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

JEE Advanced · 2020 · 27 Sep 2020 · Paper 1 · Three Dimensional Geometry

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

L : \frac{x - α}{l} = \frac{y - 1}{m} = \frac{z - γ}{- 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?

Answer: (A), (B)

The point of intersection of

L_{1}

and

L_{2}

(1, 0, 1)

.

Since L passes through

(1, 0, 1)

\frac{1 - α}{l} = \frac{- 1}{m} = \frac{1 - γ}{- 2}

.

The direction vectors of

L_{1}

and

L_{2}

are

d_{1} = \hat{i} - \hat{j} + 3 \hat{k}

and

d_{2} = - 3 \hat{i} - \hat{j} + \hat{k}

.

The acute angle bisector has direction:

\frac{d _{1}}{∣ d _{1} ∣} + \frac{d _{2}}{∣ d _{2} ∣} = \frac{i ^ - j ^ + 3 k ^}{11} + \frac{- 3 i ^ - j ^ + k ^}{11} = \frac{- 2 i ^ - 2 j ^ + 4 k ^}{11} \propto \hat{i} + \hat{j} - 2 \hat{k}

\frac{l}{1} = \frac{m}{1} = \frac{- 2}{- 2}

, giving

l = m = 1

. Thus

l + m = 2

. (B) is TRUE.

From

\frac{1 - α}{1} = - 1

, we get

α = 2

.

From

\frac{1 - γ}{- 2} = - 1

, we get

γ = - 1

.

Therefore,

α - γ = 2 - (- 1) = 3

. (A) is TRUE.

JEE Advanced · 2020 · 27 Sep 2020 · Paper 1 · One or more correct options