JEE Main · 2024 · 31 Jan 2024 · Shift 2 · Vector Algebra

$\vec{a}=3\hat{i}+2\hat{j}+\hat{k},\vec{b}=2\hat{i}-\hat{j}+3\hat{k}$, so $\vec{a}+\vec{b}=5\hat{i}+\hat{j}+4\hat{k}$.

$\vec{a}\times \vec{b}=\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 3& 2& 1\\ 2& -1& 3\end{array}\right|=\hat{i}(6+1)-\hat{j}(9-2)+\hat{k}(-3-4)=7\hat{i}-7\hat{j}-7\hat{k}$.

$2(\vec{a}\times \vec{b})+24\hat{j}-6\hat{k}=14\hat{i}+10\hat{j}-20\hat{k}$.

Let $\vec{c}=x\hat{i}+y\hat{j}+z\hat{k}$. Then

$(\vec{a}+\vec{b})\times \vec{c}=\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 5& 1& 4\\ x& y& z\end{array}\right|=(z-4y)\hat{i}+(4x-5z)\hat{j}+(5y-x)\hat{k}$.

Equating with $14\hat{i}+10\hat{j}-20\hat{k}$:

$z-4y=14,4x-5z=10,5y-x=-20$.

Also, $\vec{a}-\vec{b}+\hat{i}=(1+1)\hat{i}+3\hat{j}-2\hat{k}=2\hat{i}+3\hat{j}-2\hat{k}$, and $(\vec{a}-\vec{b}+\hat{i})\cdot \vec{c}=-3$:

$2x+3y-2z=-3$.

From the first equation, $z=14+4y$; from the third, $x=5y+20$. Substituting into the second equation confirms consistency ($10=10$) for any $y$, so use the fourth equation:

$2(5y+20)+3y-2(14+4y)=-3$

$10y+40+3y-28-8y=-3$

$5y+12=-3\Rightarrow y=-3$.

Then $x=5(-3)+20=5$, $z=14+4(-3)=2$.

$\vec{c}=(5,-3,2)$, so $|\vec{c}{|}^2=25+9+4=38$.