Question

Assertion :If the straight-lines $\overrightarrow{r}=i+2j+3k+\lambda (ai+2j+3k)$ and $\overrightarrow{r}=2i+3j+k+\mu (3i+aj+2k)$ intersect at a point then the integer a is equal to -5. Reason: Two straight lines intersect if the shortest distance between them is zero.

• A. Both Assertion and Reason are correct and Reason is the correct explanation for Assertion
• B. Both Assertion and Reason are correct but Reason is not the correct explanation for Assertion
• C. Assertion is correct but Reason is incorrect
• D. Both Assertion and Reason are incorrect

Answer 1

The correct option is A Both Assertion and Reason are correct and Reason is the correct explanation for Assertion

$\overrightarrow{\gamma }=\hat{i}+2\hat{l}+3\hat{k}+\gamma (a\hat{i}+2\hat{j}+3\hat{k})$

cortesion form of line :$\to$

$\frac{x-1}{a}=\frac{y-2}{2}=\frac{z-3}{3}=k$......(i)

and for second line

$\overrightarrow{\gamma }=2\hat{i}+3\hat{j}+8\hat{k}+\mu (3i+aj+2k)$

$\frac{x-2}{3}=\frac{y-3}{a}=\frac{x-1}{2}$.....(ii)

for intresection point from (i)

$x=ak+1,y=2k+2,z=3k+3$

now from (ii) eqn

$\Rightarrow \frac{x-2}{3}=\frac{x-1}{2}\Rightarrow \frac{ak+1}{3}=\frac{3k+3-1}{2}$

$\Rightarrow 2ak+2=9k+6$

$\Rightarrow 2k-9k=4$.....(iii)

and $\frac{y-3}{a}=\frac{z-1}{2}$

$\Rightarrow \frac{2k+2-3}{a}=\frac{3k+3-1}{2}$

$\Rightarrow \frac{2k-1}{a}=\frac{3k+2}{2}$

$\Rightarrow 4k-2=3ak+2a$

$\Rightarrow 4k-3ak=20+2$.......(iv)

from (iii) $k=\frac{4}{2a-9}$

$\Rightarrow 4\left(\frac{4}{2a-9}\right)-3a\left(\frac{4}{2a-9}\right)=2a+2$

$\Rightarrow \frac{16-12a}{2a-9}=2a+2$

$\Rightarrow 2a^2+4a-18a-18=16-12a$

$\Rightarrow 2a^2-2a-34=0$

on solving this $a=-5$ and two line intersect if the shortest distance b/w then is zero