Vectors 2 Question 13
13. Let $\overrightarrow{\mathbf{a}}$ and $\overrightarrow{\mathbf{b}}$ be two non-collinear unit vectors. If $\overrightarrow{\mathbf{u}}=\overrightarrow{\mathbf{a}}-(\overrightarrow{\mathbf{a}} \cdot \overrightarrow{\mathbf{b}}) \overrightarrow{\mathbf{b}}$ and $\overrightarrow{\mathbf{v}}=\overrightarrow{\mathbf{a}} \times \overrightarrow{\mathbf{b}}$, then $|\overrightarrow{\mathbf{v}}|$ is
(1999, 3M)
(a) $|\overrightarrow{\mathbf{u}}|$
(b) $|\overrightarrow{\mathbf{u}}|+|\overrightarrow{\mathbf{u}} \cdot \overrightarrow{\mathbf{a}}|$
(c) $|\overrightarrow{\mathbf{u}}|+|\overrightarrow{\mathbf{u}} \cdot \overrightarrow{\mathbf{b}}|$
(d) $|\overrightarrow{\mathbf{u}}|+\overrightarrow{\mathbf{u}} \cdot(\overrightarrow{\mathbf{a}}+\overrightarrow{\mathbf{b}})$
Show Answer
Answer:
Correct Answer: 13. (c)
Solution:
- Let $\theta$ be the angle between $\overrightarrow{\mathbf{a}}$ and $\overrightarrow{\mathbf{b}}$. Since, $\overrightarrow{\mathbf{a}}$ and $\overrightarrow{\mathbf{a}}$ are non-collinear vectors, then $\theta \neq 0$ and $\theta \neq \pi$.
We have, $\overrightarrow{\mathbf{a}} \cdot \overrightarrow{\mathbf{b}}=|\overrightarrow{\mathbf{a}}||\overrightarrow{\mathbf{a}}| \cos \theta$
$$ =\cos \theta \quad[\because|\overrightarrow{\mathbf{a}}|=1,|\overrightarrow{\mathbf{b}}|=1 \text {, given }] $$
Now, $\quad \overrightarrow{\mathbf{u}}=\overrightarrow{\mathbf{a}}-(\overrightarrow{\mathbf{a}} \cdot \overrightarrow{\mathbf{b}}) \overrightarrow{\mathbf{b}} \Rightarrow|\overrightarrow{\mathbf{u}}|=|\overrightarrow{\mathbf{a}}-(\overrightarrow{\mathbf{a}} \cdot \overrightarrow{\mathbf{b}}) \overrightarrow{\mathbf{b}}|$
$$ \Rightarrow \quad|\overrightarrow{\mathbf{u}}|^{2}=|\overrightarrow{\mathbf{a}}-(\overrightarrow{\mathbf{a}} \cdot \overrightarrow{\mathbf{b}}) \overrightarrow{\mathbf{b}}|^{2} $$
$\Rightarrow \quad|\overrightarrow{\mathbf{u}}|^{2}=|\overrightarrow{\mathbf{a}}-\cos \theta \overrightarrow{\mathbf{b}}|^{2}$
$\Rightarrow \quad|\overrightarrow{\mathbf{u}}|^{2}=|\overrightarrow{\mathbf{a}}|^{2}+\cos ^{2} \theta|\overrightarrow{\mathbf{b}}|^{2}-2 \cos \theta(\overrightarrow{\mathbf{a}} \cdot \overrightarrow{\mathbf{b}})$
$\Rightarrow \quad|\overrightarrow{\mathbf{u}}|^{2}=1+\cos ^{2} \theta-2 \cos ^{2} \theta$
$\Rightarrow \quad|\overrightarrow{\mathbf{u}}|^{2}=1-\cos ^{2} \theta \Rightarrow|\overrightarrow{\mathbf{u}}|^{2}=\sin ^{2} \theta$
Also, $\quad \overrightarrow{\mathbf{v}}=\overrightarrow{\mathbf{a}} \times \overrightarrow{\mathbf{b}}$
[given]
$\Rightarrow \quad|\overrightarrow{\mathbf{v}}|^{2}=|\overrightarrow{\mathbf{a}} \times \overrightarrow{\mathbf{b}}|^{2} \Rightarrow|\overrightarrow{\mathbf{v}}|^{2}=|\overrightarrow{\mathbf{a}}|^{2}|\overrightarrow{\mathbf{b}}|^{2} \cdot \sin ^{2} \theta$
$\Rightarrow \quad|\overrightarrow{\mathbf{v}}|^{2}=\sin ^{2} \theta \quad \therefore|\overrightarrow{\mathbf{u}}|^{2}=|\overrightarrow{\mathbf{v}}|^{2}$
Now, $\quad \overrightarrow{\mathbf{u}} \cdot \overrightarrow{\mathbf{a}}=[\overrightarrow{\mathbf{a}}-(\overrightarrow{\mathbf{a}} \cdot \overrightarrow{\mathbf{b}}) \overrightarrow{\mathbf{b}}] \cdot \overrightarrow{\mathbf{a}}=\overrightarrow{\mathbf{a}} \cdot \overrightarrow{\mathbf{a}}-(\overrightarrow{\mathbf{a}} \cdot \overrightarrow{\mathbf{b}})(\overrightarrow{\mathbf{b}} \cdot \overrightarrow{\mathbf{a}})$
$$ =(\overrightarrow{\mathbf{a}})^{2}-\cos ^{2} \theta=1-\cos ^{2} \theta=\sin ^{2} \theta $$
$$ \begin{aligned} \therefore \quad|\overrightarrow{\mathbf{u}}|+|\overrightarrow{\mathbf{u}} \cdot \overrightarrow{\mathbf{a}}| & =\sin \theta+\sin ^{2} \theta \neq|\overrightarrow{\mathbf{v}}| \\ \overrightarrow{\mathbf{u}} \cdot \overrightarrow{\mathbf{b}} & =[\overrightarrow{\mathbf{a}}-(\overrightarrow{\mathbf{a}} \cdot \overrightarrow{\mathbf{b}}) \overrightarrow{\mathbf{b}}] \cdot \overrightarrow{\mathbf{b}} \\ & =\overrightarrow{\mathbf{a}} \cdot \overrightarrow{\mathbf{b}}-(\overrightarrow{\mathbf{a}} \cdot \overrightarrow{\mathbf{b}})(\overrightarrow{\mathbf{b}} \cdot \overrightarrow{\mathbf{b}})=\overrightarrow{\mathbf{a}} \cdot \overrightarrow{\mathbf{b}}-\overrightarrow{\mathbf{a}} \cdot \overrightarrow{\mathbf{b}}|\overrightarrow{\mathbf{b}}|^{2} \\ & =\overrightarrow{\mathbf{a}} \cdot \overrightarrow{\mathbf{b}}-\overrightarrow{\mathbf{a}} \cdot \overrightarrow{\mathbf{b}}=0 \end{aligned} $$
$\therefore \quad|\overrightarrow{\mathbf{u}}|+|\overrightarrow{\mathbf{u}} \cdot \overrightarrow{\mathbf{b}}|=|\overrightarrow{\mathbf{u}}|+0=|\overrightarrow{\mathbf{u}}|=|\overrightarrow{\mathbf{v}}|$
Also, $\quad \overrightarrow{\mathbf{u}} \cdot(\overrightarrow{\mathbf{a}}+\overrightarrow{\mathbf{b}})=\overrightarrow{\mathbf{u}} \cdot \overrightarrow{\mathbf{a}}+\overrightarrow{\mathbf{u}} \cdot \overrightarrow{\mathbf{b}}=\overrightarrow{\mathbf{u}} \cdot \overrightarrow{\mathbf{a}}$
$\Rightarrow|\overrightarrow{\mathbf{u}}|+\overrightarrow{\mathbf{u}} \cdot(\overrightarrow{\mathbf{a}}+\overrightarrow{\mathbf{b}})=|\overrightarrow{\mathbf{u}}|+\overrightarrow{\mathbf{u}} \cdot \overrightarrow{\mathbf{a}} \neq|\overrightarrow{\mathbf{v}}|$
