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#### Need solution for rd sharma maths class 12 chapter 24 vector or cross product Exercises Multiple choice questions question 4 maths textbook solution

Answer: $\frac{(2 \hat{i}+\hat{j}+\hat{k})}{\sqrt{6}}$

Given: A unit vector is perpendicular to the plane passing through points

$P(\hat{i}-\hat{j}+2 \hat{k}), Q(2 \hat{i}-\hat{k}), R(2 \hat{j}+\hat{k})$

Hint: You must know how to find position vectors and magnitude of vectors

Explanation:    Here

$\begin{gathered} \vec{p}=\hat{i}-\hat{j}+2 \hat{k} \\ \vec{q}=2 \hat{i}-\hat{k} \\ \vec{r}=2 \hat{j}+\hat{k} \\ \overrightarrow{P Q}=(P . V \text { of } Q)-(P . V \text { of } P) \end{gathered}$

\begin{aligned} &=\vec{q}-\vec{p} \\ &=2 \hat{i}-\hat{k}-\hat{i}+\hat{j}-2 \hat{k} \\ &=\hat{i}+\hat{j}-3 \hat{k} \\ &\overline{Q R}=(\mathrm{P} . \mathrm{V} \text { of } \mathrm{R})-(\mathrm{P} \cdot \mathrm{V} \text { of } \mathrm{Q}) \\ &=2 \hat{j}+\hat{k}-2 \hat{j}+\hat{k} \\ &=-2 \hat{i}+2 \hat{j}+2 \hat{k} \\ &\frac{ }{P R}=(\mathrm{P} . \mathrm{V} \text { of } \mathrm{R})-(\mathrm{P} . \mathrm{V} \text { of } \mathrm{P}) \end{aligned}

\begin{aligned} &=2 \hat{j}+\hat{k}-\hat{i}+\hat{j}-2 \hat{k} \\ &=-\hat{i}+3 \hat{j}-\hat{k} \end{aligned}

? Normal vector  $(\vec{n})=\overrightarrow{P Q} \times \overrightarrow{P R}$

$=\left|\begin{array}{ccc} \hat{i} & \hat{j} & \hat{k} \\ 1 & 1 & -3 \\ -1 & 3 & -1 \end{array}\right| \\$

$=\hat{i}(-1+9)-\hat{j}(-1-3)+\hat{k}(3+1) \\$

$\vec{n}=8 \hat{i}+4 \hat{j}+4 \hat{k}$

$\vec{n}|=\sqrt{(8)^{2}+(4)^{2}+(4)^{2}}=\sqrt{64+16+16}=\sqrt{96}$

$\Rightarrow \hat{n}=\frac{\vec{n}}{|\vec{n}|}=\frac{1}{\sqrt{96}}(8 \hat{i}+4 \hat{j}+4 \hat{k})$

$=\frac{1}{4 \sqrt{6}} \times 4(2 \hat{i}+\hat{j}+\hat{k})$

$=\frac{(2 \hat{i}+\hat{j}+\hat{k})}{\sqrt{6}}$