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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.
 Spring '07
 Sadler
 Determinant, Multivariable Calculus

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