MULTIVARIABLE CALCULUS PRACTICE MIDTERM 1 SOLUTIONS

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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 〈a, b, 2a + 3b〉 is perpendicular to the vector 〈4, 6, −2〉. Therefore ⎧ ⎪⎨ x = at ⎪ ⎩ is an example of such a line. y = 1 + bt z = 1 + (2a + 3b)t 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 〈1, 1, −1〉 and 〈2, −3, 4〉. So the normal vector of the plane we need is 〈1, 1, −1〉 × 〈2, −3, 4〉 = 〈1, −6, −5〉. 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| √ 2 2 + (−3) 2 + 4 2 = 11 √ 29 . 5. (15 points) Determine whether the following vectors are parallel, perpendicular or neither. Explain why. a. (4 pts) 〈2, −3, 1〉 and 〈2, 1, −1〉 Solution: Since 〈2, −3, 1〉 · 〈2, 1, −1〉 = 4 − 3 − 1 = 0, the vectors are perpendicular b. (4 pts) 2i + j − 4k and −7i − 7 2 j + 14k Solution: Since (−7i − 7 2 j + 14k) = (− 7 2 )(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) 〈1, 2, −1〉 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 = 〈2, −1, −1〉. 2
Since 〈1, 2, −1〉·n = 2−2+1 = 1 ≠ 0, the vectors 〈1, 2, −1〉 and n are not perpendicular. Furthermore, 〈1, 2, −1〉 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 = 〈−2, 2, 1〉 and b = 〈−4, 4, 3〉 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 = 〈0, 0, −2〉, b = 〈−1, 1, −1〉, and c = 〈4, −4, 0〉 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 = 〈2, 2, 0〉. 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 = 〈−1, 0, 1〉 and b = 〈2, 2, 0〉 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 so that θ = 2π/3 cos θ = a · b |a||b| = −2 √ 2 √ 8 = − 1 2 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 = 〈−2, 2, −2〉 ∣ 2 2 0∣ 3
Page 1: MULTIVARIABLE CALCULUS PRACTICE MID

MULTIVARIABLE CALCULUS PRACTICE MIDTERM 1 SOLUTIONS

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