JEE Main 2021 (Online) 16th March Evening ShiftQuestion 1
Let $\vec{a}=\hat{i}+2\hat{j}-3\hat{k}$ and $\vec{b}=2\hat{i}-3\hat{j}+5\hat{k}$. If $\vec{r}\times\vec{a}=\vec{b}\times\vec{r}$, $\vec{r}\cdot\left(\alpha\hat{i}+2\hat{j}+\hat{k}\right)=3$ and $\vec{r}\cdot\left(2\hat{i}+5\hat{j}-\alpha\hat{k}\right)=-1$, $\alpha\in\mathbf{R}$, then the value of $\alpha+|\vec{r}|^{2}$ is equal to :
Vector Algebra
Solution
$\vec r\times\vec a=\vec b\times\vec r\Rightarrow\vec r\times(\vec a+\vec b)=0$, so $\vec r=t(\vec a+\vec b)=t(3,-1,2)$. $\vec r\cdot(\alpha,2,1)=t(3\alpha-2+2)=3\alpha t=3$ and $\vec r\cdot(2,5,-\alpha)=t(6-5-2\alpha)=t(1-2\alpha)=-1$. From the first, $t=\frac1\alpha$; then $1-2\alpha=-\alpha$ gives $\alpha=1$, $t=1$. $\vec r=(3,-1,2)$ and $|\vec r|^2=14$, so $\alpha+|\vec r|^2=15$. Option (D).