Question

If the shortest distance between the lines \[ \frac{x - \lambda}{3} = \frac{y - 2}{-1} = \frac{z - 1}{1} \] and \[ \frac{x + 2}{-3} = \frac{y + 5}{2} = \frac{z - 4}{4} \] is \[ \frac{44}{\sqrt{30}}, \] then the largest possible value of $|\lambda|$ is equal to ________.

Answer 1

Correct Answer: 43

The objective is to find the maximum absolute value of \(|\lambda|\) given the equations for two lines in three-dimensional space and the specified shortest distance between them. The shortest distance between two skew lines is determined using a standard vector formula.

Concept Used:

The shortest distance (SD) between two skew lines, represented as \(L_1: \vec{r} = \vec{a_1} + t\vec{b_1}\) and \(L_2: \vec{r} = \vec{a_2} + s\vec{b_2}\), is calculated using the formula:

\[\text{SD} = \left| \frac{(\vec{a_2} - \vec{a_1}) \cdot (\vec{b_1} \times \vec{b_2})}{|\vec{b_1} \times \vec{b_2}|} \right|\]

This formula quantifies the length of the projection of the vector connecting a point on each line (\(\vec{a_2} - \vec{a_1}\)) onto the normal vector of both lines (\(\vec{b_1} \times \vec{b_2}\)).

Step-by-Step Solution:

Step 1: Extract the position vectors and direction vectors from the given line equations.

For the first line, \(L_1: \frac{x - \lambda}{3} = \frac{y - 2}{-1} = \frac{z - 1}{1}\):

• A point on the line corresponds to the position vector \(\vec{a_1} = \lambda\mathbf{i} + 2\mathbf{j} + \mathbf{k}\).
• The direction vector is \(\vec{b_1} = 3\mathbf{i} - \mathbf{j} + \mathbf{k}\).

For the second line, \(L_2: \frac{x + 2}{-3} = \frac{y + 5}{2} = \frac{z - 4}{4}\):

• A point on the line corresponds to the position vector \(\vec{a_2} = -2\mathbf{i} - 5\mathbf{j} + 4\mathbf{k}\).
• The direction vector is \(\vec{b_2} = -3\mathbf{i} + 2\mathbf{j} + 4\mathbf{k}\).

Step 2: Compute the vector difference \(\vec{a_2} - \vec{a_1}\).

\[\vec{a_2} - \vec{a_1} = (-2\mathbf{i} - 5\mathbf{j} + 4\mathbf{k}) - (\lambda\mathbf{i} + 2\mathbf{j} + \mathbf{k})\]\[\vec{a_2} - \vec{a_1} = (-2 - \lambda)\mathbf{i} - 7\mathbf{j} + 3\mathbf{k}\]

Step 3: Calculate the cross product of the direction vectors, \(\vec{b_1} \times \vec{b_2}\).

\[\vec{b_1} \times \vec{b_2} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 3 & -1 & 1 \\ -3 & 2 & 4 \end{vmatrix}\]\[= \mathbf{i}((-1)(4) - (1)(2)) - \mathbf{j}((3)(4) - (1)(-3)) + \mathbf{k}((3)(2) - (-1)(-3))\]\[= \mathbf{i}(-4 - 2) - \mathbf{j}(12 + 3) + \mathbf{k}(6 - 3)\]\[= -6\mathbf{i} - 15\mathbf{j} + 3\mathbf{k}\]

Step 4: Determine the magnitude of the cross product, \(|\vec{b_1} \times \vec{b_2}|\).

\[|\vec{b_1} \times \vec{b_2}| = \sqrt{(-6)^2 + (-15)^2 + (3)^2}\]\[= \sqrt{36 + 225 + 9} = \sqrt{270} = \sqrt{9 \times 30} = 3\sqrt{30}\]

Step 5: Compute the scalar triple product, \((\vec{a_2} - \vec{a_1}) \cdot (\vec{b_1} \times \vec{b_2})\).

\[((-2 - \lambda)\mathbf{i} - 7\mathbf{j} + 3\mathbf{k}) \cdot (-6\mathbf{i} - 15\mathbf{j} + 3\mathbf{k})\]\[= (-2 - \lambda)(-6) + (-7)(-15) + (3)(3)\]\[= 12 + 6\lambda + 105 + 9 = 126 + 6\lambda\]

Step 6: Substitute these values into the shortest distance formula and solve for \(\lambda\).

The given shortest distance is \(\frac{44}{\sqrt{30}}\).

\[\frac{44}{\sqrt{30}} = \left| \frac{126 + 6\lambda}{3\sqrt{30}} \right|\]

Simplifying the equation by multiplying both sides by \(3\sqrt{30}\) and canceling \(\sqrt{30}\) from both sides:

\[44 \times 3 = |126 + 6\lambda|\]\[132 = |126 + 6\lambda|\]

Alternatively, simplifying the fraction first:

\[\frac{44}{\sqrt{30}} = \left| \frac{126 + 6\lambda}{3\sqrt{30}} \right|\]

Cancel \(\sqrt{30}\) from both sides:

\[44 = \left| \frac{126 + 6\lambda}{3} \right|\]\[44 = |42 + 2\lambda|\]

This absolute value equation yields two distinct cases:

Case 1: \(42 + 2\lambda = 44\)

\[2\lambda = 44 - 42\]\[2\lambda = 2 \implies \lambda = 1\]

Case 2: \(42 + 2\lambda = -44\)

\[2\lambda = -44 - 42\]\[2\lambda = -86 \implies \lambda = -43\]

Step 7: Identify the largest possible value of \(|\lambda|\).

The possible values for \(\lambda\) are 1 and -43. Their absolute values are calculated as:

\[|1| = 1\]\[|-43| = 43\]

Comparing these absolute values, the larger one is 43.

Thus, the largest possible value of \(|\lambda|\) is 43.