Question

The shortest distance between the lines \(\frac{x+2}{1}=\frac{y}{-2}=\frac{z-5}{2}\) and \(\frac{x-4}{1}=\frac{y-1}{2}=\frac{z+3}{0}\) is

• A. 6
• B. 7
• C. 8
• D. 9

Answer 1

The Correct Option is D

Given: Lines are given as:

• \( \frac{x + 2}{1} = \frac{y}{-2} = \frac{z - 5}{2} \)
• \( \frac{x - 4}{1} = \frac{y - 1}{2} = \frac{z + 3}{0} \)

The formula for the shortest distance between two skew lines is:

\( d = \frac{|(a_2 - a_1, b_2 - b_1, c_2 - c_1) \cdot (\mathbf{l}_1 \times \mathbf{l}_2)|}{\sqrt{\mathbf{l}_1^2} \cdot \sqrt{\mathbf{l}_2^2}} \),

where \( \mathbf{l}_1, \mathbf{l}_2 \) are the direction vectors of the lines.

• Step 1: Identify the direction vectors:

\( \mathbf{l}_1 = \langle 1, -2, 2 \rangle, \quad \mathbf{l}_2 = \langle 1, 2, 0 \rangle \).

• Step 2: Position vector difference:

\( (a_2 - a_1, b_2 - b_1, c_2 - c_1) = \langle 4 + 2, 1 - 0, -3 - 5 \rangle = \langle 6, 1, -8 \rangle \).

• Step 3: Calculate the cross product \( \mathbf{l}_1 \times \mathbf{l}_2 \):

\( \mathbf{l}_1 \times \mathbf{l}_2 = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & -2 & 2 \\ 1 & 2 & 0 \end{vmatrix} \).

• Expand the determinant:

\( = \hat{i}((-4) - (4)) - \hat{j}((0) - (2)) + \hat{k}((2) - (-2)) \).

• Simplify:

\( = -4\hat{i} - 2\hat{j} + 4\hat{k} \).

• Step 4: Magnitude of the cross product:

\( |\mathbf{l}_1 \times \mathbf{l}_2| = \sqrt{(-4)^2 + (-2)^2 + 4^2} = \sqrt{16 + 4 + 16} = \sqrt{36} = 6 \).

• Step 5: Calculate the shortest distance:

\( d = \frac{|(6, 1, -8) \cdot (-4, -2, 4)|}{6} \).

• Dot product:

\( (6)(-4) + (1)(-2) + (-8)(4) = -24 - 2 - 32 = -58 \).

• Absolute value:

\( d = \frac{|-58|}{6} = 9 \).

Final Answer: The shortest distance is \( d = 9 \).