Question

find the angle between A=1+2j-kand
B=-i+j-2k​

Answer 1

Explanation:

Given that.

Vector A, A=i+2j-kA=i+2j−k

Vector B, B=-i+j-2kB=−i+j−2k

Let \thetaθ is the angle between A and B. It can be calculated as :

A.B=|A||B|\ cos\thetaA.B=∣A∣∣B∣ cosθ

|A|=\sqrt{1^2+2^2+(-1)^2} =\sqrt 6∣A∣=

1

2

+2

2

+(−1)

2

=

6

|B|=\sqrt{(-1)^2+1^2+(-2)^2} =\sqrt 6∣B∣=

(−1)

2

+1

2

+(−2)

2

=

6

A.B=-1+2+2=3A.B=−1+2+2=3

So,

cos\theta=\dfrac{3}{(\sqrt 6)^2}cosθ=

(

6

)

2

3

cos\theta=\dfrac{1}{2}cosθ=

2

1

\theta=60^{\circ}θ=60

∘

So, the angle between vector A and B is 60 degrees. Hence, this is the required solution.

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