Let $\vec{a}, \vec{b}$ and $\vec{c}$ be three non-zero non-coplanar vectors. Let the position vectors of four points A, $B, C$ and $D$ be $\vec{a}-\vec{b}+\vec{c}, \lambda \vec{a}-3 \vec{b}+4 \vec{c}$, $-\vec{a}+2 \vec{b}-3 \vec{c}$ and $2 \vec{a}-4 \vec{b}+6 \vec{c}$ respectively. If $\overrightarrow{AB}$, $\overrightarrow{\mathrm{AC}}$ and $\overrightarrow{\mathrm{AD}}$ are coplanar, then $\lambda$ is :

JEE Main · 2023 · 29 Jan 2023 · Shift 1 · Vector Algebra

Let

a, b

and

c

be three non-zero non-coplanar vectors. Let the position vectors of four points A,

B, C

and

D

a - b + c, λ a - 3 b + 4 c

- a + 2 b - 3 c

and

2 a - 4 b + 6 c

respectively. If

A B

AC

and

AD

are coplanar, then

λ

is :

Answer: (B)

With the given position vectors,

A B = (λ - 1) a - 2 b + 3 c

A C = - 2 a + 3 b - 4 c

and

A D = a - 3 b + 5 c

. Since

a, b, c

are non-coplanar, these three are coplanar exactly when

λ - 1 - 2 1 - 2 3 - 3 3 - 4 5 = 0.

Expanding,

(λ - 1) (15 - 12) + 2 (- 10 + 4) + 3 (6 - 3) = 3 (λ - 1) - 12 + 9 = 3 λ - 6

, which vanishes when

λ = 2

JEE Main · 2023 · 29 Jan 2023 · Shift 1 · Single correct option