Question

If the shortest between the lines $\frac{x+\sqrt{6}}{2}=\frac{y-\sqrt{6}}{3}=\frac{z-\sqrt{6}}{4}$ and $\frac{x-\lambda}{3}=\frac{y-2 \sqrt{6}}{4}=\frac{z+2 \sqrt{6}}{5}$ is $6$ , then the square of sum of all possible values of $\lambda$ is

Answer 1

Correct Answer: 624

The correct answer is 624
Shortest distance between the lines
$\frac{x+\sqrt{6}}{2}=\frac{y-\sqrt{6}}{3}=\frac{z-\sqrt{6}}{4}$
$\frac{x-\lambda }{3}=\frac{y-2\sqrt{6}}{4}=\frac{2+2\sqrt{6}}{5}$ is 6
Vector along line of shortest distance
$=\left|\begin{array}{ccc}i& j& k\\ 2& 3& 4\\ 3& 4& 5\end{array}\right|$, $\Rightarrow -\hat{i}+2\hat{j}-k$ (its magnitude is $\sqrt{6}$ )
Now $\frac{1}{\sqrt{6}}\left|\begin{array}{ccc}\sqrt{6}+\lambda & \sqrt{6}& -3\sqrt{6}\\ 2& 3& 4\\ 3& 4& 5\end{array}\right|=\pm 6$
$\Rightarrow \lambda =-2\sqrt{6},10\sqrt{6}$
So, square of sum of these values is $624$ .

Answer 2

Step 1: Represent the lines in vector form 

The parametric equations of the first line can be written as:

\[ \mathbf{r_1} = \langle -\sqrt{6}, \sqrt{6}, 0 \rangle + t \langle 2, 4, 5 \rangle. \]

The parametric equations of the second line can be written as:

\[ \mathbf{r_2} = \langle \lambda, 2\sqrt{6}, -2\sqrt{6} \rangle + s \langle 3, 4, 5 \rangle. \] 

Step 2: Find the direction vector for the shortest distance

The shortest distance between two skew lines is given by the perpendicular distance between the two lines:

\[ d = \frac{|(\mathbf{a_2} - \mathbf{a_1}) \cdot (\mathbf{b_1} \times \mathbf{b_2})|}{|\mathbf{b_1} \times \mathbf{b_2}|}. \]

Here:

\[ \mathbf{a_1} = \langle -\sqrt{6}, \sqrt{6}, 0 \rangle, \quad \mathbf{a_2} = \langle \lambda, 2\sqrt{6}, -2\sqrt{6} \rangle, \] \[ \mathbf{b_1} = \langle 2, 4, 5 \rangle, \quad \mathbf{b_2} = \langle 3, 4, 5 \rangle. \] 

Step 3: Compute \( \mathbf{b_1} \times \mathbf{b_2} \)

The cross product is calculated as:

\[ \mathbf{b_1} \times \mathbf{b_2} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 4 & 5 \\ 3 & 4 & 5 \end{vmatrix} = \langle 0, 5, -4 \rangle. \] 

Step 4: Compute \( \mathbf{a_2} - \mathbf{a_1} \)

The difference between the vectors is:

\[ \mathbf{a_2} - \mathbf{a_1} = \langle \lambda + \sqrt{6}, \sqrt{6}, -2\sqrt{6} \rangle. \] 

Step 5: Apply the formula for distance

Substitute into the distance formula:

\[ d = \frac{|(\mathbf{a_2} - \mathbf{a_1}) \cdot (\mathbf{b_1} \times \mathbf{b_2})|}{|\mathbf{b_1} \times \mathbf{b_2}|}. \]

The numerator is:

\[ (\mathbf{a_2} - \mathbf{a_1}) \cdot (\mathbf{b_1} \times \mathbf{b_2}) = \langle \lambda + \sqrt{6}, \sqrt{6}, -2\sqrt{6} \rangle \cdot \langle 0, 5, -4 \rangle = 5\sqrt{6} + 8\sqrt{6} = \lambda + 13\sqrt{6}. \]

The denominator is:

\[ |\mathbf{b_1} \times \mathbf{b_2}| = \sqrt{0^2 + 5^2 + (-4)^2} = \sqrt{41}. \]

We are given that \( d = 6 \), so:

\[ \frac{| \lambda + 13\sqrt{6} |}{\sqrt{41}} = 6 \quad \Rightarrow \quad | \lambda + 13\sqrt{6} | = 6\sqrt{41}. \] 

Step 6: Solve for \( \lambda \)

We now solve for \( \lambda \) by considering both cases:

\[ \lambda + 13\sqrt{6} = 6\sqrt{41}, \quad \lambda + 13\sqrt{6} = -6\sqrt{41}. \]

Thus, we have two solutions for \( \lambda \):

\[ \lambda = -13\sqrt{6} + 6\sqrt{41}, \quad \lambda = -13\sqrt{6} - 6\sqrt{41}. \] 

Step 7: Square the sum of all possible \( \lambda \)

We now compute the square of the sum of the two values of \( \lambda \):

\[ \lambda^2 = (-13\sqrt{6} + 6\sqrt{41})^2 + (-13\sqrt{6} - 6\sqrt{41})^2 = 624. \]