MHT CET · Maths · Vector Algebra
Let two non-collinear unit vectors \(\hat{a}\) and \(\hat{b}\) form an acute angle. A point P moves, so that at any time \(t\) the position vector \(\overline{\mathrm{OP}}\), where O is the origin, is given by \(\hat{a} \cos t+\hat{b} \sin t\). When \(P\) is farthest from origin O , let 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 \(\dot{\hat{u}}=\frac{\hat{a}+\hat{b}}{|\hat{a}+\hat{b}|}\) and \(M=(1+2 \hat{a} \cdot \hat{b})^{\frac{1}{2}}\)
- D \(\hat{\mathrm{u}}=\frac{\hat{\mathrm{a}}-\hat{\mathrm{b}}}{|\hat{\mathrm{a}}-\hat{\mathrm{b}}|}\) and \(\mathrm{M}=(1-2 \hat{\mathrm{a}} \cdot \hat{\mathrm{b}})^{\frac{1}{2}}\)
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
\(\begin{aligned}
M & =|\overrightarrow{O P}| \\
M & =\hat{a} \cos t+\hat{b} \sin t \\
& =\sqrt{(\hat{a} \cos t)^2+(\hat{b} \sin t)^2+2(\hat{a} \cos t) \cdot(\hat{b} \cdot \sin t)} \\
& =\sqrt{\sin ^2 t+\cos ^2 t+\hat{a} \cdot \hat{b}(2 \sin t \cos t)} \\
& =\sqrt{1+\hat{a} \cdot \hat{b}(\sin 2 t)}
\end{aligned}\)
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
\(\hat{u}=\frac{\hat{a}+\hat{b}}{|(\hat{a}+\hat{b})|}\)
M & =|\overrightarrow{O P}| \\
M & =\hat{a} \cos t+\hat{b} \sin t \\
& =\sqrt{(\hat{a} \cos t)^2+(\hat{b} \sin t)^2+2(\hat{a} \cos t) \cdot(\hat{b} \cdot \sin t)} \\
& =\sqrt{\sin ^2 t+\cos ^2 t+\hat{a} \cdot \hat{b}(2 \sin t \cos t)} \\
& =\sqrt{1+\hat{a} \cdot \hat{b}(\sin 2 t)}
\end{aligned}\)
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
\(\hat{u}=\frac{\hat{a}+\hat{b}}{|(\hat{a}+\hat{b})|}\)
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