The shortest distance between the lines x - 5 \over 1 = y - 2 \over 2 = z - 4 \over - 3 and x + 3 \over 1 = y + 5 \over 4 = z - 1 \over - 5 is :

3D Geometry PYQ — JEE Main 2023 · 1 Feb · Shift 1 · Q25 with Solution

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Correct answer: D

For the first line, $\frac{x-5}{1}=\frac{y-2}{2}=\frac{z-4}{-3}=t$ so a point on it is $A(5,2,4)$ and its direction vector is $\vec d_1=(1,2,-3).$

For the second line, $\frac{x+3}{1}=\frac{y+5}{4}=\frac{z-1}{-5}=s$ so a point on it is $B(-3,-5,1)$ and its direction vector is $\vec d_2=(1,4,-5).$

The shortest distance is $D=\frac{|\overrightarrow{AB}\cdot (\vec d_1\times \vec d_2)|}{|\vec d_1\times \vec d_2|}.$

Here, $\overrightarrow{AB}=B-A=(-3-5,-5-2,1-4)=(-8,-7,-3).$

\begin{vmatrix} \hat i & \hat j & \hat k \\ 1 & 2 & -3 \\ 1 & 4 & -5 \end{vmatrix}$$ $$=\hat i\,(2\cdot(-5)-(-3)\cdot4)-\hat j\,(1\cdot(-5)-(-3)\cdot1)+\hat k\,(1\cdot4-2\cdot1)$$ $$=\hat i(-10+12)-\hat j(-5+3)+\hat k(4-2)$$ $$=(2,2,2).$$ Hence, $$|\vec d_1\times \vec d_2|=\sqrt{2^2+2^2+2^2}=\sqrt{12}=2\sqrt3.$$ --- 4. **Compute the scalar triple product** $$\overrightarrow{AB}\cdot (\vec d_1\times \vec d_2)=(-8,-7,-3)\cdot(2,2,2)$$ $$= -16-14-6=-36.$$ So, $$|\overrightarrow{AB}\cdot (\vec d_1\times \vec d_2)|=36.$$ --- 5. **Find the distance** $$D=\frac{36}{2\sqrt3}=\frac{18}{\sqrt3}=6\sqrt3.$$ --- 6. **Check options** The shortest distance is $$\boxed{6\sqrt3}$$ which corresponds to **Option D**. --- 7. **Comparison with stored answer** Stored correct answer: **D** Our derived answer: **D** So the derived answer agrees with the stored answer.

View written solutionFree

Correct answer: D

For the first line, $\frac{x-5}{1}=\frac{y-2}{2}=\frac{z-4}{-3}=t$ so a point on it is $A(5,2,4)$ and its direction vector is $\vec d_1=(1,2,-3).$

For the second line, $\frac{x+3}{1}=\frac{y+5}{4}=\frac{z-1}{-5}=s$ so a point on it is $B(-3,-5,1)$ and its direction vector is $\vec d_2=(1,4,-5).$

The shortest distance is $D=\frac{|\overrightarrow{AB}\cdot (\vec d_1\times \vec d_2)|}{|\vec d_1\times \vec d_2|}.$

Here, $\overrightarrow{AB}=B-A=(-3-5,-5-2,1-4)=(-8,-7,-3).$

\begin{vmatrix} \hat i & \hat j & \hat k \\ 1 & 2 & -3 \\ 1 & 4 & -5 \end{vmatrix}$$ $$=\hat i\,(2\cdot(-5)-(-3)\cdot4)-\hat j\,(1\cdot(-5)-(-3)\cdot1)+\hat k\,(1\cdot4-2\cdot1)$$ $$=\hat i(-10+12)-\hat j(-5+3)+\hat k(4-2)$$ $$=(2,2,2).$$ Hence, $$|\vec d_1\times \vec d_2|=\sqrt{2^2+2^2+2^2}=\sqrt{12}=2\sqrt3.$$ --- 4. **Compute the scalar triple product** $$\overrightarrow{AB}\cdot (\vec d_1\times \vec d_2)=(-8,-7,-3)\cdot(2,2,2)$$ $$= -16-14-6=-36.$$ So, $$|\overrightarrow{AB}\cdot (\vec d_1\times \vec d_2)|=36.$$ --- 5. **Find the distance** $$D=\frac{36}{2\sqrt3}=\frac{18}{\sqrt3}=6\sqrt3.$$ --- 6. **Check options** The shortest distance is $$\boxed{6\sqrt3}$$ which corresponds to **Option D**. --- 7. **Comparison with stored answer** Stored correct answer: **D** Our derived answer: **D** So the derived answer agrees with the stored answer.