JEE Advanced 2020 · Paper 1 · Mathematics Q11

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MathematicsMultiple correct+4 / -2Question 47 of 54

Let $L_1$ and $L_2$ be the following straight lines.

$L_1: \frac{x-1}{1} = \frac{y}{-1} = \frac{z-1}{3}\quad\text{and}\quad L_2: \frac{x-1}{-3} = \frac{y}{-1} = \frac{z-1}{1}.$

Suppose the straight line

$L: \frac{x-\alpha}{l} = \frac{y-1}{m} = \frac{z-\gamma}{-2}$

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?

\alpha - \gamma = 3

l + m = 2

\alpha - \gamma = 1

l + m = 0

Answer and worked solution

Official answer: A, B

Intersection. Both lines pass through $(1,0,1)$ (parameter 0).

Bisector direction. The directions are $\vec d_1=(1,-1,3)$ and $\vec d_2=(-3,-1,1)$ , both of length $\sqrt{11}$ . Since $\vec d_1\cdot\vec d_2=1>0$ , the angle between them is acute, and its bisector is
$\vec d_1+\vec d_2=(-2,-2,4)\parallel(1,1,-2).$
This lies in the plane of the two lines.

Matching $L$ . $L$ has direction $(l,m,-2)$ , so $l=m=1$ and $l+m=2$ .

Through the intersection point.
$\frac{1-\alpha}1=\frac{0-1}1=\frac{1-\gamma}{-2}\implies\alpha=2,\ \gamma=-1.$
So $\alpha-\gamma=3$ .

(A), (B)

Topics:Equation of a line in space Angles and distances

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JEE Advanced 2020 · Paper 1, Mathematics question 11