Question

Find the value of $p$ if the shortest distance between the lines \[ \vec{r} = (\hat{i} + 2\hat{j} + \hat{k}) + \lambda(\hat{i} - \hat{j} + \hat{k}) \quad \text{and} \quad \vec{r} = (p\hat{i} - \hat{j} - \hat{k}) + \mu(2\hat{i} + \hat{j} + 2\hat{k}) \] is exactly $\frac{3}{\sqrt{2}}$ units.

Hint:

An absolute value equation $|x| = c$ always splits into two cases: $x = c$ and $x = -c$. Forgetting the negative case means losing half of the valid solutions!

Answer 1

Concept: This problem requires using the shortest distance formula for skew lines in vector form to solve for an unknown coordinate parameter $p$. The formula is: \[ d = \frac{\left| (\vec{a_2} - \vec{a_1}) \cdot (\vec{b_1} \times \vec{b_2}) \right|}{\left| \vec{b_1} \times \vec{b_2} \right|} \] We extract the constant and direction vectors, find their cross and dot products as functions of $p$, and set the expression equal to the given distance value $\frac{3}{\sqrt{2}}$ to solve for $p$.

Step 1: Extracting vectors and computing the cross product $(\vec{b_1} \times \vec{b_2})$.
From the given equations, let us identify the component vectors: \[ \vec{a_1} = \hat{i} + 2\hat{j} + \hat{k}, \quad \vec{b_1} = \hat{i} - \hat{j} + \hat{k} \] \[ \vec{a_2} = p\hat{i} - \hat{j} - \hat{k}, \quad \vec{b_2} = 2\hat{i} + \hat{j} + 2\hat{k} \] The cross product of the direction vectors is calculated using the matrix determinant method: \[ \vec{b_1} \times \vec{b_2} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} 1 & -1 & 1 2 & 1 & 2 \end{vmatrix} \] Expanding along the first row: \[ \vec{b_1} \times \vec{b_2} = \hat{i}(-2 - 1) - \hat{j}(2 - 2) + \hat{k}(1 - (-2)) = -3\hat{i} + 3\hat{k} \] The magnitude of this cross product vector is: \[ \left| \vec{b_1} \times \vec{b_2} \right| = \sqrt{(-3)^2 + 0^2 + 3^2} = \sqrt{18} = 3\sqrt{2} \]

Step 2: Computing the displacement vector $(\vec{a_2} - \vec{a_1})$ as a function of $p$.
Subtract vector $\vec{a_1}$ from $\vec{a_2}$: \[ \vec{a_2} - \vec{a_1} = (p - 1)\hat{i} + (-1 - 2)\hat{j} + (-1 - 1)\hat{k} \] \[ \vec{a_2} - \vec{a_1} = (p - 1)\hat{i} - 3\hat{j} - 2\hat{k} \]

Step 3: Setting up the dot product equation for the numerator.
Compute the scalar dot product of the displacement vector and the cross product vector: \[ (\vec{a_2} - \vec{a_1}) \cdot (\vec{b_1} \times \vec{b_2}) = \left[ (p - 1)\hat{i} - 3\hat{j} - 2\hat{k} \right] \cdot \left[ -3\hat{i} + 0\hat{j} + 3\hat{k} \right] \] Multiply the corresponding components together: \[ = -3(p - 1) + (-3)(0) + (-2)(3) = -3p + 3 + 0 - 6 = -3p - 3 \]

Step 4: Substituting terms into the distance formula to solve for $p$.
Substitute the expressions into the shortest distance formula and set it equal to $\frac{3}{\sqrt{2}}$: \[ \frac{|-3p - 3|}{3\sqrt{2}} = \frac{3}{\sqrt{2}} \] Cancel out the common denominator term $\sqrt{2}$ from both sides: \[ \frac{|-3p - 3|}{3} = 3 \implies |-3p - 3| = 9 \] Factor out the constant magnitude 3: \[ 3|p + 1| = 9 \implies |p + 1| = 3 \] This absolute value equation yields two separate linear equations: \[ p + 1 = 3 \implies p = 2 \] \[ p + 1 = -3 \implies p = -4 \] Thus, the valid values for parameter $p$ are $2$ or $-4$.