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Mathematics Question on Product of Two Vectors

Which one of the following is the unit vector perpendicular to both a⃗=−i^+j^+k^\vec{a}=-\hat{i}+\hat{j}+\hat{k} and b⃗=i^−j^+k^\vec{b}=\hat{i}-\hat{j}+\hat{k} ?

A

i^+j^2\frac{\hat{i} + \hat{j}}{\sqrt{2}}

B

k^\hat{k}

C

j^+k^2\frac{\hat{j} + \hat{k}}{\sqrt{2}}

D

i^−j^2\frac{\hat{i} - \hat{j}}{\sqrt{2}}

Answer

i^+j^2\frac{\hat{i} + \hat{j}}{\sqrt{2}}

Explanation

Solution

According to question a=−i^+j^+k^a=-\hat{i}+\hat{j}+\hat{k} and b=i^−j^+k^b=\hat{i}-\hat{j}+\hat{k}
Then , a×b=∣i^j^k^ −111 1−11∣a \times b = \begin{vmatrix}\hat{i}&\hat{j}&\hat{k}\\\ -1&1&1\\\ 1&-1&1\end{vmatrix}
=i^[1+1]−j^[−1−1]+k^[1−1]=2(i^+j^)=\hat{i}[1+1]-\hat{j}[-1-1]+\hat{k}[1-1]=2(\hat{i}+\hat{j})
and ∣a×b∣=4+4=22|a \times b|=\sqrt{4+4}=2 \sqrt{2}
∴\therefore Required unit vector
=±2(i^+j^)22=\pm \frac{2(\hat{i}+\hat{j})}{2 \sqrt{2}}
=±i^+j^2=\pm \frac{\hat{i}+\hat{j}}{\sqrt{2}}