Question

Find angle between vector A=-2i+4j and vector B=3i-4j+2k

Answer 1

Given,

a⃗ =(2,2,−2)

b⃗ =(3,−4,0)

Therefore,

a⃗ ⋅b⃗ =axbx+ayby+azbz

a⃗ ⋅b⃗ =6−8+0=−2

∥a⃗ ∥=22+22+(−2)2−−−−−−−−−−−−−√=6–√

∥b⃗ ∥=32+(−4)2−−−−−−−−−√=5

From the dot product definition, we know that:

cosθ=a⃗ ⋅b⃗ ∥a⃗ ∥∥b⃗ ∥

Where, θ is the angle between the vectors.

Therefore,

cosθ=−26√×5

θ=arccos(−26√×5)

θ=99.39∘

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