JEE Main 2021 — Mathematics Vectors & 3D Geometry

2021integermedium

A line $l$ passing through origin is perpendicular to the lines

${l}_{1}:\vec{r}=(3+t)\hat{i}+(-1+2t)\hat{j}+(4+2t)\hat{k}$

${l}_{2}:\vec{r}=(3+2s)\hat{i}+(3+2s)\hat{j}+(2+s)\hat{k}$

If the co-ordinates of the point in the first octant on ${l}_{2}$ at a distance of $\sqrt{17}$ from the point of intersection of $l$ and ${l}_{1}$ are $(a,b,c)$ , then $18(a+b+c)$ is equal to ___ .

Official previous-year question

Held on 25 Feb 2021 · Verified 6 Jul 2026.

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Solution

${l}_{1}:\vec{r}=(3+t)\hat{i}+(-1+2t)\hat{j}+(4+2t)\hat{k}$

${l}_{2}:\vec{r}=(3+2s)\hat{i}+(3+2s)\hat{j}+(2+s)\hat{k}$

$\mathrm{DR}$ of ${l}_{1}\equiv (1,2,2)$

$\mathrm{DR}$ of ${l}_{2}\equiv (2,2,1)$

${l}_{1}\times {l}_{2}=|\begin{matrix}i & j & k \\ 1 & 2 & 2 \\ 2 & 2 & 1\end{matrix}|$

$=i(2-4)-j(1-4)+k(2-4)$

$=-2i+3j-2k$

Hence, $\mathrm{DR}$ of $l$ ( line $\perp$ to ${l}_{1}&{l}_{2}$ )

$=(-2,3,-2)$

$\therefore l:\vec{r}=-2\mu \hat{i}+3\mu \hat{j}-2\mu \hat{k}$

for intersection of $l&{l}_{1}$

$3+t=-2\mu$

$-1+2t=3\mu$

$4+2t=-2\mu$

$\Rightarrow t=-1&\mu =-1$

$\therefore$ Point of intersection $P\equiv (2,-3,2)$

Let point on ${l}_{2}$ be $Q(3+2s,3+2s,2+s)$

Given $PQ=\sqrt{17}\Rightarrow {(PQ)}^{2}=17$

$\Rightarrow {(2s+1)}^{2}+{(6+2s)}^{2}+{(s)}^{2}=17$

$\Rightarrow 9{s}^{2}+28s+20=0$

$\Rightarrow s=-2,-\frac{10}{9}$

$s\neq -2$ as point lies on ${1}^{st}$ octant.

$\therefore a=3+2(-\frac{10}{9})=\frac{7}{9}$

$b=3+2(-\frac{10}{9})=\frac{7}{9}$

$c=2+(-\frac{10}{9})=\frac{8}{9}$

$\therefore 18(a+b+c)=18(\frac{22}{9})=44$

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Solution

${l}_{1}:\vec{r}=(3+t)\hat{i}+(-1+2t)\hat{j}+(4+2t)\hat{k}$

${l}_{2}:\vec{r}=(3+2s)\hat{i}+(3+2s)\hat{j}+(2+s)\hat{k}$

$\mathrm{DR}$ of ${l}_{1}\equiv (1,2,2)$

$\mathrm{DR}$ of ${l}_{2}\equiv (2,2,1)$

${l}_{1}\times {l}_{2}=|\begin{matrix}i & j & k \\ 1 & 2 & 2 \\ 2 & 2 & 1\end{matrix}|$

$=i(2-4)-j(1-4)+k(2-4)$

$=-2i+3j-2k$

Hence, $\mathrm{DR}$ of $l$ ( line $\perp$ to ${l}_{1}&{l}_{2}$ )

$=(-2,3,-2)$

$\therefore l:\vec{r}=-2\mu \hat{i}+3\mu \hat{j}-2\mu \hat{k}$

for intersection of $l&{l}_{1}$

$3+t=-2\mu$

$-1+2t=3\mu$

$4+2t=-2\mu$

$\Rightarrow t=-1&\mu =-1$

$\therefore$ Point of intersection $P\equiv (2,-3,2)$

Let point on ${l}_{2}$ be $Q(3+2s,3+2s,2+s)$

Given $PQ=\sqrt{17}\Rightarrow {(PQ)}^{2}=17$

$\Rightarrow {(2s+1)}^{2}+{(6+2s)}^{2}+{(s)}^{2}=17$

$\Rightarrow 9{s}^{2}+28s+20=0$

$\Rightarrow s=-2,-\frac{10}{9}$

$s\neq -2$ as point lies on ${1}^{st}$ octant.

$\therefore a=3+2(-\frac{10}{9})=\frac{7}{9}$

$b=3+2(-\frac{10}{9})=\frac{7}{9}$

$c=2+(-\frac{10}{9})=\frac{8}{9}$

$\therefore 18(a+b+c)=18(\frac{22}{9})=44$

JEE Main 2021 — Mathematics Vectors & 3D Geometry

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JEE Main 2021 — Mathematics Vectors & 3D Geometry

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

Keep practicing Vectors & 3D Geometry