MULTIVARIABLE CALCULUS PRACTICE MIDTERM 1 SOLUTIONS

MULTIVARIABLE CALCULUS PRACTICE MIDTERM 1 SOLUTIONS

1.

(5 points) If ~v × w ~ = h1, 1, −1i and ~v · w ~ = 2, find the angle between ~v and w. ~

Solution: Let θ be the angle in question. By using the formula for the magnitude of cross product (Theorem 3b on page 758), we obtain √ |~v ||w| ~ sin(θ) = |~v × w| ~ = |h1, 1, −1i| = 3. We also have |~v ||w| ~ cos(θ) = ~v · w ~ = 2. It follows that tan(θ) = √ answer is θ = arctan 3/2.

sin(θ) cos(θ)

√

=

3 2 .

Therefore the

2.

(10 points) For which values of parameter a, the vectors ~u = h1, a, 2i and ~v = ha, 4, 4i are (a) parallel? (b) orthogonal? (c) make 45◦ angle? Solution: (a) ~u and ~v are parallel if and only if ~u × ~v = ~0. We compute ~u × ~v = h1, a, 2i × ha, 4, 4i = h4a − 8, 2a − 4, 4 − a2 i. This vector is zero if and only if a = 2. Answer: a = 2. (b) ~u and ~v are orthogonal if and only if ~u · ~v = a + 4a + 8 = 5a + 8 = 0. Answer: a = − 85 . (c) If ~u and ~v make 45◦ angle, then by the formula ~v · w ~ = |~v ||w| ~ cos(θ), we have p p 5a + 8 = 5 + a2 a2 + 32 cos(45◦ ). This implies a4 − 13a2 − 160a + 32 = 0. This cannot be solved by hand, but using a computer one finds that the solutions are approximately a ≈ 0.197 and a ≈ 6.171 Since calculators will not be allowed on the exam, any actual exam problem will not require extensive computation.

3.

(15 points) a. (5 pts) b. (5 pts) c. (5 pts)

Find an equation of the following lines (if they exist!): The line through A = (2, 4, 2), B = (3, 7, −2), C = (1, 3, 3). The line that passes through the points and (3, 7, 0) and (−5, 5, 1). The line through the point (0, 1, 1) and perpendicular to the vector h4, 6, −2i. −−→ −→ (a) We have AB = h1, 3, −4i and AC = h−1, −1, 1i. Since h1, 3, −4i × h−1, −1, 1i = h1, −3, −2i = 6 ~0, −−→ −→ we conclude that AB is not parallel to AC. Hence, there is no line through the three points (b) The direction vector is h−5, 5, 1i − h3, 7, 0i = h−8, −2, 1i. By taking the point (3, 7, 0) on the line, we obtain the parametric equation of the line   x = 3 − 8t y = 7 − 2t   z=t 1

Note that there are other possible answers. For example, by taking the point (−5, 5, 1) on the line, we obtain another parametric equation of the line   x = −5 − 8t y = 5 − 2t   z=t (c) There are infinitely many such lines. For any a, b such that two numbers a, b, not both equal to zero, the non-zero vector ha, b, 2a + 3bi is perpendicular to the vector h4, 6, −2i. Therefore   x = at y = 1 + bt   z = 1 + (2a + 3b)t is an example of such a line.

4.

(15 points) a. (10 pts) Find the plane perpendicular to the planes x + y − z = 1 and 2x − 3y + 4z = 5 and passing through the point P = (1, 0, −2). b. (5 pts) What is the distance from P to the plane 2x − 3y + 4z = 5? Solution: (a) The normal vector of the plane we need is perpendicular to the normal vectors of the two given planes. The normal vectors of the two given planes are h1, 1, −1i and h2, −3, 4i. So the normal vector of the plane we need is h1, 1, −1i × h2, −3, 4i = h1, −6, −5i. Thus the answer is (x − 1) − 6y − 5(z + 2) = 0; or after simplification x − 6y − 5z − 11 = 0 (b) The distance is given by the distance formula (Formula (9.5.7) on page 776): |2 × 1 − 3 × 0 + 4 × (−2) − 5| 11 p =√ . 2 2 2 29 2 + (−3) + 4

5.

(15 points) Determine whether the following vectors are parallel, perpendicular or neither. Explain why. a. (4 pts) h2, −3, 1i and h2, 1, −1i Solution: Since h2, −3, 1i · h2, 1, −1i = 4 − 3 − 1 = 0, the vectors are perpendicular

b. (4 pts) 2i + j − 4k and −7i − 27 j + 14k Solution: Since (−7i − 72 j + 14k) = (− 27 )(2i + j − 4k), the vectors are parallel c. (4 pts) a and a × b + a × c, where a, b and c are arbitrary vectors. Solution: Since a · (a × b + a × c) = a · (a × b) + a · (a × c) = 0 + 0 = 0, the vectors are perpendicular d. (3 pts) h1, 2, −1i and n, where n is any vector perpendicular to the plane defined by 2x − y − z = 1 Solution: Any vector perpendicular to the plane 2x − y − z = 1 is parallel to the normal vector of the plane n = h2, −1, −1i. 2

Since h1, 2, −1i·n = 2−2+1 = 1 6= 0, the vectors h1, 2, −1i and n are not perpendicular. Furthermore, h1, 2, −1i is not a scalar multiplie of n. Therefore, the vectors are neither parallel nor perpendicular

6.

(15 points) a. (8 pts) Determine whether the three points (1, −5, 2), (−1, −3, 3) and (−3, −1, 5) lie on the same line.

Solution: Let a = h−2, 2, 1i and b = h−4, 4, 3i be the vectors originating at the first point and ending at the second point and third point, respectively. The three points lie on the same line if and only if the area of the parallelogram they span is 0, which is true if and only if a × b = 0. We compute i j k a × b = −2 2 1 = (6 − 4)i − (−6 − (−4))j + (−8 − (−8))k = 2i + 2j −4 4 3 Since a × b is non-zero, we conclude that the four points do not lie on the same line b. (7 pts) Determine whether the four points (1, 1, 0), (1, 1, −2), (0, 2, −1) and (5, −3, 0) lie on the same plane. Solution: Let a = h0, 0, −2i, b = h−1, 1, −1i, and c = h4, −4, 0i be the vectors from the first point to the second, third and fourth, respectively. The four points lie on the same plane if and only if the volume of the parallelepiped that they span is 0. The volume of the parallelepiped is |(a × b) · c)| which we will now compute. First i j k a × b = 0 0 −2 = 2i + 2j −1 1 −1 Therefore |(a × b) · c| = |8 − 8| = 0 so the four points do lie on the same plane Alternatively, one can find the plane containing three of the points and check whether the fourth point lies in this plane. For example, the plane containing the first three points has normal vector a × b = h2, 2, 0i. Hence the plane through the first three points is 2(x − 1) + 2(y − 1) = 0 or x + y = 2. The fourth point (5, −3, 0) lies in this plane. Hence the four points are coplanar.

7.

(15 points) Let a = h−1, 0, 1i and b = h2, 2, 0i be vectors.

a. (8 pts) Find the angle between a and b. Solution: Since a · b = |a||b| cos θ where 0 ≤ θ ≤ π is the angle between a and b, we have −2 1 a·b = √ √ =− cos θ = |a||b| 2 2 8 so that θ = 2π/3 b. (7 pts) Find two unit vectors orthogonal to a and b. Solution: We compute that i j k a × b = −1 0 1 = −2i + 2j − 2k = h−2, 2, −2i 2 2 0 3

√ √ 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 √

h−

8.

√ √ 3 3 3 3 , 3 ,− 3 i

√

and

h

√ √ 3 3 3 3 ,− 3 , 3 i

(15 points) Let P be the plane perpendicular to h1, 2, 3i 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) perpendicular?

Is the plane P parallel to the plane defined by 2x + 3y − 4z = 2? Is it

Solution: Plane P has normal vector n1 = h1, 2, 3i while the plane 2x + 3y − 4z = 2 has normal vector n2 = h2, 3, −4i. Since n1 and n2 are not scalar multiplies, the planes are not parallel. Since n1 · n2 = 2 + 6 − 12 = −4 6= 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 = h3, 0, −1i. Since v · n1 = 3 − 3 = 0, v is perpendicular to n1 . 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 n1 = h1, 1, 0i and the second plane has normal vector n2 = h0, 1, 1i. Therefore, the line of intersection has the direction vector v = n1 × n2 . We compute i j k n1 × n2 = 1 1 0 = i − j + k = h1, −1, 1i 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