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 problem asks for the largest possible value of \(|\lambda|\) given the equations of two lines in 3D space and the shortest distance between them. The shortest distance between two skew lines is given by a standard formula involving vector operations.

Concept Used:

The shortest distance (SD) between two skew lines, \(L_1: \vec{r} = \vec{a_1} + t\vec{b_1}\) and \(L_2: \vec{r} = \vec{a_2} + s\vec{b_2}\), is given by 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 calculates the length of the projection of the vector connecting a point on each line (\(\vec{a_2} - \vec{a_1}\)) onto the vector normal to both lines (\(\vec{b_1} \times \vec{b_2}\)).

Step-by-Step Solution:

Step 1: Identify 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 is \((\lambda, 2, 1)\), so the position vector is \(\vec{a_1} = \lambda\mathbf{i} + 2\mathbf{j} + \mathbf{k}\).
• 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 is \((-2, -5, 4)\), so the position vector is \(\vec{a_2} = -2\mathbf{i} - 5\mathbf{j} + 4\mathbf{k}\).
• The direction vector is \(\vec{b_2} = -3\mathbf{i} + 2\mathbf{j} + 4\mathbf{k}\).

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

\[ \vec{a_2} - \vec{a_1} = (-2 - \lambda)\mathbf{i} + (-5 - 2)\mathbf{j} + (4 - 1)\mathbf{k} \] \[ \vec{a_2} - \vec{a_1} = (-2 - \lambda)\mathbf{i} - 7\mathbf{j} + 3\mathbf{k} \]

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: Calculate 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: Calculate 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 results into the shortest distance formula and solve for \(\lambda\).

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

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

We can cancel \(\sqrt{30}\) from both sides:

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

This absolute value equation leads to two possible cases:

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

\[ 2\lambda = 2 \implies \lambda = 1 \]

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

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

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

The two possible values for \(\lambda\) are 1 and -43. We find their absolute values:

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

The largest of these two values is 43.

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

Answer 2

Let
\[\vec{a}_1 = \lambda \hat{i} + 2 \hat{j} + \hat{k}\]
\[\vec{a}_2 = -2 \hat{i} - 5 \hat{j} + 4 \hat{k}\]
\[\vec{p} = 3 \hat{i} - \hat{j} + \hat{k}\]
\[\vec{q} = 3 \hat{i} + 2 \hat{j} + 4 \hat{k}\]
Now, the vector \( \vec{a}_1 - \vec{a}_2 \) is given by:
\[(\lambda + 2) \hat{i} + 7 \hat{j} - 3 \hat{k}\]
Calculate \( \vec{p} \times \vec{q} \):
\[\vec{p} \times \vec{q} = \begin{vmatrix}\hat{i} & \hat{j} & \hat{k} \\3 & -1 & 1 \\3 & 2 & 4\end{vmatrix}= -6 \hat{i} - 15 \hat{j} + 3 \hat{k}\]
The shortest distance \( d \) is given by:
\[d = \frac{|(\vec{a}_1 - \vec{a}_2) \cdot (\vec{p} \times \vec{q})|}{|\vec{p} \times \vec{q}|}\]
Substitute the values:
\[d = \frac{| -6\lambda - 12 - 105 + 9|}{\sqrt{(-6)^2 + (-15)^2 + 3^2}} = \frac{|6\lambda + 126|}{3 \sqrt{30}}\]
Equating, we get:
\[\frac{44}{\sqrt{30}} = \frac{|6\lambda + 126|}{3 \sqrt{30}}\]
\[132 = |6\lambda + 126|\]
Solving for \( \lambda \):
\[\lambda = 1 \quad \text{or} \quad \lambda = -43\]
Thus, the largest possible value of \( |\lambda| \) is:
\[|\lambda| = 43\]