12.4_ Cross Product-solutions

# 1 6 0 but then i p q p r

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Unformatted text preview: = −1 0 6 k. 6 Consequently, ∆P QR has In this case, |a × b|2 = 49 . area = 15 . Consequently, v=± 3 6 2 i− j− k 7 7 7 . keywords: vectors, cross product area, triangle, parallelogram stiurca (mas7745) – 12.4: Cross Product – meth – (91825) 005 10.0 points Find a vector v orthogonal to the plane through the points P (2, 0, 0), Q(0, 4, 0), R(0, 0, 5) . 3 3. volume = 25 4. volume = 21 5. volume = 22 correct 1. v = 5, 10, 8 Explanation: For the parallelopiped determined by vectors a, b, and c its 2. v = 20, 2, 8 volume = |a · (b × c)| . But 3. v = 4, 10, 8 31 22 a · ( b × c) = 5. v = 20, 10, 8 correct 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 the plane. =3 23 44 − −2 3 24 4. v = 20, 5, 8 4 2 3 2 4 −2 2 2 2 4 . Consequently, Here volume = 22 . − → P Q = −2, 4, 0 , − → P R = −2, 0, 5 . keywords: determinant, cross product, scalar triple product, parallelopiped, volume, Consequently, − − →− → v = P Q × P R = 20, 10, 8 is othogonal to the plane through P, Q and R. 006 10.0 points Compute the volume of the parallelopiped determined by the vectors a= 3, 1, −2 , b= 007 Which of the following statements are true for all vectors a, b = 0? A. a × b + b × a = 0, B. |a × b|2 + |a · b|2 = |a|2|b|2 , C. if a × b = 0, then a 2, 2, 3 , 1. A and B only and c = 2, 4, 4 . 10.0 points 2. C only 1. volume = 23 3. A only 2. volume = 24 4. none of them b. stiurca (mas7745) – 12.4: Cross Product – meth – (91825) 4 so if a = 0 and b = 0, then 5. all of them correct |a × b| = 0 =⇒ sin θ = 0 . 6. B only Thus θ = 0, π . In this case, a is parallel to b. 7. A and C only keywords: 8. B and C only 008 Explanation: 10.0 points Compute the volume of the parallelopiped with adjacent edges A. TRUE: if a = a1 , a2 , a3 , PQ, b = b1 , b2 , b3 , PR, PS determined by vertices then a2 b2 a×b = a3 b3 i− a1 a3 b1 b3 j+ a1 a2 b1 b2 P (1, −1, 1) , k, Q(4, −5, −1) , R(2, 3, 2) , S (2, 3, 3) . 1. volume = 15 while b2 b3 a2 b×a = a3 i− b1 b3 a1 a3 j+ b1 b2 a1 a2 k. 2. volume = 14 3. volume = 16 correct On the other hand, for a 2 × 2 determinant, a b c d = ad − cb = − c d a b 4. volume = 13 . Consequently, a × b = −b × a . 5. volume = 12 Explanation: The parallelopiped is determined by the vectors − − → a = P Q = 3, −4, −2 , − → b = PR = |a × b| = |a||b| sin θ , while 1, 4, 1 , − → c = PS = B. TRUE: if θ, 0 ≤ θ ≤ π , is the angle between a, and b, then 1, 4, 2 . Thus its volume is given in terms of a scalar triple product by a · b = |a||b| cos θ . Thus V = | a · ( b × c) | . But 2 |a × b| + |a · b| 2 = |a|2 |b|2(sin2 θ + cos2 θ ) = |a|2 |b|2 . 3 −4 =3 4 1 4 2 1 4 1 1 a · ( b × c) = C. TRUE: if θ, 0 ≤ θ ≤ π , is the angle between a, and b, then |a × b| = |a||b| sin θ , −2 4 2 +4 11 12 −2 14 14 . stiurca (mas7745) – 12.4: Cross Product – meth – (91825) Consequently, volume = 16 . keywords: determinant, cross product, vector product, scalar triple product, parallelopiped, volume, 5...
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