Asked by Aman
Find the angles between the following pairs of vectors
(1)3i+2j-6k,4i-3j+k
(2)2i-3j+k,3i-j-2k
(1)3i+2j-6k,4i-3j+k
(2)2i-3j+k,3i-j-2k
Answers
Answered by
Reiny
1.
your 2 vectors are [3,2,-6] and [4,-3,1]
using the dot product
12 - 6 - 6 = √(9+4+36) √(16+9+1) cosØ, where Ø is the angle between them
0 = .... , no point to go further.
I can see cosØ = 0
Ø = 90° or π/2 radians
2.
6 + 3 - 2 = √(4+9+1) √(9+1+4) cosØ
7 = √14√14cosØ
7 = 14cosØ
cosØ = 1/2
Ø = 60° or π/3
your 2 vectors are [3,2,-6] and [4,-3,1]
using the dot product
12 - 6 - 6 = √(9+4+36) √(16+9+1) cosØ, where Ø is the angle between them
0 = .... , no point to go further.
I can see cosØ = 0
Ø = 90° or π/2 radians
2.
6 + 3 - 2 = √(4+9+1) √(9+1+4) cosØ
7 = √14√14cosØ
7 = 14cosØ
cosØ = 1/2
Ø = 60° or π/3