Let two non-collinear unit vectors $\hat{a}$ and $\hat{b}$ form an acute angle. A point P moves, so that at…
- $\hat{\mathrm{u}}=\frac{\hat{\mathrm{a}}+\hat{\mathrm{b}}}{|\hat{\mathrm{a}}+\hat{\mathrm{b}}|}$ and $\mathrm{M}=(1+\hat{\mathrm{a}} \cdot \hat{\mathrm{b}})^{\frac{1}{2}}$
- $\hat{\mathrm{u}}=\frac{\hat{\mathrm{a}}-\hat{\mathrm{b}}}{|\hat{\mathrm{a}}-\hat{\mathrm{b}}|}$ and $\mathrm{M}=(1+\hat{\mathrm{a}} \cdot \hat{\mathrm{b}})^{\frac{1}{2}}$
- $\quad \dot{\hat{u}}=\frac{\hat{a}+\hat{b}}{|\hat{a}+\hat{b}|}$ and $M=(1+2 \hat{a} \cdot \hat{b})^{\frac{1}{2}}$
- $\quad \hat{\mathrm{u}}=\frac{\hat{\mathrm{a}}-\hat{\mathrm{b}}}{|\hat{\mathrm{a}}-\hat{\mathrm{b}}|}$ and $\mathrm{M}=(1-2 \hat{\mathrm{a}} \cdot \hat{\mathrm{b}})^{\frac{1}{2}}$
Solution
Maximum value of $\sin 2 t=1$ $\begin{array}{ll} \therefore & 2 t=\sin ^{-1}(1) \\ \therefore & t=\frac{\pi}{4} \end{array}$ $\therefore \quad M=\sqrt{1+\hat{a} \cdot \hat{b}(1)}=(1+\hat{a} \cdot \hat{b})^{\frac{1}{2}}$
Now, $\hat{\mathrm{u}}=\frac{\overline{\mathrm{OP}}}{|\overrightarrow{\mathrm{OP}}|}$ $=\frac{\hat{a} \cos t+\hat{b} \sin t}{|\hat{a} \sin t+\hat{b} \sin t|}=\frac{\hat{a}\left(\frac{1}{\sqrt{2}}\right)+\hat{b}\left(\frac{1}{\sqrt{2}}\right)}{\left|\hat{a}\left(\frac{1}{\sqrt{2}}\right)+\hat{b}\left(\frac{1}{\sqrt{2}}\right)\right|}$
Unit vector of OP is $\hat{u}=\frac{\hat{a}+\hat{b}}{|(\hat{a}+\hat{b})|}$
Asked in: MHT CET 2024 (10 May Shift 2)