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Unformatted text preview: =
−1
0 6
k.
6 Consequently, ∆P QR has In this case,
a × b2 = 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 × b2 + a · b2 = a2b2 ,
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 = ab 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 = ab cos θ .
Thus V =  a · ( b × c)  .
But 2 a × b + a · b 2 = a2 b2(sin2 θ + cos2 θ ) = a2 b2 . 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 = ab 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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 Summer '13

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