JEE Main 2024 — Mathematics Vectors & 3D Geometry

2024integermedium

If ${d}_{1}$ is the shortest distance between the lines $x+1=2y=-12z,x=y+2=6z-6$ and ${d}_{2}$ is the shortest distance between the lines $\frac{x-1}{2}=\frac{y+8}{-7}=\frac{z-4}{5},\frac{x-1}{2}=\frac{y-2}{1}=\frac{z-6}{-3}$ , then the value of $\frac{32\sqrt{3}{d}_{1}}{{d}_{2}}$ is :

Official previous-year question

Held on 30 Jan 2024 · Verified 6 Jul 2026.

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Solution

Let,

${L}_{1}:\frac{x+1}{1}=\frac{y}{\frac{1}{2}}=\frac{z}{\frac{-1}{12}},$

And ${L}_{2}:\frac{x}{1}=\frac{y+2}{1}=\frac{z-1}{\frac{1}{6}}$

Now, given ${d}_{1}=$ shortest distance between ${L}_{1}$ and ${L}_{2}$

Now using the formula of shortest distance we get,

${d}_{1}=|\frac{({\vec{a}}_{2}-{\vec{a}}_{1})\cdot ({\vec{b}}_{1}\times {\vec{b}}_{2})}{|({\vec{b}}_{1}\times {\vec{b}}_{2})|}|$

$\Rightarrow {d}_{1}=|\frac{(\hat{i}-2\hat{j}+\hat{k})\cdot ((\hat{i}+\frac{1}{2}\hat{j}-\frac{1}{12}\hat{k})\times (\hat{i}+\hat{j}+\frac{1}{6}\hat{k}))}{|(\hat{i}+\frac{1}{2}\hat{j}-\frac{1}{12}\hat{k})\times (\hat{i}+\hat{j}+\frac{1}{6}\hat{k})|}|$

$\Rightarrow {d}_{1}=|\frac{(\hat{i}-2\hat{j}+\hat{k})\cdot (\frac{\hat{i}}{6}-\frac{\hat{j}}{4}+\frac{\hat{k}}{2})}{|(\frac{\hat{i}}{6}-\frac{\hat{j}}{4}+\frac{\hat{k}}{2})|}|$

$\Rightarrow {d}_{1}=|\frac{\frac{1}{6}+\frac{1}{2}+\frac{1}{2}}{\sqrt{\frac{49}{144}}}|$

$\Rightarrow {d}_{1}=|\frac{\frac{7}{6}}{\frac{7}{12}}|=2$

Now, let

${L}_{3}:\frac{x-1}{2}=\frac{y+8}{-7}=\frac{z-4}{5},{L}_{4}:\frac{x-1}{2}=\frac{y-2}{1}=\frac{z-6}{-3}$

Similarly finding ${d}_{2}=$ shortest distance between ${L}_{3}$ & ${L}_{4}$ we get,

${d}_{2}=|\frac{(10\hat{j}+2\hat{k})\cdot ((2\hat{i}-7\hat{j}+5\hat{k})\times (2\hat{i}+\hat{j}-3\hat{k}))}{|((2\hat{i}-7\hat{j}+5\hat{k})\times (2\hat{i}+\hat{j}-3\hat{k}))|}|$

$\Rightarrow {d}_{2}=|\frac{(10\hat{j}+2\hat{k})\cdot (16\hat{i}+16\hat{j}+16\hat{k})}{|(16\hat{i}+16\hat{j}+16\hat{k})|}|$

$\Rightarrow {d}_{2}=|\frac{160+32}{16\sqrt{3}}|$

$\Rightarrow {d}_{2}=|\frac{12}{\sqrt{3}}|$

Hence, $\frac{32\sqrt{3}{d}_{1}}{{d}_{2}}=\frac{32\sqrt{3}\times 2}{\frac{12}{\sqrt{3}}}=16$

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Solution

Let,

${L}_{1}:\frac{x+1}{1}=\frac{y}{\frac{1}{2}}=\frac{z}{\frac{-1}{12}},$

And ${L}_{2}:\frac{x}{1}=\frac{y+2}{1}=\frac{z-1}{\frac{1}{6}}$

Now, given ${d}_{1}=$ shortest distance between ${L}_{1}$ and ${L}_{2}$

Now using the formula of shortest distance we get,

${d}_{1}=|\frac{({\vec{a}}_{2}-{\vec{a}}_{1})\cdot ({\vec{b}}_{1}\times {\vec{b}}_{2})}{|({\vec{b}}_{1}\times {\vec{b}}_{2})|}|$

$\Rightarrow {d}_{1}=|\frac{(\hat{i}-2\hat{j}+\hat{k})\cdot (\frac{\hat{i}}{6}-\frac{\hat{j}}{4}+\frac{\hat{k}}{2})}{|(\frac{\hat{i}}{6}-\frac{\hat{j}}{4}+\frac{\hat{k}}{2})|}|$

$\Rightarrow {d}_{1}=|\frac{\frac{1}{6}+\frac{1}{2}+\frac{1}{2}}{\sqrt{\frac{49}{144}}}|$

$\Rightarrow {d}_{1}=|\frac{\frac{7}{6}}{\frac{7}{12}}|=2$

Now, let

${L}_{3}:\frac{x-1}{2}=\frac{y+8}{-7}=\frac{z-4}{5},{L}_{4}:\frac{x-1}{2}=\frac{y-2}{1}=\frac{z-6}{-3}$

Similarly finding ${d}_{2}=$ shortest distance between ${L}_{3}$ & ${L}_{4}$ we get,

${d}_{2}=|\frac{(10\hat{j}+2\hat{k})\cdot ((2\hat{i}-7\hat{j}+5\hat{k})\times (2\hat{i}+\hat{j}-3\hat{k}))}{|((2\hat{i}-7\hat{j}+5\hat{k})\times (2\hat{i}+\hat{j}-3\hat{k}))|}|$

$\Rightarrow {d}_{2}=|\frac{(10\hat{j}+2\hat{k})\cdot (16\hat{i}+16\hat{j}+16\hat{k})}{|(16\hat{i}+16\hat{j}+16\hat{k})|}|$

$\Rightarrow {d}_{2}=|\frac{160+32}{16\sqrt{3}}|$

$\Rightarrow {d}_{2}=|\frac{12}{\sqrt{3}}|$

Hence, $\frac{32\sqrt{3}{d}_{1}}{{d}_{2}}=\frac{32\sqrt{3}\times 2}{\frac{12}{\sqrt{3}}}=16$

JEE Main 2024 — Mathematics Vectors & 3D Geometry

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JEE Main 2024 — Mathematics Vectors & 3D Geometry

Did you get this right?

Solution

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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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