#### Need solution for rd sharma maths class 12 chapter 24 vector or cross product Exercises Multiple choice questions question 7 maths textbook solution

Answer: $\hat{i}$

Given: $\vec{a}=\hat{i}+\hat{j}-\hat{k} ; \vec{b}=-\hat{i}+2 \hat{j}+2 \hat{k} ; \vec{c}=-\hat{i}+2 \hat{j}-\hat{k}$

Hint: You must know about normal to the two vectors

Explanation: Let

$\begin{gathered} \vec{d}=\vec{a}+\vec{b} \\ \vec{e}=\vec{b}-\vec{c} \\ \vec{d}=\vec{a}+\vec{b}=(\hat{i}+\hat{j}-\hat{k})+(-\hat{i}+2 \hat{j}+2 \hat{k}) \\ =0 \hat{i}+3 \hat{j}+\hat{k}=3 \hat{j}+\hat{k} \\ \vec{e}=\vec{b}-\vec{c}=-\hat{i}+2 \hat{j}+2 \hat{k}-(-\hat{i}+2 \hat{j}-\hat{k}) \\ =0 \hat{i}+4 \hat{j}+3 \hat{k}=4 \hat{j}+3 \hat{k} \end{gathered}$

A unit vector normal to the unit vector is $\frac{\overrightarrow{d}\times \overrightarrow{e}}{\left |\overrightarrow{d}\times \overrightarrow{e} \right |}$

Now,

\begin{aligned} &\vec{d} \times \vec{e}=\left|\begin{array}{ccc} \hat{i} & \hat{j} & \hat{k} \\ 0 & 3 & 1 \\ 0 & 4 & 3 \end{array}\right|=\hat{i}(9-4)=5 \hat{i} \\ &|\vec{d} \times \vec{e}|=\sqrt{(5)^{2}+(0)^{2}+(0)^{2}}=\sqrt{25}=5 \end{aligned}

Unit vector = $\frac{\overrightarrow{d}\times \overrightarrow{e}}{\left |\overrightarrow{d}\times \overrightarrow{e} \right |}=\frac{5\hat{i}}{5}=\hat{i}$