This preview shows page 1. Sign up to view the full content.
Unformatted text preview: × P R ,
∆P QR has
area =
Now
−
−
→
PQ = 1−
−
→−
→
P Q × P R  .
2 −5, 5, 0 , −
→
PR = a × b = a b sin θ
where θ, 0 ≤ θ ≤ π , is the angle between a
and b. Consequently,
length = 24 sin(π ) = 0 . keywords: vector, cross product, length,
011 10.0 points Determine the scalar triple product, V , of
the vectors
a = i − j + 2 k, −2, 1, 0 . b = i − j + 2k , and But then
i
−
−
→−
→
P Q × P R = −5
−2 jk
−5 5
k.
50=
−2 1
10 1. V = −2
2. V = 0 correct Consequently, ∆P QR has
area = c = −2 i − 2 j + k . 5
2 . 3. V = 1
4. V = −1 keywords: vectors, cross product area, triangle, parallelogram
010 10.0 points Determine the length of the cross product
of a, b when a = 6, b = 4 and the angle
between a, b is π .
√
1. length = 12 2 5. V = −3 Explanation:
The scalar triple product is given by
a · ( b × c) = = −1
−2 1
1
−2 1
2
+
−2
1 −1
−1
−2 2
2
1 1
2
+2
−2
1 −1
.
−2 mehmood (ajm4462) – Homework 12.4 – karakurt – (56295)
Consequently, the scalar triple product of a, b
and c is
V = a · ( b × c) = 0 . 6 1. volume = 12 correct
2. vo...
View Full
Document
 Spring '10
 odell

Click to edit the document details