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JEE Main · 2025 · 03 Apr 2025 · Shift 1 · Vector Algebra

Let

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

and

\hat{d}

be a unit vector such that

a \times \hat{d} = b \times \hat{d}

and

c \cdot \hat{d} = 1

, If

c

is perpendicular to

a

, then

∣3 λ \hat{d} + μ c ∣^{2}

is equal to .

Answer: 5

Solution

From

a \times \hat{d} = b \times \hat{d}

,

(a - b) \times \hat{d} = 0

, so

\hat{d}

is parallel to

a - b = (- 2, - 1, 2)

, whose length is

3

; thus

\hat{d} = \pm \frac{1}{3} (- 2, - 1, 2)

. Since

c = λ \hat{j} + μ \hat{k}

is perpendicular to

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

,

λ + μ = 0

, so

μ = - λ

,

c = λ (\hat{j} - \hat{k})

and

∣ c ∣^{2} = 2 λ^{2}

. The condition

c \cdot \hat{d} = 1

gives

\mp λ = 1

, so

λ^{2} = 1

. Then

∣3 λ \hat{d} + μ c ∣^{2} = 9 λ^{2} ∣ \hat{d} ∣^{2} + μ^{2} ∣ c ∣^{2} + 6 λ μ (\hat{d} \cdot c) = 9 λ^{2} + 2 λ^{4} - 6 λ^{2} = 3 λ^{2} + 2 λ^{4} .

With

λ^{2} = 1

this equals

3 + 2 = 5

.

JEE Main · 2025 · 03 Apr 2025 · Shift 1 · Integer answer

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