Math 2433 Homework #10 - OU

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Find a non-zero vector orthogonal to the plane through the point P(−1,3,1), ... Is the line through (4,1,−1) and (2,5,3) perpendicular to the line through (−3,2,0).
Math 2433 Homework #10 Solutions Important: I will only provide helpful hints and not detailed solutions, but you have to write all the steps involved along with careful explanations. Section 13.4. 12. Find the vector (i + j) × (i − j) using the properties of the cross product. Solution: We have (i + j) × (i − j) = (i + j) × i − (i + j) × j = j × i − i × j = −k − k = −2k. 20. Find two unit vectors orthogonal to both a = i + j + k and b = 2i + k. Solution: The cross product a × b is i j k a × b = 1 1 1 = i + j − 2k. 2 0 1 Since |a × b| =



6, a vector orthogonal to a and b is 1 2 1 √ i + √ j − √ k. 6 6 6

Similarly, b × a = −i − j = 2k and thus another unit vector which is orthogonal to both a and b is 1 1 2 − √ i − √ j + √ k. 6 6 6 28. Find the area of the parallelogram with vertices K(1, 2, 3), L(1, 3, 6), M (3, 8, 6) and N (3, 7, 3). −→

Solution: We have KL=< 0, 1, 3 > and i j −→ −→ KL × KM = 0 1 2 6

−→

KM =< 2, 6, 3 > so k 3 = −15i + 6j − 2k. 3

and the area of the parallelogram is the magnitude of this vector. √ √ area = 225 + 36 + 4 = 265. 1

2

Solutions

30. Find a non-zero vector orthogonal to the plane through the point P (−1, 3, 1), Q(0, 5, 2) and R(4, 3, −1). −→

−→

Solution: We have P Q=< 1, 2, 1 > and P R=< 5, 0, −2 >. A non-zero vector orthogonal to the plane through the points is i j k −→ −→ 1 = −4i + 7j − 10k. P Q × P R= 1 2 5 0 −2 34. Find the volume of the parallelepiped determined by the vectors a = i + j − k, b = i − j + k, c = −i + j + k. Solution: The volume of the parallelepiped determined by the vectors is the magnitude of the scalar triple product. 1 1 −1 1 = −2 − 2 a · (b × c) = a × b = 1 −1 −1 1 1 so the volume is V = |a · (b × c)| = 4. 38. Use the scalar triple product to determine whether the points A(1, 3, 2), B(3, −1, 6), C(5, 2, 0) and D(3, 6, −4) lie in the same plane. −→

−→

−→

Solution: The points are in the same plane if AB ·(AC × AD) = 0. You will have to check this. Section 13.5 4. Find a vector equation and parametric equations for the line through the point (0, 14, −10) and parallel to the line x = −1 + 2t, y = 6 − 3t and z = 3 + 9t. Solution: The given line has a direction given by < 2, −3, 9 >. A vector equation of the line through (0, 14, −10) and parallel to the given line is r = h0, 14, −10i + t h2, −3, 9i . The parametric equations are x = 2t, y = 14 − 3t, z = −10 + 9t. 10. Find parametric equations and symmetric equations for the line through (2, 1, 0) and perpendicular to both i + j and j + k.

Math 2433 Homework #10

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Solution: A direction perpendicular to i + j and j + k is (i + j) × (j + k) = k − j + i =< 1, −1, 1 > . The parametric equations are x = 2 + t, y = 1 − t, z = t and the symmetric equations are x−2 y−1 z = = . 1 −1 1 14. Is the line through (4, 1, −1) and (2, 5, 3) perpendicular to the line through (−3, 2, 0) and (5, 1, 4)? Solution: The directions vectors of the two lines are < −2, 4, 4 > and < 8, −1, 4 >. We calculate the dot product < −2, 4, 4 > · < 8, −1, 4 >= −16 − 4 + 16 = −4 6= 0 so the vectors are not orthogonal, hence the lines are not orthogonal. 16. (a) Find parametric equations of the line through (2, 4, 6) that is perpendicular to the plane x − y + 3z = 7.

(b) In what points does the line intersect the coordinate planes?

Solution: A normal vector to the plane is < 1, −1, 3 > and a non-zero vector orthogonal to it is, for example, < −1, 2, 1 >. Hence, parametric equations of the line through (2, 4, 6) which is orthogonal to the given plane are x = 2 − t, y = 4 + 2t, z = 6 + t. The line intersects the x − y plane when z = 0, or t = −6. So the point of intersection is (8, −8, 0). You can find the intersection with the other two coordinate axes easily. 20. Determine whether the line L1 and L2 are parallel, skew or perpendicular, If they intersect, find the point of intersection. L1 : x = 1 + 2t, y = 3t, z = 2 − t

L2 : x = −1 + s, y = 4 + s, z = 1 + 3s

Solution: Direction vectors for L1 and L2 are < 2, 3, −1 > and < 1, 1, 3 > respectively. Since they are not multiples of each other, the lines are not parallel. To check if they intersect, we equate the x and y coordinates of each line to get s − 2t = 2 and s − 3t = −4. Solving for s and t we get s = 14 and t = 6. Substituting in the expressions for the z coordinate on each line, we get z = 2 − t = −4 6= 1 + 3s = 1 + 42 = 43. Since the z coordinates don’t agree, the lines do not intersect. Hence, they are skew.