Question

The shortest distance between the lines $\vec{r}=\hat{i}+\hat{j}+3\hat{k}+\lambda(2\hat{i}+2\hat{j}+\hat{k})$ and $\vec{r}=(2\mu+1)\hat{i}+(2\mu-1)\hat{j}+(\mu+1)\hat{k}$ where $\lambda$ and $\mu$ are parameters, is

• A. 1
• B. 6
• C. 3
• D. 4
• E. 2

Hint:

Logic Tip: Always quickly inspect the direction vectors before diving into the skew lines formula (the one with the determinant scalar triple product). If the lines are parallel, the scalar triple product will be zero, and you must use the parallel line distance formula instead.

Answer 1

Answer: (E)

Concept:
To find the shortest distance between two lines, first check if they are parallel or skew. If the direction vectors are parallel ($\vec{b}_1 \parallel \vec{b}_2$), the shortest distance formula is: $$d = \frac{|\vec{b} \times (\vec{a}_2 - \vec{a}_1)|}{|\vec{b}|}$$ Where $\vec{b}$ is the common direction vector, and $\vec{a}_1, \vec{a}_2$ are position vectors of points on the lines.
Step 1: Extract position and direction vectors.
Line 1: $\vec{r} = (\hat{i} + \hat{j} + 3\hat{k}) + \lambda(2\hat{i} + 2\hat{j} + \hat{k})$ $\vec{a}_1 = \hat{i} + \hat{j} + 3\hat{k}$ $\vec{b}_1 = 2\hat{i} + 2\hat{j} + \hat{k}$ Line 2: Expand and rewrite to separate the parameter $\mu$: $\vec{r} = (1\hat{i} - 1\hat{j} + 1\hat{k}) + \mu(2\hat{i} + 2\hat{j} + \hat{k})$ $\vec{a}_2 = \hat{i} - \hat{j} + \hat{k}$ $\vec{b}_2 = 2\hat{i} + 2\hat{j} + \hat{k}$ Notice that $\vec{b}_1 = \vec{b}_2 = \vec{b}$. The lines are parallel!
Step 2: Calculate the vector difference $(\vec{a}_2 - \vec{a}_1)$.
$$\vec{a}_2 - \vec{a}_1 = (\hat{i} - \hat{j} + \hat{k}) - (\hat{i} + \hat{j} + 3\hat{k})$$ $$\vec{a}_2 - \vec{a}_1 = 0\hat{i} - 2\hat{j} - 2\hat{k}$$
Step 3: Compute the cross product $\vec{b} \times (\vec{a}_2 - \vec{a}_1)$.
$$ \vec{b} \times (\vec{a}_2 - \vec{a}_1) = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 2 & 1 \\ 0 & -2 & -2 \end{vmatrix} $$ $$ = \hat{i}(-4 - (-2)) - \hat{j}(-4 - 0) + \hat{k}(-4 - 0) $$ $$ = -2\hat{i} + 4\hat{j} - 4\hat{k} $$
Step 4: Calculate the required magnitudes and final distance.
Magnitude of the cross product: $$|\vec{b} \times (\vec{a}_2 - \vec{a}_1)| = \sqrt{(-2)^2 + 4^2 + (-4)^2} = \sqrt{4 + 16 + 16} = \sqrt{36} = 6$$ Magnitude of direction vector $\vec{b}$: $$|\vec{b}| = \sqrt{2^2 + 2^2 + 1^2} = \sqrt{4 + 4 + 1} = \sqrt{9} = 3$$ Calculate the distance $d$: $$d = \frac{6}{3} = 2$$