Math 234, Practice Test #1
Math 234, Practice Test #1
Math 234, Practice Test #1
1. Find parametric equations for the line in which the planes x+2y+z = 1
and x − y + 2z = −8 intersect.
2. Compute the distance from the point (2, 2, 3) to the plane through the
points A = (0, 0, 0), B = (2, 0, −1) and C = (2, −1, 0).
3. Compute the area of the parallelogram with three of its vertices given
by
A = (2, −2, 1) , B = (3, −1, 2) and C = (3, −1, 1)
(a) xy-plane
(b) yz-plane
(c) xz-plane
1
Solutions
1. Normal vectors of the two planes are given by
−8 = x − y + 2z = 1 − 2y − z − y + 2z = 1 − 3y + z
x = 10 + 5t , y = −t , z = −9 − 3t
Comment: There are many possible solutions which are all correct (and
for which you would get full credit). For example, the vector (−5, 1, 3)
is also parallel to the line of intersection (do you know why ?). Later
on, instead of choosing y = 0 to find a point on the line we could have
chosen something else, for example y = 1. We would have obtained
x = 10 − 5y = 5 and z = −9 + 3y = −6, so that we use the point
(5, 1, −6) instead. Remember that there are many points on a line,
2
and there is no reason to prefer one over the other. The parametric
equations would then look as follows
x = 5 − 5t , y = 1 + t , z = −6 + 3t.
Although these equations are different they describe the same line, and
they are therefore also a valid solution to the problem.
2. We need to find a normal vector n to the plane, i.e. a vector perpen-
⇀ ⇀
dicular to both AB= (2, 0, −1) and AC= (2, −1, 0). We get a such a
vector by taking the cross product
⇀ ⇀
i j k
n =AB × AC= 2 0 −1 = −i − 2j − 2k
2 −1 0
and ⇀ ⇀ √
| AB × AC | = 2.
3
4. The statement is false. Pick two different vectors v and w, for example
v = (1, 0, 0) and w = (0, 1, 0). Then choose u perpendicular to both v
and w, for example u = (0, 0, 1) would do. Then
u•v =u•w =0
but v 6= w.