JEE Main 2023 — Mathematics Vectors & 3D Geometry

2023integermedium

Let ${\lambda }_{1},{\lambda }_{2}$ be the values of $\lambda$ for which the points $(\frac{5}{2},1,\lambda )$ and $(-2,0,1)$ are at equal distance from the plane $2x+3y-6z+7$ If ${\lambda }_{1}>{\lambda }_{2}$

then the distance of the point $({\lambda }_{1}-{\lambda }_{2},{\lambda }_{2},{\lambda }_{1})$ from the line $\frac{x-5}{1}=\frac{y-1}{2}=\frac{z+7}{2}$ is $______$

Official previous-year question

Held on 8 Apr 2023 · Verified 6 Jul 2026.

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Solution

Given,

${\lambda }_{1},{\lambda }_{2}$ be the values of $\lambda$ for which the points $(\frac{5}{2},1,\lambda )$ and $(-2,0,1)$ are at equal distance from the plane $2x+3y-6z+7$ ,

So, by using distance formula of a point from a plane we get,

$|\frac{-4+0-6+7}{\sqrt{{2}^{2}+{3}^{2}+{6}^{2}}}|=|\frac{5+3-6\lambda +7}{\sqrt{{2}^{2}+{3}^{2}+{6}^{2}}}|$

$\Rightarrow |\frac{-4+0-6+7}{7}|=|\frac{5+3-6\lambda +7}{7}|$

$\Rightarrow \frac{3}{7}=|\frac{15-6\lambda }{7}|$

$\Rightarrow \lambda =2$ or $3$

$\Rightarrow {\lambda }_{1}=3,{\lambda }_{2}=2$ as ${\lambda }_{1}>{\lambda }_{2}$

So, $({\lambda }_{1}-{\lambda }_{2},{\lambda }_{2},{\lambda }_{1})=(1,2,3)$

Now finding the distance of point $(1,2,3)$ from the line $\frac{x-5}{1}=\frac{y-1}{2}=\frac{z+7}{2}=\lambda$ we get,

JEE Main 2023 Mathematics, Three Dimensional Geometry — solution figure

Now form diagram we can see that, $\vec{PM}$ is perpendicular to given line,

So, by perpendicular condition we get,

$\vec{PM}\cdot (\hat{i}+2\hat{j}+2\hat{k})=0$

$\Rightarrow (\lambda +4)+2(2\lambda -1)+2(2\lambda -10)=0$

$\Rightarrow 9\lambda =18\text{or}\lambda =2$

Hence, the point $M$ will be $M(7,5,-3)$

So, the distance $|\vec{PM}|=\sqrt{{6}^{2}+{3}^{2}+{6}^{2}}=9$

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Solution

Given,

${\lambda }_{1},{\lambda }_{2}$ be the values of $\lambda$ for which the points $(\frac{5}{2},1,\lambda )$ and $(-2,0,1)$ are at equal distance from the plane $2x+3y-6z+7$ ,

So, by using distance formula of a point from a plane we get,

$|\frac{-4+0-6+7}{\sqrt{{2}^{2}+{3}^{2}+{6}^{2}}}|=|\frac{5+3-6\lambda +7}{\sqrt{{2}^{2}+{3}^{2}+{6}^{2}}}|$

$\Rightarrow |\frac{-4+0-6+7}{7}|=|\frac{5+3-6\lambda +7}{7}|$

$\Rightarrow \frac{3}{7}=|\frac{15-6\lambda }{7}|$

$\Rightarrow \lambda =2$ or $3$

$\Rightarrow {\lambda }_{1}=3,{\lambda }_{2}=2$ as ${\lambda }_{1}>{\lambda }_{2}$

So, $({\lambda }_{1}-{\lambda }_{2},{\lambda }_{2},{\lambda }_{1})=(1,2,3)$

Now finding the distance of point $(1,2,3)$ from the line $\frac{x-5}{1}=\frac{y-1}{2}=\frac{z+7}{2}=\lambda$ we get,

JEE Main 2023 Mathematics, Three Dimensional Geometry — solution figure

Now form diagram we can see that, $\vec{PM}$ is perpendicular to given line,

So, by perpendicular condition we get,

$\vec{PM}\cdot (\hat{i}+2\hat{j}+2\hat{k})=0$

$\Rightarrow (\lambda +4)+2(2\lambda -1)+2(2\lambda -10)=0$

$\Rightarrow 9\lambda =18\text{or}\lambda =2$

Hence, the point $M$ will be $M(7,5,-3)$

So, the distance $|\vec{PM}|=\sqrt{{6}^{2}+{3}^{2}+{6}^{2}}=9$

JEE Main 2023 — Mathematics Vectors & 3D Geometry

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Solution

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JEE Main 2023 — 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