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Physics Question on work, energy and power

The scalar product of two vectors A⃗=2i^+2j^−k^\vec{A} =2\hat{i}+2\hat{j}-\hat{k} and B⃗=−j^+k^,\vec{B} = -\hat{j}+\hat{k}, is given by

A

A⃗.B⃗=3\vec{A}. \vec{B} = 3

B

A⃗.B⃗=4\vec{A} .\vec{B} = 4

C

A⃗.B⃗=−4\vec{A} . \vec{B} = -4

D

A⃗.B⃗=−3 \vec{A} . \vec{B} = -3

Answer

A⃗.B⃗=−3 \vec{A} . \vec{B} = -3

Explanation

Solution

Given, A=2i^+2j^−k^A=2 \hat{i}+2 \hat{j}-\hat{k} and B=−j^+k^B=-\hat{j}+\hat{k}
Scalar product A⋅B=(2i^+2j^−k^)⋅(−j^+k^)A \cdot B=(2 \hat{i}+2 \hat{j}-\hat{k}) \cdot(-\hat{j}+\hat{k})
Using i^⋅i^=1,j^⋅j^=1,k^⋅k^=1=1\hat{i} \cdot \hat{i}=1, \hat{j} \cdot \hat{j}=1, \hat{k} \cdot \hat{k}=1=1
we have A⋅B=−2−1=−3A \cdot B=-2-1=-3