The line L_1 is parallel to the vector \overrightarrowa=-3 \hati+2 \hatj+4 \hatk and passes through the point (7,6,2) and the line L_2 is parallel to the vector \overrightarrowb=2 \hati+\hatj+3 \hatk and passes through the point (5,3,4) . The shortest distance between the lines L_1 and L_2 is :

3D Geometry PYQ — JEE Main 2025 · 2 Apr · Shift 2 · Q38 with Solution

View written solutionFree

Correct answer: A

Write the given lines in vector form

The two lines are:

$L_1: \text{ through } P_1=(7,6,2), \quad \text{direction } \vec a=(-3,2,4)$

$L_2: \text{ through } P_2=(5,3,4), \quad \text{direction } \vec b=(2,1,3)$

For two skew lines, the shortest distance is

$d=\frac{|(\overrightarrow{P_1P_2})\cdot(\vec a\times \vec b)|}{|\vec a\times \vec b|}$

Find $\vec a\times \vec b$

\begin{vmatrix} \hat i & \hat j & \hat k\\ -3 & 2 & 4\\ 2 & 1 & 3 \end{vmatrix}$$ $$=\hat i(2\cdot 3-4\cdot 1)-\hat j((-3)\cdot 3-4\cdot 2)+\hat k((-3)\cdot 1-2\cdot 2)$$ $$=\hat i(6-4)-\hat j(-9-8)+\hat k(-3-4)$$ $$=2\hat i+17\hat j-7\hat k$$ So, $$\vec a\times \vec b=(2,17,-7)$$ 3. **Find the vector joining a point on** $L_1$ **to a point on** $L_2$ Take $$\overrightarrow{P_1P_2}=P_2-P_1=(5-7,3-6,4-2)=(-2,-3,2)$$ 4. **Compute the scalar triple product** $$\overrightarrow{P_1P_2}\cdot(\vec a\times \vec b)=(-2,-3,2)\cdot(2,17,-7)$$ $$=-2\cdot 2+(-3)\cdot 17+2\cdot(-7)$$ $$=-4-51-14=-69$$ Hence, $$\left|\overrightarrow{P_1P_2}\cdot(\vec a\times \vec b)\right|=69$$ 5. **Find the magnitude of** $\vec a\times \vec b$ $$|\vec a\times \vec b|=\sqrt{2^2+17^2+(-7)^2}$$ $$=\sqrt{4+289+49}=\sqrt{342}=3\sqrt{38}$$ 6. **Shortest distance** $$d=\frac{69}{3\sqrt{38}}=\frac{23}{\sqrt{38}}$$ So the shortest distance is $$\boxed{\frac{23}{\sqrt{38}}}$$ 7. **Check options** This matches **Option A**.

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View written solutionFree

Correct answer: A

Write the given lines in vector form

The two lines are:

$L_1: \text{ through } P_1=(7,6,2), \quad \text{direction } \vec a=(-3,2,4)$

$L_2: \text{ through } P_2=(5,3,4), \quad \text{direction } \vec b=(2,1,3)$

For two skew lines, the shortest distance is

$d=\frac{|(\overrightarrow{P_1P_2})\cdot(\vec a\times \vec b)|}{|\vec a\times \vec b|}$

Find $\vec a\times \vec b$

\begin{vmatrix} \hat i & \hat j & \hat k\\ -3 & 2 & 4\\ 2 & 1 & 3 \end{vmatrix}$$ $$=\hat i(2\cdot 3-4\cdot 1)-\hat j((-3)\cdot 3-4\cdot 2)+\hat k((-3)\cdot 1-2\cdot 2)$$ $$=\hat i(6-4)-\hat j(-9-8)+\hat k(-3-4)$$ $$=2\hat i+17\hat j-7\hat k$$ So, $$\vec a\times \vec b=(2,17,-7)$$ 3. **Find the vector joining a point on** $L_1$ **to a point on** $L_2$ Take $$\overrightarrow{P_1P_2}=P_2-P_1=(5-7,3-6,4-2)=(-2,-3,2)$$ 4. **Compute the scalar triple product** $$\overrightarrow{P_1P_2}\cdot(\vec a\times \vec b)=(-2,-3,2)\cdot(2,17,-7)$$ $$=-2\cdot 2+(-3)\cdot 17+2\cdot(-7)$$ $$=-4-51-14=-69$$ Hence, $$\left|\overrightarrow{P_1P_2}\cdot(\vec a\times \vec b)\right|=69$$ 5. **Find the magnitude of** $\vec a\times \vec b$ $$|\vec a\times \vec b|=\sqrt{2^2+17^2+(-7)^2}$$ $$=\sqrt{4+289+49}=\sqrt{342}=3\sqrt{38}$$ 6. **Shortest distance** $$d=\frac{69}{3\sqrt{38}}=\frac{23}{\sqrt{38}}$$ So the shortest distance is $$\boxed{\frac{23}{\sqrt{38}}}$$ 7. **Check options** This matches **Option A**.

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