If \(|\vec{a}| = a\), then the value of \(|\vec{a} \times \hat{\imath}|^2 + |\vec{a} \times \hat{\jmath}|^2 + |\vec{a} \times \hat{k}|^2\) is
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \(2a^2\)
Let \(\vec{a} = a_1\hat{\imath} + a_2\hat{\jmath} + a_3\hat{k}\), so that \(a_1^2 + a_2^2 + a_3^2 = a^2\).
For any vectors, Lagrange’s identity gives \(|\vec{a} \times \vec{b}|^2 = |\vec{a}|^2|\vec{b}|^2 - (\vec{a} \cdot \vec{b})^2\). With the unit vectors:
\(|\vec{a} \times \hat{\imath}|^2 = a^2 - (\vec{a} \cdot \hat{\imath})^2 = a^2 - a_1^2\)
\(|\vec{a} \times \hat{\jmath}|^2 = a^2 - a_2^2\)
\(|\vec{a} \times \hat{k}|^2 = a^2 - a_3^2\)
Adding:
\(\text{Sum} = 3a^2 - (a_1^2 + a_2^2 + a_3^2)\)
\(= 3a^2 - a^2 = 2a^2\)
Tip: This is a result worth remembering: \(|\vec{a} \times \hat{\imath}|^2 + |\vec{a} \times \hat{\jmath}|^2 + |\vec{a} \times \hat{k}|^2 = 2|\vec{a}|^2\), while the matching sum of squared dot products is \(|\vec{a}|^2\).