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JEE Main · 2024 · 04 Apr 2024 · Shift 2 · Vector Algebra

Let

a = \hat{i} + \hat{j} + \hat{k}

,

b = 2 \hat{i} + 4 \hat{j} - 5 \hat{k}

and

c = x \hat{i} + 2 \hat{j} + 3 \hat{k}

,

x \in R

. If

d

is the unit vector in the direction of

b + c

such that

a \cdot d = 1

, then

(a \times b) \cdot c

is equal to

(A)9
(B)11✓ Correct
(C)3
(D)6

Answer: (B)

Solution

b + c = (2 + x) \hat{i} + 6 \hat{j} - 2 \hat{k}

, and

d = \frac{b + c}{∣ b + c ∣}

.

a \cdot (b + c) = (2 + x) + 6 - 2 = x + 6

, so

a \cdot d = 1

gives

x + 6 = ∣ b + c ∣ = (x + 2)^{2} + 36 + 4 = (x + 2)^{2} + 40

Squaring:

x^{2} + 12 x + 36 = x^{2} + 4 x + 44 \Rightarrow 8 x = 8 \Rightarrow x = 1

.

With

x = 1

:

c = \hat{i} + 2 \hat{j} + 3 \hat{k}

, and

a \times b = \hat{i} 12 \hat{j} 14 \hat{k} 1 - 5 = (- 5 - 4) \hat{i} - (- 5 - 2) \hat{j} + (4 - 2) \hat{k} = - 9 \hat{i} + 7 \hat{j} + 2 \hat{k}

(a \times b) \cdot c = - 9 (1) + 7 (2) + 2 (3) = - 9 + 14 + 6 = 11

JEE Main · 2024 · 04 Apr 2024 · Shift 2 · Single correct option

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