Question

If the lines $\frac{x-1}{2}=\frac{y+1}{3}=\frac{z}{1}\text{and}\frac{x-2k}{3}=\frac{y}{1}=\frac{z-k}{2}$ intersect at the point $(\alpha ,\beta ,\gamma )$, then

• A. The unit vector perpendicular to both lines is $\pm \frac{1}{\sqrt{3}}(\hat{i}-\hat{j}+\hat{k})$
• B. The value of k is 1
• C. The value of k is 1
• D. The unit vector perpendicular to the both the lines is $\pm \frac{1}{5\sqrt{3}}(5\hat{i}-\hat{j}-7\hat{k})$

Answer 1

Answer: (B), (D)

The unit vector perpendicular to the both the lines is $\pm \frac{1}{5\sqrt{3}}(5\hat{i}-\hat{j}-7\hat{k})$

$2t+1=3\lambda +2k$ ...(i)
$3t–1=\lambda$ ...(ii)
$t=2\lambda +k$ ...(iii)
From (i), $–2\times )$ (iii)
$1=–\lambda$
$\Rightarrow \lambda =-1$
From (ii)
$t=0$
From (i)
$1=–3+2k$
$k=2$
$\Rightarrow$ Intersection point is $(1,–1,0)$
$\overrightarrow{b_1}\times \overrightarrow{b_2}=\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 2& 3& 1\\ 3& 1& 2\end{array}\right|=5\hat{i}-\hat{j}-7\hat{k}$