For any vector {p}, the value of ≤ft[2≤ft\{ | {p} × {} | ²+ | {p} × {} | ²+ | {p} × {k} | ²\}] is

Correct Answer is : 4∣p⃗∣24 | {p} | ²4∣p∣2

Correct Answer is : 4∣p⃗∣24 | {p} | ²4∣p∣2

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COMEDK 2025 Evening Solved Paper

Question : 167 of 180

Marks: +1, -0

For any vector

\vec{p}

\left[2\left\{ | \vec{p} \times \hat{\imath} | ^2+ | \vec{p} \times \hat{\jmath} | ^2+ | \vec{p} \times \hat{k} | ^2\right\}\right]

Solution:

Concept:

Express the cross product magnitudes in component form to simplify the expression.

Explanation:

Let

\vec{p} = (p_x, p_y, p_z)

Compute cross product with

\hat{\imath}

\vec{p} \times \hat{\imath} = (0, p_z, -p_y)

Its squared magnitude:

|\vec{p} \times \hat{\imath}|^2 = p_y^2 + p_z^2

Similarly,

\vec{p} \times \hat{\jmath} = (-p_z, 0, p_x)

, so

|\vec{p} \times \hat{\jmath}|^2 = p_x^2 + p_z^2

And

\vec{p} \times \hat{k} = (p_y, -p_x, 0)

, so

|\vec{p} \times \hat{k}|^2 = p_x^2 + p_y^2

Add them: sum =

(p_y^2+p_z^2) + (p_x^2+p_z^2) + (p_x^2+p_y^2) = 2(p_x^2+p_y^2+p_z^2)

Thus

|\vec{p} \times \hat{\imath}|^2 + |\vec{p} \times \hat{\jmath}|^2 + |\vec{p} \times \hat{k}|^2 = 2|\vec{p}|^2

Now multiply by the outer factor

2

2\left\{2|\vec{p}|^2\right\} = 4|\vec{p}|^2

Answer:

4|\vec{p}|^2

, which corresponds to option A.

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