This preview shows page 1. Sign up to view the full content.
Unformatted text preview: 7
7 . Determine all unit vectors v orthogonal to
a = −3 i + 4 j + k,
1. v = ± b = −9 i + 10 j + 2 k . 3
6
2
i + j − k correct
7
7
7 4. v = 6
3
2
i− j+ k
7
7
7 Find the area of the triangle having vertices 4. area = 7
5. area = 8 correct Explanation:
The nonzero vectors orthogonal to a and b
are all of the form
v = λ(a × b) , λ = 0, with λ a scalar. The only unit vectors orthogonal to a, b are thus
a×b
.
a × b Explanation:
To use vectors we shall identify a line segment with the corresponding directed line segment.
Since the area of the parallelogram having
adjacent edges P Q and P R is given by
−
−
→−
→
P Q × P R ,
∆P QR has
area = But for the given vectors a and b, = Now
−
−
→
PQ = jk
41
10 2 −3
1
i−
−9
2 4
10 R(−1, −3) . 3. area = 9 6. v = −2 i − 3 j + 6 k a×b = Q(−2, 2) , 15
2
17
2. area =
2 3
6
2
5. v = − i − j + k
7
7
7 i
−3
−9 10.0 points 1. area = 2
6
3
i− j+ k
7
7
7 v=± 004 P (1, 3) , 2. v = 2 i − 6 j + 3 k
3. v = ± keywords: vector product, cross product, unit
vector, orthogonal, 1−
−
→−
→
P Q × P R .
2 −3, −1, 0 , −
→
PR = −2, −6, 0 . But then
−3
1
j+
−9
2 = −2 i − 3 j + 6 k .
In this case,
a × b2 = 49 . 4
k
10 i
−
−
→−
→
P Q × P R = −3
−2 j
−1
−6 k
−3
0=
−2
0 Consequently, ∆P QR has
area = 8 . −1
k.
−6 morrell (mm59638) – HW 12.4 – radin – (56025)
keywords: vectors, c...
View
Full
Document
This document was uploaded on 04/04/2014.
 Spring '14

Click to edit the document details