Vectors 2 Question 22
22. For any two vectors $\overrightarrow{\mathbf{u}}$ and $\overrightarrow{\mathbf{v}}$, prove that
(i) $|\overrightarrow{\mathbf{u}} \cdot \overrightarrow{\mathbf{v}}|^{2}+|\overrightarrow{\mathbf{u}} \times \overrightarrow{\mathbf{v}}|^{2}=|\overrightarrow{\mathbf{u}}|^{2}|\overrightarrow{\mathbf{v}}|^{2}$
(ii) $\left(1+|\overrightarrow{\mathbf{u}}|^{2}\right)\left(1+|\overrightarrow{\mathbf{v}}|^{2}\right)=|1-\overrightarrow{\mathbf{u}} \cdot \overrightarrow{\mathbf{v}}|^{2}+|\overrightarrow{\mathbf{u}}+\overrightarrow{\mathbf{v}}+(\overrightarrow{\mathbf{u}} \times \overrightarrow{\mathbf{v}})|^{2}$
$(1998,8 M)$
Show Answer
Solution:
- (i) Since, $\overrightarrow{\mathbf{u}} \cdot \overrightarrow{\mathbf{v}}=|\overrightarrow{\mathbf{u}}||\overrightarrow{\mathbf{v}}| \cos \theta$ and $\quad \overrightarrow{\mathbf{u}} \times \overrightarrow{\mathbf{v}}=|\overrightarrow{\mathbf{u}}||\overrightarrow{\mathbf{v}}| \sin \theta \hat{\mathbf{n}}$ where, $\theta$ is the angle between $\overrightarrow{\mathbf{u}}$ and $\overrightarrow{\mathbf{v}}$ and $\hat{\mathbf{n}}$ is unit vector perpendicular to the plane of $\overrightarrow{\mathbf{u}}$ and $\overrightarrow{\mathbf{v}}$.
Again,
$$ |\overrightarrow{\mathbf{u}} \cdot \overrightarrow{\mathbf{v}}|^{2}=|\overrightarrow{\mathbf{u}}|^{2}|\overrightarrow{\mathbf{v}}|^{2} \cos ^{2} \theta \text { and } $$
$$ |\overrightarrow{\mathbf{u}} \times \overrightarrow{\mathbf{v}}|^{2}=|\overrightarrow{\mathbf{u}}|^{2}|\overrightarrow{\mathbf{v}}|^{2} \sin ^{2} \theta=|\overrightarrow{\mathbf{u}}|^{2}|\overrightarrow{\mathbf{v}}|^{2} \sin ^{2} \theta $$
$\therefore|\overrightarrow{\mathbf{u}} \cdot \overrightarrow{\mathbf{v}}|^{2}+|\overrightarrow{\mathbf{u}} \times \overrightarrow{\mathbf{v}}|^{2}=|\overrightarrow{\mathbf{u}}|^{2}|\overrightarrow{\mathbf{v}}|^{2}\left(\cos ^{2} \theta+\sin ^{2} \theta\right)$
$$ =|\overrightarrow{\mathbf{u}}|^{2}|\overrightarrow{\mathbf{v}}|^{2} $$
(ii) $|\overrightarrow{\mathbf{u}}+\overrightarrow{\mathbf{v}}+(\overrightarrow{\mathbf{u}} \times \overrightarrow{\mathbf{v}})|^{2}$
$$ \begin{aligned} & =|\overrightarrow{\mathbf{u}}+\overrightarrow{\mathbf{v}}|^{2}+|\overrightarrow{\mathbf{u}} \times \overrightarrow{\mathbf{v}}|^{2}+2(\overrightarrow{\mathbf{u}}+\overrightarrow{\mathbf{v}}) \cdot(\overrightarrow{\mathbf{u}} \times \overrightarrow{\mathbf{v}}) \\ & =|\overrightarrow{\mathbf{u}}|^{2}+|\overrightarrow{\mathbf{v}}|^{2}+2 \overrightarrow{\mathbf{u}} \cdot \overrightarrow{\mathbf{v}}+|\overrightarrow{\mathbf{u}} \times \overrightarrow{\mathbf{v}}|^{2}+0 \end{aligned} $$
$[\because \overrightarrow{\mathbf{u}} \times \overrightarrow{\mathbf{v}}$ is perpendicular to the plane of $\overrightarrow{\mathbf{u}}$ and $\overrightarrow{\mathbf{v}}]$
$\therefore|\overrightarrow{\mathbf{u}}+\overrightarrow{\mathbf{v}}+(\overrightarrow{\mathbf{u}} \times \overrightarrow{\mathbf{v}})|^{2}+|1-\overrightarrow{\mathbf{u}} \cdot \overrightarrow{\mathbf{v}}|^{2}$
$=|\overrightarrow{\mathbf{u}}|^{2}+|\overrightarrow{\mathbf{v}}|^{2}+2 \overrightarrow{\mathbf{u}} \cdot \overrightarrow{\mathbf{v}}+|\overrightarrow{\mathbf{u}} \times \overrightarrow{\mathbf{v}}|^{2}+1-2 \overrightarrow{\mathbf{u}} \cdot \overrightarrow{\mathbf{v}}+|\overrightarrow{\mathbf{u}} \cdot \overrightarrow{\mathbf{v}}|^{2}$
$=|\overrightarrow{\mathbf{u}}|^{2}+|\overrightarrow{\mathbf{v}}|^{2}+1+|\overrightarrow{\mathbf{u}}|^{2}|\overrightarrow{\mathbf{v}}|^{2}$
[from Eq. (i)]
$=|\overrightarrow{\mathbf{u}}|^{2}\left(1+|\overrightarrow{\mathbf{v}}|^{2}\right)+\left(1+|\overrightarrow{\mathbf{v}}|^{2}\right)=\left(1+|\overrightarrow{\mathbf{v}}|^{2}\right)\left(1+|\overrightarrow{\mathbf{u}}|^{2}\right)$
