CUET Preparation Today
| If \(\vec{a}\) is any vector, then \(|\vec{a}\times \hat{i}|^2+|\vec{a}\times \hat{j}|^2+|\vec{a}\times \hat{k}|^2\) is equal to |
\(0\) \(a\) \(a^2\) \(2a^2\) |
| \(2a^2\) |
| Note that if \(\vec{a}=x\hat{i}+y\hat{j}+x\hat{k}\) then \(|\vec{a}\times \hat{i}|^{2}=z^{2}+y^{2},|\vec{a}\times \hat{j}|^{2}=x^{2}+z^{2},|\vec{a}\times \hat{k}|^{2}=x^{2}+y^{2}\hspace{3cm}\) \(|\vec{a}\times \hat{i}|^{2}+|\vec{a}\times \hat{j}|^{2}+|\vec{a}\times \hat{k}|^{2}=2(x^{2}+y^{2}+z^{2})=2a^{2}\) |
