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JEE Main · 2020 · 04 Sep 2020 · Shift 2 · Vector Algebra

If

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

, then the value of

∣ \hat{i} \times (a \times \hat{i}) ∣^{2} + ∣ \hat{j} \times (a \times \hat{j}) ∣^{2} + ∣ \hat{k} \times (a \times \hat{k}) ∣^{2}

is equal to:

Answer: 18

Solution

For any vector

a = x \hat{i} + y \hat{j} + z \hat{k}

the triple product gives

\hat{i} \times (a \times \hat{i}) = a - (a \cdot \hat{i}) \hat{i} = y \hat{j} + z \hat{k}

, and likewise for the other two unit vectors. The sum of the squared magnitudes is

(y^{2} + z^{2}) + (x^{2} + z^{2}) + (x^{2} + y^{2}) = 2 (x^{2} + y^{2} + z^{2}) = 2∣ a ∣^{2}

. With

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

,

∣ a ∣^{2} = 9

, so the value is

2 \times 9 = 18

.

JEE Main · 2020 · 04 Sep 2020 · Shift 2 · Integer answer

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