CUET Preparation Today
Let \(\vec{a}\) be a unit vector and \(\vec{x}\) be a vector satisfying \(\left(\vec{x}-\vec{a}\right)\cdot \left(\vec{x}+\vec{a}\right)=12\) then \(\left|\vec{x}\right|\) is |
\(\sqrt{11}\) \(\sqrt{12}\) \(\sqrt{13}\) \(\sqrt{14}\) |
\(\sqrt{13}\) |
| \(\begin{aligned}\left(\vec{x}-\vec{a}\right)\cdot \left(\vec{x}+\vec{a}\right)&=12\\|\vec{x}|^{2}-|\vec{a}|^{2}&=12\\ |\vec{x}|^{2}&=13\\ |\vec{x}|&=\sqrt{13}\end{aligned}\) |
