# 19_new - HW07 gilbert(55035 This print-out should have 18...

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HW07 – gilbert – (55035) 1 This print-out should have 18 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 1 10.0 points A triangle P QR in 3-space has vertices P ( 1 , 3 , − 2) , Q (3 , 5 , − 3) , R ( 2 , 6 , 0) . Use vectors to decide which one of the follow-ing properties the triangle has. 1. right-angled at Q 2. not right-angled at P, Q, or R 3. right-angled at P correct 4. right-angled at R Explanation: Vectors a and b are perpendicular when a · b = 0. Thus P QR will be −−→ −→ (1) right-angled at P when QP · RP = 0, −−→ −−→ (2) right-angled at Q when P Q · RQ = 0, −→ −−→ (3) right-angled at R when P R · QR = 0. But for the vertices P ( 1 , 3 , − 2) , Q (3 , 5 , − 3) , R ( 2 , 6 , 0) we see that keywords: vectors, dot product, right trian-gle, perpendicular, 002 10.0 points Find the scalar projection of b onto a when b = 2 i + 3 j − 2 k , a = 2 i − j − 2 k . 1. scalar projection = 2 2. scalar projection = 4 3 3. scalar projection = 1 4. scalar projection = 5 3 correct 5. scalar projection = 7 3 Explanation: The scalar projection of b onto a is given in terms of the dot product by comp A b = a · b . |a| Now when b = 2 i + 3 j − 2 k , a = 2 i − j − 2 k , we see that −−→ PQ = 4 , 2 , 1 , −−→ QR = 5 , 1 , 3 , h O h − O while −→ RP = 1 , 3 , 2 . h O Thus −−→ QP · −→ RP = 0 , −−→ PQ · −−→ RQ = 21 , and −→ PR · −−→ QR = 14 . Consequently, P QR is right-angled at P . a·b = 5 , |a| = q (2) 2 + ( 1) 2 + ( 2) 2 . Consequently, comp A b = 5 3 . keywords: 003 10.0 points The box shown in

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HW07 – gilbert – (55035) 2 z C B D y A x is the unit cube having one corner at the origin and the coordinate planes for three of its faces. −−→ Find the cosine of the angle θ between DA −−→ and DC . 1. cos θ = 1 2 2. cos θ = r 2 3 3. cos θ = 3 2 4. cos θ = 0 correct 5. cos θ = 1 3 6. cos θ = 1 2 Explanation: To use vectors we shall replace a line seg-ment with the corresponding directed line seg-ment. Now the angle θ between any pair of vectors u , v is given in terms of their dot product by cos θ = u · v . |u||v| On the other hand, since the unit cube has sidelength 1, A = (1 , 1 , 0) , C = (0 , 0 , 1) , −−→ while D = (1 , 0 , 1). In this case DA is a directed line segment determining the vector u = h 0 , 1 , −1 O , −−→ while DC determines v = h −1 , 0 , 0 O . For these choices of u and v, u · v = 0 = 2 cos θ . Consequently, the cosine of the angle between −−→ −−→ DA and DC is given by cos θ = u · v = 0 . |u| |v|
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## This note was uploaded on 04/04/2012 for the course M 408d taught by Professor Sadler during the Spring '07 term at University of Texas.

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19_new - HW07 gilbert(55035 This print-out should have 18...

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