Let \(\alpha \in \mathbb{R}\) and the three vectors \(\vec{a}=\alpha \hat{i}+\hat{j}+3 \hat{k}\), \(\vec{b}=2 \hat{i}+\hat{j}-\alpha \hat{k}\) and \(\vec{c}=\alpha \hat{i}-2 \hat{j}+3 \hat{k}\). Then the set \(S=\{\alpha: \vec{a}, \vec{b}\text{ and }\vec{c}\text{ are coplanar}\}\)

(A) Contains exactly two numbers only one of which is positive

(C) Contains exactly two positive numbers

JEE Main · 2019 · 12 Apr 2019 · Shift 2 · Vector Algebra

Let

α \in R

and the three vectors

a = α \hat{i} + \hat{j} + 3 \hat{k}

b = 2 \hat{i} + \hat{j} - α \hat{k}

and

c = α \hat{i} - 2 \hat{j} + 3 \hat{k}

. Then the set

S = {α : a, b and c are coplanar}

Answer: (B)

Coplanarity requires the scalar triple product to vanish, so

α 2 α 11 - 2 3 - α 3 = 0

. Expanding gives

α (3 - 2 α) - 1 (6 + α^{2}) + 3 (- 4 - α) = - 3 α^{2} - 18

. Setting

- 3 α^{2} - 18 = 0

forces

α^{2} = - 6

, which no real

α

satisfies, so the set is empty.

JEE Main · 2019 · 12 Apr 2019 · Shift 2 · Single correct option