Question

The line $l_1$ passes through the point $(2,6,2)$ and is perpendicular to the plane $2 x+y-2 z=10$. Then the shortest distance between the line $l_1$ and the line $\frac{x+1}{2}=\frac{y+4}{-3}=\frac{z}{2}$ is :

• A. $\frac{19}{3}$
• B. 7
• C. 9
• D. $\frac{13}{3}$

Answer 1

The Correct Option is C

The given line \( \ell \) passes through the point \( A(2, 6, 2) \) and is perpendicular to the plane \( 2x + y - 2z = 10 \). The equations of the two lines are:

• \( L_1: \frac{x - 2}{2} = \frac{y - 6}{1} = \frac{z - 2}{-2} \),
• \( L_2: \frac{x + 1}{2} = \frac{y + 4}{-3} = \frac{z}{2} \).

The shortest distance \( d \) between the two skew lines is given by:

\[ d = \frac{| \overrightarrow{AB} \cdot (\overrightarrow{MN}) |}{| \overrightarrow{MN} |}, \]

where:

• \( \overrightarrow{AB} \) is the vector from point \( A(2, 6, 2) \) on \( L_1 \) to point \( B(-1, -4, 0) \) on \( L_2 \),
• \( \overrightarrow{MN} \) is the cross product of the direction ratios of the two lines.

The vector \( \overrightarrow{AB} \) is:

\[ \overrightarrow{AB} = (-1 - 2) \hat{i} + (-4 - 6) \hat{j} + (0 - 2) \hat{k} = -3 \hat{i} - 10 \hat{j} - 2 \hat{k}. \]

The direction ratios of \( L_1 \) are \( 2 \hat{i} + 1 \hat{j} - 2 \hat{k} \), and the direction ratios of \( L_2 \) are \( 2 \hat{i} - 3 \hat{j} + 2 \hat{k} \). The cross product \( \overrightarrow{MN} \) is:

\[ \overrightarrow{MN} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 1 & -2 \\ 2 & -3 & 2 \end{vmatrix}. \]

Expanding the determinant gives:

\[ \overrightarrow{MN} = (-4) \hat{i} - (8) \hat{j} - (8) \hat{k}. \]

The magnitude of \( \overrightarrow{MN} \) is:

\[ |\overrightarrow{MN}| = \sqrt{(-4)^2 + (-8)^2 + (-8)^2} = \sqrt{16 + 64 + 64} = 12. \]

Now, compute the dot product \( \overrightarrow{AB} \cdot \overrightarrow{MN} \):

\[ \overrightarrow{AB} \cdot \overrightarrow{MN} = (-3)(-4) + (-10)(-8) + (-2)(-8) = 12 + 80 + 16 = 108. \]

The shortest distance \( d \) is:

\[ d = \frac{| \overrightarrow{AB} \cdot \overrightarrow{MN} |}{| \overrightarrow{MN} |} = \frac{108}{12} = 9. \]

Conclusion

The shortest distance between the two lines is:

\[ \boxed{9}. \]

Answer 2

Line $\ell$, is given by
$L_1:\frac{x-2}{2}=\frac{y-6}{1}=\frac{z-2}{-2}$
Given,
$L_2:\frac{x+1}{2}=\frac{y+4}{-3}=\frac{z}{2}$

Shortest distance $=\left|\frac{\overrightarrow{AB}\cdot \overrightarrow{MN}}{MN}\right|$
$\overrightarrow{AB}=3\hat{i}+10\hat{j}+2\hat{k}$
$\overrightarrow{MN}=\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 2& 1& -2\\ 2& -3& 2\end{array}\right|=-4\hat{i}-8\hat{j}-8\hat{k}$
$MN=\sqrt{16+64+64}=12$
$\therefore$ Shortest distance $=\left|\frac{-12-80-16}{12}\right|=9$
$\therefore$