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JEE Main · 2024 · 30 Jan 2024 · Shift 2 · Vector Algebra

Let

a = \hat{i} + α \hat{j} + β \hat{k}, α, β \in R

. Let a vector

b

be such that the angle between

a

and

b

is

\frac{π}{4}

and

∣ b ∣^{2} = 6

, If

a \cdot b = 32

, then the value of

(α^{2} + β^{2}) ∣ a \times b ∣^{2}

is equal to

(A)90✓ Correct
(B)75
(C)95
(D)85

Answer: (A)

Solution

a \cdot b = ∣ a ∣∣ b ∣ cos \frac{π}{4} = 32

. With

∣ b ∣^{2} = 6

,

∣ a ∣ \cdot 6 \cdot \frac{1}{2} = 32 \Rightarrow ∣ a ∣ 3 = 32 \Rightarrow ∣ a ∣ = 6

.

So

∣ a ∣^{2} = 1 + α^{2} + β^{2} = 6 \Rightarrow α^{2} + β^{2} = 5

.

∣ a \times b ∣^{2} = ∣ a ∣^{2} ∣ b ∣^{2} sin^{2} \frac{π}{4} = 6 \cdot 6 \cdot \frac{1}{2} = 18

.

(α^{2} + β^{2}) ∣ a \times b ∣^{2} = 5 \times 18 = 90

.

JEE Main · 2024 · 30 Jan 2024 · Shift 2 · Single correct option

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