parametric equations
x = 4 + 6t, y = 2-3t, z = -l + 2t.
Find equations of the line through P(4, 2, — 1) and parallel to the intersection .2" of the planes x — y +2z + 4 = 0
and 2x + 3y + 6z-l2 = 0.
The line «S? is perpendicular to the normal vectors of the planes, (1, — 1, 2) and (2, 3,6), and is, therefore,
parallel to their cross product (1, —1,2) x (2,3, 6) = (-12, -2, 5). Hence, the required equations are x =
4-12?, y = 2-2t, z = -l + 5t.
Find an equation of the plane through P(l, 2, 3) and parallel to the vectors (2,1, -1) and (3, 6, -2).
A normal vector to the plane is (2,1, —1) x (3,6, ^2) = (4,1,9). So, the plane has an equation of the
form 4x + y + 9z = d. Since (1,2, 3) lies in the plane, 4 + 2 +27 = d, d = 33. So, the equation is 4x +
y + 9z = 33.
Find an equation of the plane through (2, -3,2) and the line 3!determined by the planes 6x + 4y + 3z + 5 = 0
and 2* + _y + z-2 = 0.
Consider the plane (6x + 4y + 3z + 5) + k(2x + y + z - 2) = 0. This plane passes through X. We want
the point (2,-3,2) to be on the plane. So, (12- 12 + 6 + 5) + fc(4-3 + 2-2) = 0, ll + fc = 0, A: =-11.
Therefore, we get 6x + 4y + 3z + 5 - 11(2* + y + z - 2) = 0, -I6x -7y - 8z + 27 = 0, 16x + 7y + 8z -
27 = 0.
Find an equation of the plane through P 0 (2, —1, —1) and P^l, 2, 3) and perpendicular to the plane 2x + 3y —
5z-6 = 0.
is parallel to the plane. Also, the normal vector (2, 3, -5) to the plane 2x + 3y - 5z -
6 = 0 is parallel to the sought plane. Hence, a normal vector to the required plane is (-1, 3, 4) x (2,3, -5) =
(-27,3,-9). Thus, an equation of that plane has the form -27* + 3y - 9z = d. Since (1,2, 3) lies in the
plane, -27 + 6-27 = d, d=-48. Thus, we get -27* + 3y - 9z = -48, or 9* - y + 3z = 16.
Let A(l, 2,3), B(2, -1,5), and C(4,1,3) be consecutive vertices of a parallelogram ABCD (Fig. 40-9). Find
the coordinates of D.
Hence, D has coordinates
(3,4,1).
AD = BC=(2, 2, -2). OD = OA + AD = (1, 2, 3) + (2, 2, -2) = (3, 4,1).
P 0 P, = (-1,3,4)
359
40.101
40.102
40.103
40.104
40.105
40.106
40.107
40.108
40.109
Show that the line X of intersection of the planes x + y - z = 0 and x- y -5z + 7 = 0 is parallel to the line
M determined by
The line Z£ must be perpendicular to the normal vectors of the planes, (1,1, -1) and (1, -1, -5). Note that
M is parallel to the vector (3, -2,1), which is perpendicular to both (1,1, -1) and (1, -1, -5). Hence, M is
parallel to £.
Show that the second-degree equation (2x + y - z - 3) + (x + 2y - 3r + 5) =0 represents a straight line in
space.
a
2 + b
2 = 0 if and only if a = 0 and b = 0. So, the given equation is equivalent to 2x + y — z — 3 = 0
and x + 2y — 3z + 5 = 0, a pair of intersecting planes that determine a straight line.
Find cos 0, where 0 is the angle between the planes 2x — y + 1z = 3 and 3* + 2y — 6z = 7.
)is equal to the angle between the normal vectors to the planes: (2, —1,2) and (3,2, -6). So,
Find an equation of the line through P(4,2, —1) and perpendicular to the plane 6x - 3y + z = 5.
The line is parallel to the normal vector to the plane, (6, -3, 2). Hence, the line has the parametric equations
x = 4 + 6t, y = 2-3t, z = -l + 2t.
Find equations of the line through P(4, 2, — 1) and parallel to the intersection .2" of the planes x — y +2z + 4 = 0
and 2x + 3y + 6z-l2 = 0.
The line «S? is perpendicular to the normal vectors of the planes, (1, — 1, 2) and (2, 3,6), and is, therefore,
parallel to their cross product (1, —1,2) x (2,3, 6) = (-12, -2, 5). Hence, the required equations are x =
4-12?, y = 2-2t, z = -l + 5t.
Find an equation of the plane through P(l, 2, 3) and parallel to the vectors (2,1, -1) and (3, 6, -2).
A normal vector to the plane is (2,1, —1) x (3,6, ^2) = (4,1,9). So, the plane has an equation of the
form 4x + y + 9z = d. Since (1,2, 3) lies in the plane, 4 + 2 +27 = d, d = 33. So, the equation is 4x +
y + 9z = 33.
Find an equation of the plane through (2, -3,2) and the line 3!determined by the planes 6x + 4y + 3z + 5 = 0
and 2* + _y + z-2 = 0.
Consider the plane (6x + 4y + 3z + 5) + k(2x + y + z - 2) = 0. This plane passes through X. We want
the point (2,-3,2) to be on the plane. So, (12- 12 + 6 + 5) + fc(4-3 + 2-2) = 0, ll + fc = 0, A: =-11.
Therefore, we get 6x + 4y + 3z + 5 - 11(2* + y + z - 2) = 0, -I6x -7y - 8z + 27 = 0, 16x + 7y + 8z -
27 = 0.
Find an equation of the plane through P 0 (2, —1, —1) and P^l, 2, 3) and perpendicular to the plane 2x + 3y —
5z-6 = 0.
is parallel to the plane. Also, the normal vector (2, 3, -5) to the plane 2x + 3y - 5z -
6 = 0 is parallel to the sought plane. Hence, a normal vector to the required plane is (-1, 3, 4) x (2,3, -5) =
(-27,3,-9). Thus, an equation of that plane has the form -27* + 3y - 9z = d. Since (1,2, 3) lies in the
plane, -27 + 6-27 = d, d=-48. Thus, we get -27* + 3y - 9z = -48, or 9* - y + 3z = 16.
Let A(l, 2,3), B(2, -1,5), and C(4,1,3) be consecutive vertices of a parallelogram ABCD (Fig. 40-9). Find
the coordinates of D.
Hence, D has coordinates
(3,4,1).
AD = BC=(2, 2, -2). OD = OA + AD = (1, 2, 3) + (2, 2, -2) = (3, 4,1).
P 0 P, = (-1,3,4)