Answers!
Compute the angle between two vectors 3i + 4j – k and 2i – j + k.
solution:
Let
a⃗ = 3i + 4j – k and
b⃗ = 2i – j + k
The dot product is defined as
a⃗ .b⃗ = (3i + 4j – k).(2i – j + k)
= (3)(2) + (4)(-1) + (-1)(1)
= 6-4-1
= 1
Thus, a⃗ .b⃗ = 1
The Magnitude of vectors is given by
|a⃗ |=(32+42+(−1)2)−−−−−−−−−−−−−−√=26−−√=5.09
|b⃗ |=(22+(−1)2+12)−−−−−−−−−−−−−−√=6–√=2.45
The angle between the two vectors is
θ=cos−1a⃗ .b⃗ |a⃗ ||b⃗ |
θ=cos−11(5.09)(2.45)
θ=cos−1112.47
θ=cos−1(0.0802)
θ=85.39∘
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Answer the question...
Compute the angle between two vectors 3i + 4j – k and 2i – j + k.
solution:
Let
a⃗ = 3i + 4j – k and
b⃗ = 2i – j + k
The dot product is defined as
a⃗ .b⃗ = (3i + 4j – k).(2i – j + k)
= (3)(2) + (4)(-1) + (-1)(1)
= 6-4-1
= 1
Thus, a⃗ .b⃗ = 1
The Magnitude of vectors is given by
|a⃗ |=(32+42+(−1)2)−−−−−−−−−−−−−−√=26−−√=5.09
|b⃗ |=(22+(−1)2+12)−−−−−−−−−−−−−−√=6–√=2.45
The angle between the two vectors is
θ=cos−1a⃗ .b⃗ |a⃗ ||b⃗ |
θ=cos−11(5.09)(2.45)
θ=cos−1112.47
θ=cos−1(0.0802)
θ=85.39∘
Comment