HW09-solutions

# HW09-solutions

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Unformatted text preview: or cos 2θ = 1. C. TRUE: if a = a1 , a2 , a3 , b = b1 , b2 , b3 , geyer (dag2798) – HW09 – rusin – (55735) then 3 Consequently, a×b = a2 a3 b2 b3 i− a1 a3 b1 b3 j+ a1 a2 b1 b2 − − →− → v = P Q × P R = 8, 6, 12 k, is othogonal to the plane through P, Q and R. while b×a = b2 b3 a2 a3 i− b1 b3 a1 a3 j+ b1 b2 a1 a2 On the other hand, for a 2 × 2 determinant, a b c d = ad − cb = − c d a b 006 k. 10.0 points Compute the volume of the parallelopiped with adjacent edges PQ, . PR, PS determined by vertices P (1, −1, 1) , a × b = −b × a . Q(5, −2, −3) , R(4, 1, −2) , Consequently, S (3, 2, −2) . 1. volume = 10 2. volume = 14 keywords: 3. volume = 13 005 10.0 points Find a vector v orthogonal to the plane through the points P (3, 0, 0), Q(0, 4, 0), R(0, 0, 2) . 1. v = 2, 6, 12 4. volume = 12 5. volume = 11 correct Explanation: The parallelopiped is determined by the vectors − − → a = P Q = 4, −1, −4 , 2. v = 4, 6, 12 − → b = PR = 3, 2, −3 , 3. v = 8, 6, 12 correct − → c = PS = 2, 3, −3 . 4. v = 8, 3, 12 Thus its volume is given in terms of a scalar triple product by 5. v = 8, 2, 12 V = | a · ( b × c) | . Explanation: Because the plane through P , Q, R con− − → − → tains the vectors P Q and P R, any vector v orthogonal to both of these vectors (such as their cross product) must therefore be orthogonal to t...
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## This note was uploaded on 05/12/2013 for the course M 408d taught by Professor Sadler during the Spring '07 term at University of Texas at Austin.

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