The sum of all values of $\alpha$, for which the points whose position vectors $\hat{i}-2 \hat{j}+3 \hat{k}$, $2 \hat{i}-3 \hat{j}+4 \hat{k}$, $(\alpha+1) \hat{i}+2 \hat{k}$ and $9 \hat{i}+(\alpha-8) \hat{j}+6 \hat{k}$ are coplanar, is equal to

JEE Main · 2023 · 06 Apr 2023 · Shift 2 · Vector Algebra

The sum of all values of

α

, for which the points whose position vectors

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

2 \hat{i} - 3 \hat{j} + 4 \hat{k}

(α + 1) \hat{i} + 2 \hat{k}

and

9 \hat{i} + (α - 8) \hat{j} + 6 \hat{k}

are coplanar, is equal to

Answer: (D)

Let the four points be

A (1, - 2, 3)

B (2, - 3, 4)

C (α + 1, 0, 2)

, and

D (9, α - 8, 6)

. For coplanarity, the vectors

A B

A C

, and

A D

must be coplanar, i.e., their scalar triple product is zero. We have

A B = (1, - 1, 1)

A C = (α, 2, - 1)

, and

A D = (8, α - 6, 3)

. The determinant is

1 α 8 - 1 2 α - 6 1 - 1 3 = 1 (6 - (- (α - 6))) - (- 1) (3 α - (- 8)) + 1 (α (α - 6) - 16) = (6 + α - 6) + (3 α + 8) + (α^{2} - 6 α - 16) = α^{2} - 2 α - 8 = 0.

Solving:

(α - 4) (α + 2) = 0

, so

α = 4

α = - 2

. The sum is

4 + (- 2) = 2

JEE Main · 2023 · 06 Apr 2023 · Shift 2 · Single correct option