QJEEM24Mabaad: JEE Main 2024 Physics - Class 11 JEEM24Mabaad Solution: Let and be three given vectors. If is a vector such that and ,…

$\begin{aligned}& \vec{a}=9 \hat{i}-13 \hat{j}+25 \hat{k} \\& \vec{b}=3 \hat{i}+7 \hat{j}-13 \hat{k} \\& \vec{c}=17 \hat{i}-2 \hat{j}+\hat{k} \\& \vec{r} \times \vec{a}=(\vec{b}+\vec{c}) \times \vec{a} \\& (\vec{r}-(\vec{b}+\vec{c})) \times \vec{a}=0 \\& \Rightarrow \vec{r}=(\vec{b}+\vec{c})+\lambda \vec{a} \\& \vec{r}=(20 \hat{i}+5 \hat{j}-12 \hat{k})+\lambda(9 \hat{i}-13 \hat{j}+25 \hat{k}) \\& =(20+9 \lambda) \hat{i}+(5-13 \lambda) \hat{j}+(25 \lambda-12) \hat{k}\end{aligned}$

Now $\vec{r} \cdot(\vec{b}-\vec{c})=0$

$\vec{r} \cdot(-14 \hat{i}+9 \hat{j}-14 \hat{k})=0$

Now

$\begin{aligned}& -14(20+9 \lambda)+9(5-13 \lambda)-14(25 \lambda-12)=0 \\& -593 \lambda-67=0 \\& \lambda=-\frac{67}{593} \\& \therefore \vec{r}=(\vec{b}+\vec{c})-\frac{67}{593} \vec{a} \\& \frac{|593 \vec{r}+67 \vec{a}|^2}{|593|^2}=|\vec{b}+\vec{c}|^2=|20 \hat{i}+5 \hat{j}-12 \hat{k}|^2 \\& =569\end{aligned}$