Let O be the origin. Let… - JEE Main 2021 Mathematics PYQ

JEE Main 2021 — Mathematics Vectors & 3D Geometry

2021mcqhard

Let $O$ be the origin. Let $\vec{OP}=x\hat{i}+y\hat{j}-\hat{k}$ and $\vec{OQ}=-\hat{i}+2\hat{j}+3x\hat{k},x,y\in R,x>0,$ be such that $|\vec{PQ}|=\sqrt{20}$ and the vector $\vec{OP}$ is perpendicular to $\vec{OQ}.$ If $\vec{OR}=3\hat{i}+z\hat{j}-7\hat{k},z\in R,$ is coplanar with $\vec{OP}$ and $\vec{OQ},$ then the value of ${x}^{2}+{y}^{2}+{z}^{2}$ is equal to

Official previous-year question

Held on 17 Mar 2021 · Verified 6 Jul 2026.

Options

A
$7$
B
$9$
C
$2$
D
$1$

Did you get this right?

Solution

If two vectors ${a}_{1}\hat{i}+{b}_{1}\hat{j}+{c}_{1}\hat{k}$ and ${a}_{2}\hat{i}+{b}_{2}\hat{j}+{c}_{2}\hat{k}$ are perpendicular, then ${a}_{1}{a}_{2}+{b}_{1}{b}_{2}+{c}_{1}{c}_{2}=0.$

Given $\vec{OP}\perp \vec{OQ}$

$\Rightarrow -x+2y-3x=0$

$\Rightarrow y=2x...(i)$

Also, $\vec{PQ}=\vec{OQ}-\vec{OP}$

$\Rightarrow \vec{PQ}=(-1-x)\hat{i}+(2-y)\hat{j}+(3x+1)\hat{k}$

And $|\vec{PQ}|=\sqrt{20}$

$\Rightarrow \sqrt{{(-1-x)}^{2}+{(2-y)}^{2}+{(3x+1)}^{2}}=\sqrt{20}$

$\Rightarrow (x+1{)}^{2}+(y-2{)}^{2}+(1+3x{)}^{2}=20$

Put the value from equation $(i),$ to get

$(x+1{)}^{2}+(2x-2{)}^{2}+(1+3x{)}^{2}=20$

$\Rightarrow {x}^{2}+2x+1+4{x}^{2}-8x+4+1+6x+9{x}^{2}=20$

$\Rightarrow 14{x}^{2}+6=20$

$\Rightarrow 14{x}^{2}=14$

$\Rightarrow x=\pm 1,$ but given $x>0$

$\Rightarrow x=1$ and $y=2.$

Now $\vec{OP},\vec{OQ},\vec{OR}$ are coplanar and we know that the three vectors ${a}_{1}\hat{i}+{b}_{1}\hat{j}+{c}_{1}\hat{k},{a}_{2}\hat{i}+{b}_{2}\hat{j}+{c}_{2}\hat{k}$ and ${a}_{3}\hat{i}+{b}_{3}\hat{j}+{c}_{3}\hat{k}$ are coplanar, then $|\begin{matrix}{a}_{1} & {b}_{1} & {c}_{1} \\ {a}_{2} & {b}_{2} & {c}_{2} \\ {a}_{3} & {b}_{3} & {c}_{3}\end{matrix}|=0$

$\Rightarrow |\begin{matrix}x & y & -1 \\ -1 & 2 & 3x \\ 3 & z & -7\end{matrix}|=0$

$\Rightarrow |\begin{matrix}1 & 2 & -1 \\ -1 & 2 & 3 \\ 3 & z & -7\end{matrix}|=0$

$\Rightarrow 1(-14-3z)-2(7-9)-1(-z-6)=0$

$\Rightarrow z=-2$

$\therefore {x}^{2}+{y}^{2}+{z}^{2}=1+4+4=9.$

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Solution

If two vectors ${a}_{1}\hat{i}+{b}_{1}\hat{j}+{c}_{1}\hat{k}$ and ${a}_{2}\hat{i}+{b}_{2}\hat{j}+{c}_{2}\hat{k}$ are perpendicular, then ${a}_{1}{a}_{2}+{b}_{1}{b}_{2}+{c}_{1}{c}_{2}=0.$

Given $\vec{OP}\perp \vec{OQ}$

$\Rightarrow -x+2y-3x=0$

$\Rightarrow y=2x...(i)$

Also, $\vec{PQ}=\vec{OQ}-\vec{OP}$

$\Rightarrow \vec{PQ}=(-1-x)\hat{i}+(2-y)\hat{j}+(3x+1)\hat{k}$

And $|\vec{PQ}|=\sqrt{20}$

$\Rightarrow \sqrt{{(-1-x)}^{2}+{(2-y)}^{2}+{(3x+1)}^{2}}=\sqrt{20}$

$\Rightarrow (x+1{)}^{2}+(y-2{)}^{2}+(1+3x{)}^{2}=20$

Put the value from equation $(i),$ to get

$(x+1{)}^{2}+(2x-2{)}^{2}+(1+3x{)}^{2}=20$

$\Rightarrow {x}^{2}+2x+1+4{x}^{2}-8x+4+1+6x+9{x}^{2}=20$

$\Rightarrow 14{x}^{2}+6=20$

$\Rightarrow 14{x}^{2}=14$

$\Rightarrow x=\pm 1,$ but given $x>0$

$\Rightarrow x=1$ and $y=2.$

$\Rightarrow |\begin{matrix}x & y & -1 \\ -1 & 2 & 3x \\ 3 & z & -7\end{matrix}|=0$

$\Rightarrow |\begin{matrix}1 & 2 & -1 \\ -1 & 2 & 3 \\ 3 & z & -7\end{matrix}|=0$

$\Rightarrow 1(-14-3z)-2(7-9)-1(-z-6)=0$

$\Rightarrow z=-2$

$\therefore {x}^{2}+{y}^{2}+{z}^{2}=1+4+4=9.$

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Options

Did you get this right?

Solution

Your note

JEE Main Vectors & 3D Geometry in other years

More JEE Main Vectors & 3D Geometry questions

Other Mathematics topics for JEE Main

Algebra

Calculus

Coordinate Geometry

Probability & Statistics

Trigonometry

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