Question

Given Ā= 2 i-+.6K, B = 2j-k find A.B​

Answer 1

Step-by-step explanation:

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Answer 2

Answer:

Given vectors are

A=2i+3j+k and B=-3i+6k.

Cosine of angle θ between vectors A and B is given by

cosθ=(A.B)/(|A||B|)

Calculate A.B as

A.B=(2i+3j+k).(-3i+6k)

=-6(i.i)+12(i.k)-9(j.i)+18(j.k)-3(k.i)+

6(k.k)

=-6+0-0+0-0+6

=-6+6=0

Thus cosθ=0=>θ=90°.

Hence given vectors are orthogonal to each other (angle between them is 90°).

Hope you got your point......

Step-by-step explanation:

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