MHT CET · Maths · Vector Algebra
Let two non-collinear vectors \(\hat{a}\) and \(\hat{b}\) form an acute angle. A point \(\mathrm{P}\) moves, so that at any time \(t\) the position vector \(\overline{\mathrm{OP}}\), where \(\mathrm{O}\) is origin, is given by \(\hat{a} \sin t+\hat{b} \cos t\), when \(P\) is farthest from origin \(\mathrm{O}\), let \(\mathrm{M}\) be the length of \(\overline{\mathrm{OP}}\) and \(\hat{\mathrm{u}}\) be the unit vector along \(\overline{\mathrm{OP}}\), then
- A \(\hat{\mathrm{u}}=\frac{\hat{\mathrm{a}}+\hat{\mathrm{b}}}{|\hat{\mathrm{a}}+\hat{\mathrm{b}}|}\) and \(\mathrm{M}=(1+\hat{\mathrm{a}} \cdot \hat{\mathrm{b}})^{\frac{1}{2}}\)
- B \(\hat{\mathrm{u}}=\frac{\hat{\mathrm{a}}-\hat{\mathrm{b}}}{|\hat{\mathrm{a}}-\hat{\mathrm{b}}|}\) and \(\mathrm{M}=(1+\hat{\mathrm{a}} \cdot \hat{\mathrm{b}})^{\frac{1}{2}}\)
- C \(\hat{\mathrm{u}}=\frac{\hat{\mathrm{a}}+\hat{\mathrm{b}}}{|\hat{\mathrm{a}}+\hat{\mathrm{b}}|}\) and \(\mathrm{M}=(1+\hat{\mathrm{2a}} \cdot \hat{\mathrm{b}})^{\frac{1}{2}}\)
- D \(\hat{\mathrm{u}}=\frac{\hat{\mathrm{a}}-\hat{\mathrm{b}}}{|\hat{\mathrm{a}}-\hat{\mathrm{b}}|}\) and \(\mathrm{M}=(1-\hat{\mathrm{2a}} \cdot \hat{\mathrm{b}})^{\frac{1}{2}}\)
Answer & Solution
Correct Answer
(A) \(\hat{\mathrm{u}}=\frac{\hat{\mathrm{a}}+\hat{\mathrm{b}}}{|\hat{\mathrm{a}}+\hat{\mathrm{b}}|}\) and \(\mathrm{M}=(1+\hat{\mathrm{a}} \cdot \hat{\mathrm{b}})^{\frac{1}{2}}\)
Step-by-step Solution
Detailed explanation
\(\begin{aligned}
& \mathrm{M}=|\overrightarrow{\mathrm{OP}}| \\
& M=\sqrt{(\hat{a} \sin t+\hat{b} \cos t)^2} \\
& =\sqrt{(\hat{a} \sin t)^2+(\hat{b} \cos t)^2+2(\hat{a} \sin t) \cdot(\hat{b} \cos t)} \\
& =\sqrt{\sin ^2 t+\cos ^2 t+\hat{a} \cdot \hat{b}(2 \sin t \cos t)} \\
& =\sqrt{1+\hat{\mathrm{a}} \cdot \hat{\mathrm{b}}(\sin 2 \mathrm{t})} \\
& \text { Maximum value of } \sin 2 t=1 \\
& \therefore \quad 2 \mathrm{t}=\sin ^{-1}(1) \\
& \therefore \quad \mathrm{t}=\frac{\pi}{4} \\
& \therefore \quad \mathrm{M}=\sqrt{1+\hat{\mathrm{a}} \cdot \hat{\mathrm{b}}(1)} \\
& =(1+\hat{a} \cdot \hat{b})^{\frac{1}{2}} \\
& \text { Now, } \hat{\mathrm{u}}=\frac{\overline{\mathrm{OP}}}{|\overline{\mathrm{OP}}|} \\
& =\frac{\hat{a} \sin t+\hat{b} \cos t}{|\hat{a} \sin t+\hat{b} \cos t|} \\
& =\frac{\hat{a}\left(\frac{1}{\sqrt{2}}\right)+\hat{b}\left(\frac{1}{\sqrt{2}}\right)}{\left|\hat{a}\left(\frac{1}{\sqrt{2}}\right)+\hat{b}\left(\frac{1}{\sqrt{2}}\right)\right|}\\
&
\end{aligned}\)
Unit vector of OP is
\(\hat{\mathrm{u}}=\frac{\hat{\mathrm{a}}+\hat{\mathrm{b}}}{|(\hat{\mathrm{a}}+\hat{\mathrm{b}})|}\)
& \mathrm{M}=|\overrightarrow{\mathrm{OP}}| \\
& M=\sqrt{(\hat{a} \sin t+\hat{b} \cos t)^2} \\
& =\sqrt{(\hat{a} \sin t)^2+(\hat{b} \cos t)^2+2(\hat{a} \sin t) \cdot(\hat{b} \cos t)} \\
& =\sqrt{\sin ^2 t+\cos ^2 t+\hat{a} \cdot \hat{b}(2 \sin t \cos t)} \\
& =\sqrt{1+\hat{\mathrm{a}} \cdot \hat{\mathrm{b}}(\sin 2 \mathrm{t})} \\
& \text { Maximum value of } \sin 2 t=1 \\
& \therefore \quad 2 \mathrm{t}=\sin ^{-1}(1) \\
& \therefore \quad \mathrm{t}=\frac{\pi}{4} \\
& \therefore \quad \mathrm{M}=\sqrt{1+\hat{\mathrm{a}} \cdot \hat{\mathrm{b}}(1)} \\
& =(1+\hat{a} \cdot \hat{b})^{\frac{1}{2}} \\
& \text { Now, } \hat{\mathrm{u}}=\frac{\overline{\mathrm{OP}}}{|\overline{\mathrm{OP}}|} \\
& =\frac{\hat{a} \sin t+\hat{b} \cos t}{|\hat{a} \sin t+\hat{b} \cos 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|}\\
&
\end{aligned}\)
Unit vector of OP is
\(\hat{\mathrm{u}}=\frac{\hat{\mathrm{a}}+\hat{\mathrm{b}}}{|(\hat{\mathrm{a}}+\hat{\mathrm{b}})|}\)
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