Question

Assertion: The pair of lines given by $\vec{ r }=\hat{ i }-\hat{ j }+\lambda(2 i + k )$ and $\vec{ r }=2 \hat{ i }-\hat{ k }+\mu( i +\hat{ j }- k )$ intersect.
Reason: Two lines intersect each other, if they are not parallel and shortest distance $=0$.

• A. Assertion is correct, Reason is correct; Reason is a correct explanation for assertion.
• B. Assertion is correct, Reason is correct; Reason is not a correct explanation for Assertion
• C. Assertion is correct, Reason is incorrect
• D. Assertion is incorrect, Reason is correct.

Answer 1

Answer: (A)

Here, $\vec{ a }_{1}=\hat{ i }-\hat{ j }, \vec{ b }_{1}=2 \hat{ i }+\hat{ k }$
$\vec{ a }_{2}=2 \hat{ i }-\hat{ k }, \vec{ b }_{2}=\hat{ i }+\hat{ j }-\hat{ k }$
$\because \vec{ b }_{1} \neq \lambda \vec{ b }_{2}$, for any scalar $\lambda$
$\therefore $ Given lines are not parallel.
$\vec{a}_2 - \vec{a}_1 = ( 2\hat{i} - \hat{k} )- (\hat{i} - \hat{j}) = \hat{i} + \hat{j} - \hat{k}$
$\vec{ b }_{1} \times \vec{ b }_{2} =\begin{vmatrix}
\hat{ i } & \hat{ j } & \hat{ k } \\
2 & 0 & 1 \\
1 & 1 & -1
\end{vmatrix}$
$=\hat{ i }(0-1)-\hat{ j }(-2-1)+\hat{ k }(2-0) $
$=-\hat{ i }+3 \hat{ j }+2 \hat{ k } $
$\left|\vec{ b }_{1} \times \vec{ b }_{2}\right|=\sqrt{(-1)^{2}+(3)^{2}+(2)^{2}} $
$= \sqrt{1 + 9 + 4}= \sqrt{14}$
$SD = \frac{\left(\vec{ a }_{2}- \vec{ a} _{1}\right)\cdot\left(\vec{ b }_{2}-\vec{ b }_{1}\right)}{\left|\vec{ b }_{1} \times \vec{ b }_{2}\right|}| $
$=\left|\frac{(\hat{ i }+ j -\hat{ k }) \cdot(-\hat{ i }+3 j +2 \hat{ k })}{\sqrt{14}}\right| $
$=\left|\frac{-1+3-2}{\sqrt{14}}\right|=0$
Hence, two lines intersect each other.
Two lines intersect each other,
if they are not parallel and shortest distance $=0$.