Question

If the shortest distance between the lines \[ \frac{x - \lambda}{2} = \frac{y - 4}{3} = \frac{z - 3}{4} \] and \[ \frac{x - 2}{4} = \frac{y - 4}{6} = \frac{z - 7}{8} \] is \(\frac{13}{\sqrt{29}}\), then a value of \(\lambda\) is:

• A. \(-\frac{13}{25}\)
• B. \(\frac{13}{25}\)
• C. 1
• D. -1

Answer 1

The Correct Option is C

\( r_1 = (\lambda \hat{i} + 4\hat{j} + 3\hat{k}) + \alpha (2\hat{i} + 3\hat{j} + 4\hat{k}) \)

\( r_2 = (2\hat{i} + 4\hat{j} + 7\hat{k}) + \beta (2\hat{i} + 3\hat{j} + 4\hat{k}) \)

\( \mathbf{b} = 2\hat{i} + 3\hat{j} + 4\hat{k} \)

\( \mathbf{a}_1 = \lambda \hat{i} + 4\hat{j} + 3\hat{k} \)

\( \mathbf{a}_2 = 2\hat{i} + 4\hat{j} + 7\hat{k} \)

Shortest distance: \( \dfrac{|\mathbf{b} \times (\mathbf{a}_2 - \mathbf{a}_1)|}{|\mathbf{b}|} = \dfrac{13}{\sqrt{29}} \)

\( |(2\hat{i} + 3\hat{j} + 4\hat{k}) \times ((2 - \lambda)\hat{i} + 4\hat{k})| = \dfrac{13}{\sqrt{29}} \)

\( |- 8\hat{i} - 3(2 - \lambda)\hat{k} + 12\hat{i} + 4(2 - \lambda)\hat{j}| = 13 \)

\( |12\hat{i} - 4\hat{j} + (3\lambda - 6)\hat{k}| = 13 \)

\( 144 + 16 \lambda^2 + (3\lambda - 6)^2 = 169 \)

\( 16\lambda^2 + (3\lambda - 6)^2 = 25 \)

\( \Rightarrow \lambda = 1 \)

Answer 2

Let the parametric equations of the two lines be:

\(\vec{r}_1 = (\lambda \hat{i} + 4 \hat{j} + 3 \hat{k}) + \alpha (2 \hat{i} + 3 \hat{j} + 4 \hat{k})\),

\(\vec{r}_2 = (2 \hat{i} + 6 \hat{j} + 7 \hat{k}) + \beta (2 \hat{i} + 3 \hat{j} + 4 \hat{k})\).

Here, the direction vector for both lines is:

\(\vec{b} = 2 \hat{i} + 3 \hat{j} + 4 \hat{k}\),

and the position vectors of points on the lines are:

\(\vec{a}_1 = \lambda \hat{i} + 4 \hat{j} + 3 \hat{k}\), \(\vec{a}_2 = 2 \hat{i} + 6 \hat{j} + 7 \hat{k}\).

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

\(\text{Shortest distance} = \frac{|\vec{b} \times (\vec{a}_2 - \vec{a}_1) \cdot \vec{b}|}{|\vec{b}|} = \frac{13}{\sqrt{29}}\).

Substituting \(\vec{a}_2 - \vec{a}_1\):

\(\vec{a}_2 - \vec{a}_1 = (2 - \lambda) \hat{i} + 2 \hat{j} + 4 \hat{k}\).

The cross product \(\vec{b} \times (\vec{a}_2 - \vec{a}_1)\) simplifies as follows:

\(\vec{b} \times (\vec{a}_2 - \vec{a}_1) = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 3 & 4 \\ 2 - \lambda & 2 & 4 \end{vmatrix}\).

Solving this determinant gives:

\(\vec{b} \times (\vec{a}_2 - \vec{a}_1) = (-8 \hat{j} + 12 \hat{i} + 4 (2 - \lambda) \hat{j})\).

Taking the magnitude and using the formula for shortest distance:

\(\text{Shortest distance} = \frac{|(-8 \hat{j} + 12 \hat{i})|}{13}\).

Finally solving the quadratic for \(\lambda = 1\).