JEE Main · 2023 · 01 Feb 2023 · Shift 2 · Three Dimensional Geometry

The point of intersection C of the plane

8 x + y + 2 z = 0

and the line joining the points

A (- 3, - 6, 1)

and

B (2, 4, - 3)

divides the line segment AB internally in the ratio

k : 1

. If

a, b, c

(

∣ a ∣, ∣ b ∣, ∣ c ∣

are coprime) are the direction ratios of the perpendicular from the point C on the line

\frac{1 - x}{1} = \frac{y + 4}{2} = \frac{z + 2}{3}

, then

∣ a + b + c ∣

is equal to ______.

Answer: 10

Solution

Points on line

A B

are

(- 3 + 5 t, - 6 + 10 t, 1 - 4 t)

. Substituting into

8 x + y + 2 z = 0

gives

- 28 + 42 t = 0

, so

t = \frac{2}{3}

and

C = (\frac{1}{3}, \frac{2}{3}, - \frac{5}{3})

. Writing a point on the given line as

D = (1 - μ, - 4 + 2 μ, - 2 + 3 μ)

, the vector

C D = (\frac{2}{3} - μ, - \frac{14}{3} + 2 μ, - \frac{1}{3} + 3 μ)

. Requiring

C D

perpendicular to the line's direction

(- 1, 2, 3)

gives

14 μ - 11 = 0

, so

μ = \frac{11}{14}

and

C D = \frac{1}{42} (- 5, - 130, 85)

, giving direction ratios

(- 1, - 26, 17)

after removing the common factor. Therefore

∣ a + b + c ∣ = ∣ - 1 - 26 + 17∣ = 10

JEE Main · 2023 · 01 Feb 2023 · Shift 2 · Integer answer