Question

Let $L_1$ and $L_2$ be the following straight lines. $L_1:\frac{x-1}{1}=\frac{y}{1}=\frac{z-1}{3}$ and $L_2:\frac{x-1}{-3}=\frac{y}{-1}=\frac{z-1}{1}$. Suppose the straight line is $L:\frac{x-\alpha }{l}=\frac{y-1}{m}=\frac{z-\gamma }{-2}$. Lies in the plane containing $L_1$ and $L_2$, and passes through the point of intersection of $L_1$ and

$L_2$. If the line $L$ bisects the acute angle between the lines $L_1$ and $L_2$, then which of the following statements is/are TRUE?

• A. $\alpha -\gamma =3$
• B. $l+m=2$
• C. $\alpha -\gamma =1$
• D. $l+m=0$

Answer 1

Answer: (A), (B)

$l+m=2$

Explanation for the correct options.

Given: The straight lines as $L_1:\frac{x-1}{1}=\frac{y}{1}=\frac{z-1}{3}$ and $L_2:\frac{x-1}{-3}=\frac{y}{-1}=\frac{z-1}{1}$.

If the line $L:\frac{x-\alpha }{l}=\frac{y-1}{m}=\frac{z-\gamma }{-2}$ lies on the plane which contains $L_1$ and $L_2$.

Now find the direction ratio of $L_1$.

$\begin{array}{rcl}Dr\left(L_1\right)& =& \frac{\hat{i}-\hat{j}+3\hat{k}}{\sqrt{11}}+\frac{-3\hat{i}-\hat{j}+\hat{k}}{\sqrt{11}}\\ & =& \frac{-2\hat{i}-2\hat{j}+4\hat{k}}{\sqrt{11}}\end{array}$

Let $L_1$ pass through $L$, then we get

$L:\frac{x-1}{-2}=\frac{y-0}{-2}=\frac{z-1}{4}$

Let us divide it by $-2$.

$L:\frac{x-1}{1}=\frac{y-0}{1}=\frac{z-1}{-2}$

From the given condition $L:\frac{x-\alpha }{l}=\frac{y-1}{m}=\frac{z-\gamma }{-2}$ we get the values of $x$, $y$, and $z$ as

$y=1;x=2;z=-1$

Then the obtained points are $\left(2,1,-1\right)$.

Now using the points the line is obtained as,

$L:\frac{x-2}{-1}=\frac{y-1}{1}=\frac{z+1}{-2}$.

Comparing the above line with given condition $L:\frac{x-\alpha }{l}=\frac{y-1}{m}=\frac{z-\gamma }{-2}$ we get

$\alpha =2;\gamma =-1;l=1;m=1$.

Option (A)

$\begin{array}{rcl}\alpha -\gamma & =& 3\\ 2-\left(-1\right)& =& 3\\ 2+1& =& 3\\ 3& =& 3\end{array}$

Option (A) satisfied.

Option (B)

$\begin{array}{rcl}l+m& =& 2\\ 1+1& =& 2\\ 2& =& 2\end{array}$

Option (B) is also satisfied.

Option (C)

$\begin{array}{rcl}\alpha -\gamma & =& 1\\ 2-\left(-1\right)& =& 1\\ 2+1& =& 1\\ 3& \ne & 1\end{array}$

Option (C) does not satisfy.

Option (D)

$\begin{array}{rcl}l+m& =& 0\\ 1+1& =& 0\\ 2& \ne & 0\end{array}$

Option (D) does not satisfy.

Therefore, the correct statements are $\alpha -\gamma =3$ and $l+m=2$.

Hence, the correct option are (A) and (B).