Let the image of the point ( $1,0,7$ ) in the line $\frac{x}{1}=\frac{y-1}{2}=\frac{z-2}{3}$ be the point $(\alpha, \beta, \gamma)$. Then which one of the following points lies on the line passing through ( $\alpha, \beta, \gamma$ ) and making angles $\frac{2 \pi}{3}$ and $\frac{3 \pi}{4}$ with $y$-axis and $z$-axis respectively and an acute angle with x -axis ?

JEE Main · 2024 · 27 Jan 2024 · Shift 2 · Three Dimensional Geometry

Let the image of the point (

1, 0, 7

) in the line

\frac{x}{1} = \frac{y - 1}{2} = \frac{z - 2}{3}

be the point

(α, β, γ)

. Then which one of the following points lies on the line passing through (

α, β, γ

) and making angles

\frac{2 π}{3}

and

\frac{3 π}{4}

with

y

-axis and

z

-axis respectively and an acute angle with x -axis ?

Answer: (C)

Line

L : \frac{x}{1} = \frac{y - 1}{2} = \frac{z - 2}{3} = λ

; a general point on it is

M (λ, 1 + 2 λ, 2 + 3 λ)

.

For

M

to be the foot of the perpendicular from

P (1, 0, 7)

P M

must be perpendicular to the direction

(1, 2, 3)

P M = (λ - 1, 1 + 2 λ, 3 λ - 5),

(λ - 1) (1) + (1 + 2 λ) (2) + (3 λ - 5) (3) = 0 \Rightarrow 14 λ - 14 = 0 \Rightarrow λ = 1.

M = (1, 3, 5)

, and the image

Q = 2 M - P = (2 - 1, 6 - 0, 10 - 7) = (1, 6, 3)

. Thus

(α, β, γ) = (1, 6, 3)

.

The required line through

(1, 6, 3)

makes angle

\frac{2 π}{3}

with the y-axis and

\frac{3 π}{4}

with the z-axis, so its direction cosines satisfy

m = cos \frac{2 π}{3} = - \frac{1}{2}, n = cos \frac{3 π}{4} = - \frac{1}{2} .

l^{2} = 1 - m^{2} - n^{2} = 1 - \frac{1}{4} - \frac{1}{2} = \frac{1}{4} \Rightarrow l = \frac{1}{2} (taking l > 0 for an acute angle with the x-axis) .

The line is

r = (1, 6, 3) + t (\frac{1}{2}, - \frac{1}{2}, - \frac{1}{2})

. Taking

t = 4

(1 + 2, 6 - 2, 3 - 22) = (3, 4, 3 - 22),

which lies on this line.

JEE Main · 2024 · 27 Jan 2024 · Shift 2 · Single correct option