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#### Provide solution for RD Sharma Maths Class 12 Chapter 24 Vector or Cross Product exercise Very Short Answer Question, question 16.

$\frac{\hat{i}-\hat{\jmath}+\hat{k}}{\sqrt{3}}$

HINT:

Just find the cross product of vectors.

GIVEN:

Given, two vectors,      $\left ( \widehat{i}+\widehat{j} \right )$ and $\left ( \widehat{j}+\widehat{k} \right )$

SOLUTION:

\begin{aligned} &\text { Let } \vec{a}=\hat{\imath}+\hat{\jmath}+0 \hat{k} \\ &\qquad \vec{b}=0 \hat{\imath}+\hat{\jmath}-\hat{k} \\ &\text { Now } \vec{a} \times \vec{b}=\left|\begin{array}{lll} \hat{\imath} & \hat{\jmath} & \hat{k} \\ 1 & 1 & 0 \\ 0 & 1 & 1 \end{array}\right| \\ &\quad=\hat{\imath}(1-0)-\hat{\jmath}(1-0)+\hat{k}(1-0) \\ &\quad=-\hat{\imath}-\hat{\jmath}+\hat{k} \end{aligned}

Now unit vector perpendicular to $\overrightarrow{a}\times \overrightarrow{b}$ is $\widehat{n}$

\begin{aligned} \therefore \hat{n} &=\frac{(\vec{a} \times \vec{b})}{|\vec{a} \times \vec{b}|} \\ &=\frac{\hat{\imath}-\hat{\jmath}+\hat{k}}{\sqrt{(1)^{2}+(-1)^{2}+(1)^{2}}} \\ &=\frac{\hat{i}-\hat{\jmath}+\hat{k}}{\sqrt{3}} \end{aligned}