#### Explain solution RD Sharma class 12 chapter Vector or Cross Product exercise 24.1 question 28 maths

$\frac{1}{3}(2 \hat{i}-2 \hat{j}-\hat{k})$

Hint:

To solve this we use determinant method

Given:

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

Solution:

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

Let

\begin{aligned} &\vec{d}=\vec{a}+\vec{b} \\\\ &\vec{d}=4 \hat{i}+4 \hat{j}+0 \hat{k} \end{aligned}

And

\begin{aligned} &\vec{e}=\vec{a}-\vec{b} \\\\ &=2 \hat{i}+0 \hat{j}+4 \hat{k} \end{aligned}

Let  $\vec{f}$ be any vector perpendicular to both  $\vec{d} \; \& \; \vec{e}$, hence parallel to $\vec{d} \times \vec{e}$

\begin{aligned} &\therefore \vec{f}=\vec{d} \times \vec{e} \\\\ &=\left|\begin{array}{ccc} \hat{i} & \hat{j} & \hat{k} \\ 4 & 4 & 0 \\ 2 & 0 & 4 \end{array}\right| \\\\ &=16 \hat{i}-16 \hat{j}-8 \hat{k} \end{aligned}

\begin{aligned} &=\frac{16\hat {i}-16 \hat{j}-8 \hat {k}}{\sqrt{(16)^{2}+(-16)^{2}+(-8)^{2}}} \\\\ &=\frac{16}{24} \hat{i}-\frac{16}{24} \hat{j}-\frac{8}{24} \hat{k} \\\\ &=\frac{1}{3}(2 \hat{i}-2 \hat{j}-\hat{k}) \end{aligned}