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JEE Main · 2024 · 01 Feb 2024 · Shift 1 · Vector Algebra

Let

a = - 5 \hat{i} + \hat{j} - 3 \hat{k}, b = \hat{i} + 2 \hat{j} - 4 \hat{k}

and

c = ((((a \times b) \times \hat{i}) \times \hat{i}) \times \hat{i})

. Then

c \cdot (- \hat{i} + \hat{j} + \hat{k})

is equal to

(A)-12✓ Correct
(B)-10
(C)-13
(D)-15

Answer: (A)

Solution

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

,

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

.

a \times b = \hat{i} - 5 1 \hat{j} 12 \hat{k} - 3 - 4 = \hat{i} (- 4 + 6) - \hat{j} (20 + 3) + \hat{k} (- 10 - 1) = 2 \hat{i} - 23 \hat{j} - 11 \hat{k}

For any vector

v = (v_{1}, v_{2}, v_{3})

,

v \times \hat{i} = (0, v_{3}, - v_{2})

(the

x

-component vanishes and

(v_{2}, v_{3}) \to (v_{3}, - v_{2})

). Applying this repeatedly to

u = a \times b = (2, - 23, - 11)

:

u \times \hat{i} = (0, - 11, 23)

(u \times \hat{i}) \times \hat{i} = (0, 23, 11)

c = ((u \times \hat{i}) \times \hat{i}) \times \hat{i} = (0, 11, - 23) = 11 \hat{j} - 23 \hat{k}

c \cdot (- \hat{i} + \hat{j} + \hat{k}) = 0 (- 1) + 11 (1) + (- 23) (1) = 11 - 23 = - 12

JEE Main · 2024 · 01 Feb 2024 · Shift 1 · Single correct option

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