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JEE Main · 2023 · 30 Jan 2023 · Shift 2 · Vector Algebra

Let

λ \in R, a = λ \hat{i} + 2 \hat{j} - 3 \hat{k}, b = \hat{i} - λ \hat{j} + 2 \hat{k}

.

If

((a + b) \times (a \times b)) \times (a - b) = 8 \hat{i} - 40 \hat{j} - 24 \hat{k}

, then

∣ λ (a + b) \times (a - b) ∣^{2}

is equal to

(A)140✓ Correct
(B)132
(C)144
(D)136

Answer: (A)

Solution

By the identity

(X \times Y) \times Z = (X \cdot Z) Y - (Y \cdot Z) X

with

X = a + b, Y = a \times b, Z = a - b

, and since

(a \times b) \cdot (a - b) = 0

, the left side reduces to

((a + b) \cdot (a - b)) (a \times b) = (∣ a ∣^{2} - ∣ b ∣^{2}) (a \times b)

. Here

∣ a ∣^{2} - ∣ b ∣^{2} = (λ^{2} + 13) - (λ^{2} + 5) = 8

, so

8 (a \times b) = 8 \hat{i} - 40 \hat{j} - 24 \hat{k}

and

a \times b = \hat{i} - 5 \hat{j} - 3 \hat{k}

. Computing the cross product directly, its

\hat{i}

-component

4 - 3 λ = 1

gives

λ = 1

. With

λ = 1

,

a + b = 2 \hat{i} + \hat{j} - \hat{k}

and

a - b = 3 \hat{j} - 5 \hat{k}

, so

(a + b) \times (a - b) = - 2 \hat{i} + 10 \hat{j} + 6 \hat{k}

and

λ (a + b) \times (a - b)^{2} = 4 + 100 + 36 = 140

.

JEE Main · 2023 · 30 Jan 2023 · Shift 2 · Single correct option

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