For any vector \(\vec{a}\), 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) → \(2|\vec{a}|^2\)
Let \(\vec{a} = x\hat{\imath}+y\hat{\jmath}+z\hat{k}\). Using \(\hat{\jmath}\times\hat{\imath} = -\hat{k}\), \(\hat{k}\times\hat{\imath} = \hat{\jmath}\) and \(\hat{\imath}\times\hat{\imath} = \vec{0}\):
\(\vec{a}\times\hat{\imath} = z\hat{\jmath}-y\hat{k}\)
\(|\vec{a}\times\hat{\imath}|^2 = y^2+z^2\)
In the same way, crossing with \(\hat{\jmath}\) drops the y-component and crossing with \(\hat{k}\) drops the z-component:
\(|\vec{a}\times\hat{\jmath}|^2 = z^2+x^2\)
\(|\vec{a}\times\hat{k}|^2 = x^2+y^2\)
Adding, each square appears twice:
\(2(x^2+y^2+z^2) = 2|\vec{a}|^2\)