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 equal to:
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → \(2|\vec a|^2\)
Let \(\vec a = a_1\hat{\imath} + a_2\hat{\jmath} + a_3\hat{k}\). Using the cross products of unit vectors (\(\hat{\imath}\times\hat{\jmath} = \hat{k}\), \(\hat{\jmath}\times\hat{k} = \hat{\imath}\), \(\hat{k}\times\hat{\imath} = \hat{\jmath}\), and the cross product of a vector with itself is \(\vec 0\)):
\(\vec a\times\hat{\imath} = a_3\hat{\jmath} - a_2\hat{k}\)
\(|\vec a\times\hat{\imath}|^2 = a_2^2 + a_3^2\)
\(\vec a\times\hat{\jmath} = a_1\hat{k} - a_3\hat{\imath}\)
\(|\vec a\times\hat{\jmath}|^2 = a_1^2 + a_3^2\)
\(\vec a\times\hat{k} = a_2\hat{\imath} - a_1\hat{\jmath}\)
\(|\vec a\times\hat{k}|^2 = a_1^2 + a_2^2\)
Adding:
\(2(a_1^2 + a_2^2 + a_3^2) = 2|\vec a|^2\)
Shortcut: By Lagrange’s identity, \(|\vec a\times\hat{\imath}|^2 = |\vec a|^2 - (\vec a\cdot\hat{\imath})^2 = |\vec a|^2 - a_1^2\), and the same for \(\hat{\jmath}\) and \(\hat{k}\). The sum is \(3|\vec a|^2 - |\vec a|^2 = 2|\vec a|^2\).
NCERT Exemplar:
Vector Algebra
