Two adjacent sides of parallelogram are given by

$\overrightarrow{AB}=2 i+10j+11k \quad \text{and} \\ \overrightarrow{AD}=-i+2j+2k$

The side $A D$ is rotated by an acute angle $\theta$ in the plane of parallelogram so that $A D$ becomes $A D^{\prime}$.

If $A D^{\prime}$ makes a right angle with side $A B$, then what is the value of the cosine of the angle $\theta$ ?

$\cos \theta=\text { ? }$

$\overrightarrow{A B} \cdot \overrightarrow{A D}=\overrightarrow{A B}|| \overrightarrow{A D} \mid \cos (\eta)$

$\cos \eta=\frac{\overrightarrow{A B} \cdot \overrightarrow{A D}}{|\overrightarrow{A B}||\overrightarrow{A D}|}=\frac{40}{\sqrt{225} \sqrt{9}}=\frac{40}{45}$

$=\frac{8}{9}$

$\boxed{\cos \eta=\frac{8}{9}} , \quad \cos ^2 \eta+\sin ^2 \eta=1\$

$\eta+\theta=90^{\circ}$

$\cos (\theta)=\cos \left(90^{\circ}-\eta\right)=\sin (\eta)=\frac{\sqrt{17}}{9}$

Given 3 points whose position vectors are

$x i+y j+z k,$

$i+z j \quad \&-i-j \text {. }$

Find the condition for which these points are collinear.

$\overrightarrow{O A}=x i+y j+z k$,

$\overrightarrow{O B}=i+z j$,

$\overrightarrow{O C}=-i-j$,

$\overrightarrow{A B}=\overrightarrow{O B}-\overrightarrow{D A}$, $=(1-x) i+(z-y) j-z k$,

$\overrightarrow{A C}=\overrightarrow{O C}-\overrightarrow{O A}$, $=(-1-x) i+(-1-y) j-z k$,

Find the cosine of the angle between vectors $\vec{p} \& \vec{q}$ such that

$2 \vec{p}+\vec{q}=i+j \quad \& \vec{p}+2 \vec{q}=i-j .$

Solution:

$3 \vec{p}+3 \vec{q}=2 i \quad \vec{p}-\vec{q}=2 j$

$\vec{p}+\vec{q}=\frac{2}{3} i$ and $2 \vec{p}=\frac{2}{3} i+2 j \quad 2 \vec{q}=\frac{2}{3} i-2 j$

$4 \vec{p} \cdot \vec{q}=\frac{4}{9}-4=\frac{-32}{9} \Rightarrow \vec{p} \cdot \vec{q}=\frac{-8}{9}$

$|\vec{p}|^2=\frac{4}{36} \times 10 |\vec{p}|=\frac{\sqrt{10}}{3}$

$|\vec{q}|^2=\frac{10}{q} |\vec{q}|=\frac{\sqrt{10}}{3}$

$\cos \theta =\frac{\vec{p} \cdot \vec{q}}{|\vec{p}||\vec{q}|}=\frac{-81 d}{10 / x}=\frac{-4}{5} .$

$\theta =\cos ^{-1}\left(\frac{-4}{5}\right)$

Determine the values of $c$ such that for all real $x$, the vectors $c x i-6 j+3 k \quad$ and $x_i$+2 j+2 c x k make an obtuse angle with each other.

Solution:

If $\theta$ is an angle between these vectors

$\cos \theta$

$=\frac{\left(c x_i-6 j+3 k\right) \cdot\left(x_i+2 j+2 c x k\right)}{\sqrt{x^2+4+4 c^2 x^2} \sqrt{c^2 x^2+36+9}}$

$=\frac{c x^2+6 c x-12}{\sqrt{\left(1+4 c^2\right) x^2+4} \sqrt{c^2 x^2+45}}$.

$\cos \theta<0, \quad|\cos \theta| \leqslant 1$

$\frac{c x^2-12+6 c x}{\sqrt{\left(1+4 c^2\right) x^2+4} \sqrt{c^2 x^2+45}}<0, \forall x \in \mathbb{R}$ .

$c x^2+6 c x-12<0$

$c < 0 ~ \text{and} ~ (6 c)^2-4 c(-12) < 0$

$\Leftrightarrow c < 0 ~ \text{and} ~ 36 c^2+48 c <0$

$\Leftrightarrow c < 0 ~ \text{and} ~ 12 c(3 c+4) <0$

$\Leftrightarrow c < 0 ~ \text{and} ~ 3 c+4>0$

$\Leftrightarrow c < 0 ~ \text{and} ~ c > -\frac{4}{3}$

$\Leftrightarrow \frac{4}{3} < c < 0 .$

If $\vec{a} ~\& ~ \vec{b}$ are two unit vectors such that $\vec{a}+2 \vec{b} ~\& ~ 5 \vec{a}-4 \vec{b}$ are perpendicular to each other then find the angle between $\vec{a} ~ \& ~ \vec{b}$.

Solution:

$(\vec{a}+2 \vec{b}) \cdot(5 \vec{a}-4 \vec{b})=0 . |\vec{a}|=1-|\vec{b}|$

$\Leftrightarrow 5 \vec{a} \cdot \vec{a}-8 \vec{b} \cdot \vec{b}+(10-4) \vec{a} \cdot \vec{b}=0 \vec{a} \cdot \vec{a}=1$

$\Leftrightarrow 5-8+6 \vec{a} \cdot \vec{b}=0$

$\Leftrightarrow \vec{a} \cdot \vec{b}=\frac{3}{6}=\frac{1}{2}$

$\Leftrightarrow \frac{1}{\left|\vec{a}\right|}\frac{1}{\left|\vec{b}\right|} \cos \theta=\frac{1}{2}$

$\Leftrightarrow \theta=\cos ^{-1}(1 / 2)=60^{\circ}$

If $\vec{a}, \vec{b} ~ \& ~ \vec{c}$ are unit vectors, then find $|\vec{a}-\vec{b}|^2+|\vec{b}-\vec{c}|^2+|\vec{c}-\vec{a}|^2 \leqslant$ ?

Solution:

$0 \leqslant|\vec{a}-\vec{b}|^2=(\vec{a}-\vec{b}) \cdot(\vec{a}-\vec{b})$

$=|\vec{a}|^2+|\vec{b}|^2-2 \vec{a} \cdot \vec{b}$

$=2(1-\vec{a} \cdot \vec{b})$

$1-\vec{a} \cdot \vec{b} \geq 0 \quad-1 \leqslant \vec{a} \cdot \vec{b} \leqslant 1$

$|\vec{a}-\vec{b}|^2+|\vec{b}-\vec{c}|^2+|\vec{c}-\vec{a}|^2$

$=2\left(|\vec{a}|^2+|\vec{b}|^2+|\vec{c}|^2\right)$ \

$-2(\vec{a} \cdot \vec{b}+\vec{b} \cdot \vec{c}+\vec{c} \cdot \vec{a})$

Solution:

$\begin{aligned} & \vec{a} \cdot \vec{b}=0 \Rightarrow \quad \vec{a} \perp \vec{b} \\ & \vec{a} \cdot \vec{c}=0 \Rightarrow \vec{a} \perp \vec{c} \quad \vec{b} \times \vec{c} \\ & \vec{a} | \vec{b} \times \vec{c} \Rightarrow \boxed{\vec{a}=k(\vec{b} \times \vec{c})} \\ & 1=|\vec{a}|=|k(\vec{b} \times \vec{c})|=|k||\vec{b} \times \vec{c}| \end{aligned}$

$\vec{b}$ & $\vec{c}$ angle is $\pi / 6$

$|\vec{b} \times \vec{c}|=|\vec{b}||\vec{c}| \sin \frac{\pi}{6} =\sin \frac{\pi}{6} = \frac{1}{2}$

$1=|k| \frac{1}{2}$

$|k|=2$

$k= \pm 2$

$\vec{a}= \pm 2(\vec{b} \times \vec{c}) .$