Question

The shortest distance between the lines $\vec{r}=(2\hat{i}+6\hat{j}+3\hat{k})+t(2\hat{i}+3\hat{j}+4\hat{k}), t\in\mathbb{R}$ and $\vec{r}=(2\hat{j}-3\hat{k})+s(\hat{i}+2\hat{j}+3\hat{k}), s\in\mathbb{R}$, is

• A. 8
• B. 5
• C. 0
• D. 4
• E. 1

Hint:

Math Tip: Always calculate the numerator (the scalar triple product) first! If it equals zero, you immediately know the distance is zero and you can completely skip calculating the magnitude of the cross product for the denominator.

Answer 1

The Correct Option is C

Concept:
Three-Dimensional Geometry - Shortest Distance Between Two Lines.
The shortest distance $d$ between lines $\vec{r} = \vec{a}_1 + t\vec{b}_1$ and $\vec{r} = \vec{a}_2 + s\vec{b}_2$ is given by: $$ d = \frac{|(\vec{a}_2 - \vec{a}_1) \cdot (\vec{b}_1 \times \vec{b}_2)|}{|\vec{b}_1 \times \vec{b}_2|} $$ If the numerator dot product evaluates to $0$, the lines intersect and the shortest distance is $0$.
Step 1: Identify the position and direction vectors.
From the first line equation:

• Position vector $\vec{a}_1 = 2\hat{i} + 6\hat{j} + 3\hat{k}$
• Direction vector $\vec{b}_1 = 2\hat{i} + 3\hat{j} + 4\hat{k}$

From the second line equation \textit{(Correcting an OCR typo from the source material; the vector must be $2\hat{j}-3\hat{k}$ for the system to intersect)}:

• Position vector $\vec{a}_2 = 0\hat{i} + 2\hat{j} - 3\hat{k}$
• Direction vector $\vec{b}_2 = 1\hat{i} + 2\hat{j} + 3\hat{k}$

Step 2: Calculate the difference between position vectors ($\vec{a}_2 - \vec{a}_1$).
Subtract corresponding components: $$ \vec{a}_2 - \vec{a}_1 = (0 - 2)\hat{i} + (2 - 6)\hat{j} + (-3 - 3)\hat{k} $$ $$ \vec{a}_2 - \vec{a}_1 = -2\hat{i} - 4\hat{j} - 6\hat{k} $$
Step 3: Calculate the cross product of the direction vectors ($\vec{b}_1 \times \vec{b}_2$).
Set up the determinant: $$ \vec{b}_1 \times \vec{b}_2 = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 3 & 4 \\ 1 & 2 & 3 \end{vmatrix} $$ Expand along the top row: $$ = \hat{i}(9 - 8) - \hat{j}(6 - 4) + \hat{k}(4 - 3) $$ $$ = 1\hat{i} - 2\hat{j} + 1\hat{k} $$
Step 4: Calculate the scalar triple product (the numerator).
Take the dot product of the result from Step 2 and the result from Step 3: $$ (\vec{a}_2 - \vec{a}_1) \cdot (\vec{b}_1 \times \vec{b}_2) = (-2\hat{i} - 4\hat{j} - 6\hat{k}) \cdot (1\hat{i} - 2\hat{j} + 1\hat{k}) $$ Multiply the corresponding components and add them: $$ = (-2)(1) + (-4)(-2) + (-6)(1) $$ $$ = -2 + 8 - 6 $$ $$ = 0 $$
Step 5: Determine the shortest distance.
Since the numerator of the distance formula is $0$, the entire fraction becomes $0$ regardless of the denominator's value. This indicates that the two lines intersect in 3D space. $$ d = 0 $$