Vector Algebra

$\left(2\widehat{i}+6\widehat{j}+27\widehat{k}\right)*\left(\widehat{i}+\lambda \widehat{j}+\mu \widehat{k}\right)=\overrightarrow{0}$

$\begin{array}{l}\Rightarrow \widehat{i}\left(6\mu -27\lambda \right)-\widehat{j}\left(2\mu -27\right)+\widehat{k}\left(2\lambda -6\right)\\ =0\widehat{i}+0\widehat{j}+0\widehat{k}\end{array}$

On comparing both side components,

$\begin{array}{l}6\mu -27\lambda =0,\\ 2\mu -27=0\\ 2\mu =27\\ \mu =\frac{27}{2},\\ 2\lambda -6=0\\ 2\lambda =6\\ \lambda =\frac{6}{2}=3\end{array}$

$\therefore$ Therefore, the value of $\mu =\frac{27}{2}$ and $\lambda =3$

Show that

$\begin{array}{l}\left(\overrightarrow{a}-\overrightarrow{b}\right)*\left(\overrightarrow{a}+\overrightarrow{b}\right)=2\left(\overrightarrow{a}*\overrightarrow{b}\right)\\ \left(\overrightarrow{a}-\overrightarrow{b}\right)*\left(\overrightarrow{a}+\overrightarrow{b}\right)\\ =\overrightarrow{a}\left(\overrightarrow{a}+\overrightarrow{b}\right)-\overrightarrow{b}\left(\overrightarrow{a}+\overrightarrow{b}\right)\\ =\overrightarrow{a}*\overrightarrow{a}+\overrightarrow{a}*\overrightarrow{b}-\overrightarrow{b}*\overrightarrow{a}-\overrightarrow{b}*\overrightarrow{b}\\ =0+\overrightarrow{a}*\overrightarrow{b}-\overrightarrow{b}*\overrightarrow{a}-0\\ =\overrightarrow{a}*\overrightarrow{b}+\overrightarrow{a}*\overrightarrow{b}\left[\overrightarrow{a}*\overrightarrow{b}=-\overrightarrow{b}*\overrightarrow{a}\right]\\ =2(\overrightarrow{a}*\overrightarrow{b})\end{array}$

Let $\overrightarrow{a}=\left(a_1,a_2,a_3\right)$ as component

We know,

$\overrightarrow{a}$ is a unit vector, $\left|\overrightarrow{a}\right|=1$

Given that,

$\overrightarrow{a}$ marks angles $\frac{\pi }{3}$ with $\widehat{i}$ , $\frac{\pi }{4}$ with $\widehat{j}$ and $\theta$ with $\widehat{k}$ acute angle.

Now,

$cos\frac{\pi }{3}=\frac{a_1}{\left|\overrightarrow{a}\right|}\Rightarrow \frac{1}{2}=a_1\left[\left|\overrightarrow{a}\right|=1\right]\\ cos\frac{\pi }{4}=\frac{a_2}{\left|\overrightarrow{a}\right|}\Rightarrow 1/\surd 2$
$\begin{array}{l}\frac{\mathrm{}}{}=a_2\\ cos\theta =\frac{a_3}{\left|\overrightarrow{a}\right|}\Rightarrow a_3=cos\theta \end{array}$

We know,

$\left|\overrightarrow{a}\right|=1$

Given,

$\begin{array}{l}\overrightarrow{a}=3\widehat{i}+2\widehat{j}+2\widehat{k}\\ \overrightarrow{b}=\widehat{i}+2\widehat{j}-2\widehat{k}\\ \overrightarrow{a}+\overrightarrow{b}=4\widehat{i}+4\widehat{j},\overrightarrow{a}-\overrightarrow{b}=2\widehat{i}+4\widehat{j}\end{array}$

A vector which is perpendicular to both $\overrightarrow{a}+\overrightarrow{b}$ and $\overrightarrow{a}-\overrightarrow{b}$ is given by

Say

Therefore, the unit vector is

$\begin{array}{l}\frac{\overrightarrow{c}}{\left|\overrightarrow{c}\right|}=\pm \frac{16\widehat{i}-16\widehat{j}-8\widehat{k}}{24}\\ =\pm \frac{16}{24}\widehat{i}\pm \frac{16}{24}\widehat{j}\pm \frac{8}{24}\widehat{k}\\ =\pm \frac{2}{3}\widehat{i}\pm \frac{2}{3}\widehat{j}\pm \frac{1}{3}\stackrel{}{k}\end{array}$

Kindly go through the solution

Vector $\lambda \overrightarrow{a}$ is a unit vector if $\left|\lambda \overrightarrow{a}\right|=1$

Now,

$\begin{array}{l}\left|\lambda \overrightarrow{a}\right|=1\\ \left|\lambda \right|\left|\overrightarrow{a}\right|=1\\ \left|\overrightarrow{a}\right|=\frac{1}{\left|\lambda \right|}\left[\lambda \ne 0\right]\\ a=\frac{1}{\left|\lambda \right|}\left[\left|\overrightarrow{a}\right|=a\right]\end{array}$

$\therefore$ Therefore, vectar $\lambda \overrightarrow{a}$ is a unit vector if a= $\frac{1}{\left|\lambda \right|}$ .

Option (D)is correct.

Let vector $2\widehat{i}-\widehat{j}+\widehat{k},\widehat{i}-3\widehat{j}-5\widehat{k}$ and $3\widehat{i}-4\widehat{j}-4\widehat{k}$ be position vector of point A, B, C respectively.

So,

$\begin{array}{l}O\overrightarrow{A}=2\widehat{i}-\widehat{j}+\widehat{k}\\ O\overrightarrow{B}=\widehat{i}-3\widehat{j}-5\widehat{k}\\ O\overrightarrow{C}=3\widehat{i}-4\widehat{j}-4\widehat{k}\end{array}$

Now, vectors $A\overrightarrow{B},B\overrightarrow{C}$ and $A\overrightarrow{C}$ represents the sides of $?ABC$ .

Hence,

Given, point are

Vertices of $?ABC$ are given as

$\begin{array}{l}A\left(1,2,3\right),B\left(-1,0,0\right),C\left(0,1,2\right)\\ \end{array}$

$\therefore ?ABC$ is the angle between the vectors $B\overrightarrow{A}$ and $B\overrightarrow{C}$

Consider

$\begin{array}{l}\overrightarrow{a}=2\widehat{i}+4\widehat{j}+3\widehat{k}\\ \overrightarrow{b}=3\widehat{i}+3\widehat{j}-6\widehat{k}\end{array}$ and

Then,

$\begin{array}{l}\overrightarrow{a}.\overrightarrow{b}=2.3+4.3+3.(-6)\\ =6+12-18=0\end{array}$

Therefore, the converse of the given statement need not be true.