Question

Let $\overrightarrow{a}=\hat{i}+5\hat{j}+\alpha \hat{k},\overrightarrow{b}=\hat{i}+3\hat{j}+\beta \hat{k}$ and $\overrightarrow{c}=-\hat{i}+2\hat{j}-3\hat{k}$ be three vectors such that, $|\overrightarrow{b}\times \overrightarrow{c}|=5\sqrt{3}$ and $\overrightarrow{a}$ is perpendicular to $\overrightarrow{b}$. Then the greatest amongst the values of $|\overrightarrow{a}{|}^2$ is

Answer 1

Given: $\overrightarrow{a}=\hat{i}+5\hat{j}+\alpha \hat{k},\overrightarrow{b}=\hat{i}+3\hat{j}+\beta \hat{k}$
and $\overrightarrow{c}=-\hat{i}+2\hat{j}-3\hat{k}$
Now,
$\overrightarrow{b}\times \overrightarrow{c}=\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 1& 3& \beta \\ -1& 2& -3\end{array}\right|$
$\Rightarrow \overrightarrow{b}\times \overrightarrow{c}=(-9-2\beta )\hat{i}+(3-\beta )\hat{j}+5\hat{k}$
$\Rightarrow |\overrightarrow{b}\times \overrightarrow{c}|=(9+2\beta {)}^2+(3-\beta {)}^2+25$
$\therefore (9+2\beta {)}^2+(3-\beta {)}^2+25=5\sqrt{3}$
$\Rightarrow {\beta }^2+6\beta +8=0$
$\Rightarrow \beta =-2,-4$
Now, $\overrightarrow{a}$ is perpendicular to $\overrightarrow{b}$
$\Rightarrow \overrightarrow{a}\cdot \overrightarrow{b}=0$
$\Rightarrow (\hat{i}+5\hat{j}+\alpha \hat{k})\cdot (\hat{i}+3\hat{j}+\beta \hat{k})=0$
$\Rightarrow 1+15+\alpha \beta =0$
$\Rightarrow \alpha =-\frac{16}{\beta }$
Now,
$|\overrightarrow{a}{|}^2=1+25+{\alpha }^2$
$\Rightarrow |\overrightarrow{a}{|}^2=26+{\left(\frac{-16}{\beta }\right)}^2$
$\Rightarrow |\overrightarrow{a}{|}^2=26+\frac{256}{{\beta }^2}$
$\Rightarrow |\overrightarrow{a}{|}_{max}^2=26+\frac{256}{4}$
$\Rightarrow |\overrightarrow{a}{|}_{max}^2=26+64$
$\Rightarrow |\overrightarrow{a}{|}_{max}^2=90$