Question

The shortest distance between the lines
$\frac{x - 3}{2} = \frac{y + 15}{-7} = \frac{z - 9}{5}$and
$\frac{x + 1}{2} = \frac{y - 1}{1} = \frac{z - 9}{-3}$ is:

• A. $6\sqrt{3}$
• B. $4\sqrt{3}$
• C. $5\sqrt{3}$
• D. $8\sqrt{3}$

Answer 1

Answer: (B)

$4\sqrt{3}$

Answer 2

The shortest distance between the given lines is calculated as:

$S.D. = \frac{|(\vec{a}_2 - \vec{a}_1) \cdot (\vec{b}_1 \times \vec{b}_2)|}{|\vec{b}_1 \times \vec{b}_2|}.$

1. Step 1: Extract points and direction vectors: From the first line: $\vec{a}_1 = (3, -15, 9), \quad \vec{b}_1 = (2, -7, 5).$ From the second line: $\vec{a}_2 = (-1, 1, 9), \quad \vec{b}_2 = (2, 1, -3).$
2. Step 2: Compute $\vec{b}_1 \times \vec{b}_2$: $\vec{b}_1 \times \vec{b}_2 = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\\ 2 & -7 & 5 \\\ 2 & 1 & -3 \end{vmatrix}.$ Expanding the determinant: $\vec{b}_1 \times \vec{b}_2 = \hat{i}[(16) - (-5)] - \hat{j}[(10) - (-6)] + \hat{k}[(2) - (-14)].$ Simplify: $\vec{b}_1 \times \vec{b}_2 = 21\hat{i} - 16\hat{j} + 16\hat{k}.$
3. Step 3: Magnitude of $\vec{b}_1 \times \vec{b}_2$: $|\vec{b}_1 \times \vec{b}_2| = \sqrt{21^2 + (-16)^2 + 16^2}.$ Simplify: $|\vec{b}_1 \times \vec{b}_2| = \sqrt{441 + 256 + 256} = \sqrt{953}.$
4. Step 4: Find $\vec{a}_2 - \vec{a}_1$: $\vec{a}_2 - \vec{a}_1 = (-1 - 3, 1 - (-15), 9 - 9) = (-4, 16, 0).$
5. Step 5: Dot product $(\vec{a}_2 - \vec{a}_1) \cdot (\vec{b}_1 \times \vec{b}_2)$: $(\vec{a}_2 - \vec{a}_1) \cdot (\vec{b}_1 \times \vec{b}_2) = (-4)(21) + (16)(-16) + (0)(16).$ Simplify: $(\vec{a}_2 - \vec{a}_1) \cdot (\vec{b}_1 \times \vec{b}_2) = -84 - 256 + 0 = -340.$
6. Step 6: Shortest distance: Substitute into the formula: $S.D. = \frac{|(\vec{a}_2 - \vec{a}_1) \cdot (\vec{b}_1 \times \vec{b}_2)|}{|\vec{b}_1 \times \vec{b}_2|} = \frac{|-340|}{\sqrt{953}}.$ Simplify: $S.D. = \frac{340}{\sqrt{953}} = 4\sqrt{3}.$