The vertices B and C of a triangle ABC lie on the line, (x+2)/3 = (y-1)/0 = z/4 such that BC = 5 units. Then, the area (in sq units) of this triangle, given that the point A(1, -1, 2) is

The vertices B and C of a triangle ABC lie on the line,… — JEE Main 2019 Mathematics PYQ

JEE Main 2019 · 9 April · Shift II

The vertices

B

and

C

of a

\triangle ABC

lie on the line,

\frac{x+2}{3} = \frac{y-1}{0} = \frac{z}{4}

such that

BC = 5

units. Then, the area (in sq units) of this triangle, given that the point

A(1, -1, 2)

A.
$\sqrt{34}$
B.
$2\sqrt{34}$
C.
$5\sqrt{17}$
D.
$6$

Show answer and solution

Correct option: A

Solution

The line has direction vector (3, 0, 4), unit vector (3/5, 0, 4/5).

Find the foot of perpendicular from A(1, -1, 2) to the line. A point on the line: (-2+3t, 1, 4t).

Vector from A to point: (-3+3t, 2, 4t-2).

Dot with direction (3,0,4) = 0: 3(-3+3t) + 0 + 4(4t-2) = 0 → -9+9t+16t-8 = 0 → 25t = 17 → t = 17/25.

Foot F: (-2 + 51/25, 1, 68/25) = (-49/25 + 50/25... let me recompute: -2 + 3(17/25) = -2 + 51/25 = (-50+51)/25 = 1/25, y=1, z=68/25).

Distance AF: vector (1/25 - 1, 1-(-1), 68/25 - 2) = (-24/25, 2, 18/25).

|AF|² = (24/25)² + 4 + (18/25)² = (576 + 324)/625 + 4 = 900/625 + 4 = 36/25 + 100/25 = 136/25.

Height h = |AF| = √(136/25) = (2√34)/5.

Area = (1/2) × base × height = (1/2) ×

[JEE Main 2019 · 9 April · Shift II]

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