Maths

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

51. The given system of inequality is

x+2y≤ 10- (1)

x+y≥ 1 - (2)

x – y ≤ 0 - (3)

x≥ 0 and y≥ 0 - (4)

The corresponding equation of (1), (2) and (3) are

x + 2y = 10

and x + y =1

and x – y = 0

Putting (2,0)= (x, y) in inequality (1), (2) and (3),

2+2 * 0 ≤ 10 =>  2≤ 10 is true.

and 2+0 ≥ 1 =>  2 ≥ 1 is true.

and 2 – 0 ≤ 0  => 2 ≤ 0 is false.

So, the solution of inequality (1) and (2) is the plane that includes point (2,0) whereas the solution of inequality (3) is the plane which includes point (2, 0)

∴ The shaded region represents the solution of the given system of inequality.

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.