If \( \vec{a} \) is any vector, then 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) → \( 2|\vec{a}|^2 \)
By Lagrange's identity with \( |\hat{\imath}|=1 \), \( |\vec{a}\times\hat{\imath}|^2=|\vec{a}|^2-(\vec{a}\cdot\hat{\imath})^2 \), and similarly for \( \hat{\jmath} \) and \( \hat{k} \).
Let \( \vec{a}=a_1\hat{\imath}+a_2\hat{\jmath}+a_3\hat{k} \), so \( \vec{a}\cdot\hat{\imath}=a_1 \), \( \vec{a}\cdot\hat{\jmath}=a_2 \), \( \vec{a}\cdot\hat{k}=a_3 \).
\( \displaystyle \text{Sum}=3|\vec{a}|^2-(a_1^2+a_2^2+a_3^2) \)
\( \displaystyle =3|\vec{a}|^2-|\vec{a}|^2 \)
\( \displaystyle =2|\vec{a}|^2 \)
Shortcut: Put \( \vec{a}=\hat{\imath} \): the sum is \( 0+|\hat{k}|^2+|-\hat{\jmath}|^2=2 \), while \( |\vec{a}|^2=1 \). So the sum is \( 2|\vec{a}|^2 \).
Tip: \( |\vec{a}\times\hat{\imath}| \) is not \( |\vec{a}| \): the cross product removes the component of \( \vec{a} \) along \( \hat{\imath} \), which is why the total is \( 2|\vec{a}|^2 \) and not \( 3|\vec{a}|^2 \).
NCERT Exemplar:
Vector Algebra
