MULTIVARIABLE CALCULUS PRACTICE MIDTERM 1 SOLUTIONS

The length of a × b is ||a × b|| = √ 4 + 4 + 4 = 2 √ 3. Both ±(a × b)/|a × b| are unit vectors
perpendicular to a and b. Therefore the answer is
√ √ √ √ √ √
〈− 3
3 , 3
3 , − 3
3 〉 and 〈 3
3 , − 3
3 , 3
3 〉
8. (15 points) Let P be the plane perpendicular to 〈1, 2, 3〉 passing through the point (1, 0, 1).
a. (5 pts) Find the scalar equation for the plane P .
Solution: The scalar equation is (x − 1) + 2y + 3(z − 1) = 0 or x + 2y + 3z = 4.
b. (5 pts) Is the plane P parallel to the plane defined by 2x + 3y − 4z = 2 Is it
perpendicular
Solution: Plane P has normal vector n 1 = 〈1, 2, 3〉 while the plane 2x + 3y − 4z = 2 has normal
vector n 2 = 〈2, 3, −4〉. Since n 1 and n 2 are not scalar multiplies, the planes are not parallel.
Since n 1 · n 2 = 2 + 6 − 12 = −4 ≠ 0, the planes are not perpendicular.
c. (5 pts) Does the line given by the parameterization x(t) = 3t + 1, y(t) = 3 and
z(t) = −t + 3 intersect the plane P
Solution: The line is in the direction of the vector v = 〈3, 0, −1〉. Since v · n 1 = 3 − 3 = 0, v
is perpendicular to n 1 . Therefore either the line lies on the plane or is parallel to the plane but
does not intersect. By setting t = 0, we see that the point (3, 3, 3) is on the line. But this point
does not line on the plane as it does not satisfy the equation x + 2y + 3z = 4. Therefore, the
answer is No.
9. (15 points)
a. (10 pts) Find parametric equations for the line of intersection of the planes x + y = 1
and y + z = 1.
Solution: The point (0, 1, 0) is on both planes (one could use other points, such as (1, 0, 1) and
this would lead to a slightly different parametric equation of the line.)
The first plane has normal vector n 1 = 〈1, 1, 0〉 and the second plane has normal vector n 2 =
〈0, 1, 1〉. Therefore, the line of intersection has the direction vector v = n 1 × n 2 . We compute
i j k
n 1 × n 2 =
1 1 0
= i − j + k = 〈1, −1, 1〉
∣0 1 1∣ This gives the parametric equation for the line of intersection:
x(t) = t
y(t) = 1 − t
z(t) = t
b. (5 pts) Find symmetric equations in x, y and z for the same line.
Solution: By solving for t, we get the symmetric equation:
x = 1 − y = z
4