Question

The lines $\vec{r} = (\hat{i} + \hat{j} - \hat{k}) + \lambda(3\hat{i} - \hat{j})$ and $\vec{r} = (4\hat{i} - \hat{k}) + \mu(2\hat{i} + 3\hat{k})$ are \dots}

• A. intersecting but not perpendicular
• B. perpendicular
• C. parallel
• D. skew lines

Hint:

To quickly test if lines intersect, equate the simplest coordinates first (like $y$ and $z$ here, which were trivial to solve) and use the remaining coordinate strictly as a "verification check" for the parameters.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
Two lines are skew if they are not parallel and do not intersect. Parallelism is checked via direction vectors; intersection is checked by the Shortest Distance (S.D.) formula.

Step 2: Formula Application:
$L_1: \vec{a}_1 = \hat{i}+\hat{j}-\hat{k}, \vec{b}_1 = 3\hat{i}-\hat{j}$ $L_2: \vec{a}_2 = 4\hat{i}-\hat{k}, \vec{b}_2 = 2\hat{i}+3\hat{k}$ S.D. $= \frac{|(\vec{a}_2 - \vec{a}_1) \cdot (\vec{b}_1 \times \vec{b}_2)|}{|\vec{b}_1 \times \vec{b}_2|}$.

Step 3: Explanation:
$\vec{b}_1 \times \vec{b}_2 = (-3-0)\hat{i} - (9-0)\hat{j} + (0-(-2))\hat{k} = -3\hat{i} - 9\hat{j} + 2\hat{k}$. $\vec{a}_2 - \vec{a}_1 = (4-1)\hat{i} + (0-1)\hat{j} + (-1-(-1))\hat{k} = 3\hat{i} - \hat{j} + 0\hat{k}$. Dot product $= (3)(-3) + (-1)(-9) + (0)(2) = -9 + 9 + 0 = 0$. Since the S.D. numerator is 0, the lines actually intersect. Since $\vec{b}_1 \cdot \vec{b}_2 = 6 + 0 + 0 = 6 \neq 0$, they are not perpendicular.

Step 4: Final Answer:
The lines are intersecting but not perpendicular. (Correcting initial assessment to Option A).