QJEEAdv12Ma85c1: JEE Advanced 2012 Physics - Class 11 JEEAdv12Ma85c1 Solution: If and are vectors such that and then a possible value of is

$\begin{aligned}& \text { Given, } \vec{a} \times(2 \hat{i}+3 \hat{j}+4 \hat{k})=(2 \hat{i}+3 \hat{j}+4 \hat{k}) \times \vec{b} \\& \begin{aligned}& \Rightarrow \vec{a} \times(2 \hat{i}+3 \hat{j}+4 \hat{k})-(2 \hat{i}+3 \hat{j}+4 \hat{k}) \times \vec{b}=0 \\& \Rightarrow \vec{a} \times(2 \hat{i}+3 \hat{j}+4 \hat{k})+\vec{b} \times(2 \hat{i}+3 \hat{j}+4 \hat{k})=0 \\& \therefore(\overrightarrow{\mathrm{A}} \times \overrightarrow{\mathrm{B}}=-\overrightarrow{\mathrm{B}} \times \overrightarrow{\mathrm{A}}) \\& \Rightarrow \quad (\vec{a}+\vec{b}) \times(2 \hat{i}+3 \hat{j}+4 \hat{k})=0\end{aligned}\end{aligned}$

Hence, $\vec{a}+\vec{b}$ is collinear to $2 \hat{i}+3 \hat{j}+4 \hat{k}$

Let $\vec{a}+\vec{b}=\lambda(2 \hat{i}+3 \hat{j}+4 \hat{k})\quad \text{... (i)}$

$\begin{array}{l}\Rightarrow & |\vec{a}+\vec{b}|=|\lambda| \sqrt{2^2+3^2+4^2} \\\Rightarrow & \sqrt{29}=|\lambda| \sqrt{29} \quad \text { (Given, }|\vec{a}+\vec{b}|=\sqrt{29}) \\\Rightarrow & |\lambda|=1 \\\Rightarrow & |\lambda|= \pm 1 \\\therefore & \vec{a}+\vec{b}= \pm(2 \hat{i}+3 \hat{j}+4 \hat{k})\end{array}$

$\begin{aligned}\text { Now, } &(\vec{a}+\vec{b}) \cdot(-7 \hat{i}+2 \hat{j}+3 \hat{k}) \\&= \pm(2 \hat{i}+3 \hat{j}+4 \hat{k}) \cdot(-7 \hat{i}+2 \hat{j}+3 \hat{k}) \\&= \pm(-14+6+12) \\&= \pm 4\end{aligned}$