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Given, A(1,2,-3) and (-1,-2,1)

Now,

$\begin{array}{l}\left|\overrightarrow{AB}\right|=(-1-1)\hat{i}+(-2-2)\hat{j}+\left(1-(-3)\right)\hat{k}\\ =-2\hat{i}-4\hat{j}+4\hat{k}\end{array}$

Then,$\begin{array}{l}\\ \end{array}$

Let, l, m, n be direction cosine,

$l=\frac{x}{\left|\overrightarrow{AB}\right|}=\frac{-2}{6}=\frac{-1}{3};m=\frac{y}{\left|\overrightarrow{AB}\right|}=\frac{-4}{6}=\frac{-2}{3};n=\frac{z}{\left|\overrightarrow{AB}\right|}=\frac{4}{6}=\frac{2}{3}$

Therefore, direction cosine of $\overrightarrow{AB}$ are $\left(\frac{-1}{3},\frac{-2}{3},\frac{2}{3}\right)$

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Let $\overrightarrow{a}=\hat{i}+2\hat{j}+3\hat{k}$

Let,  $\overrightarrow{a}=2\hat{i}-3\hat{j}+4\hat{k}\&\overrightarrow{b}=-4\hat{i}+\widehat{6j}-8\hat{k}$

It is seen that

$\begin{array}{l}\overrightarrow{b}=-4\hat{i}+\widehat{6j}-8\hat{k}\\ =-2\left(2\hat{i}-3\hat{j}+4\hat{k}\right)\\ =-2\overrightarrow{a}\\ \therefore \overrightarrow{b}=\lambda \overrightarrow{a}\end{array}$

Where,  $\lambda =-2$

Therefore, we can say that the given vector are collinear.

45. The given system of inequality is

5x+4y≤ 20 - (1)

x≥ 1 - (2)

and  y≥ 2 - (3)

The equation of inequality (1) is 5x+4y=20.

Putting (x, y)= (0,0) in inequality (1) we get,

5 × 0+4 × 0 ≤ 20   => 0 ≤ 20 which is true.

So, the solution region of inequality (1) includes the plane with origin (0,0).

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

Kindly go through the solution

Given,  $\overrightarrow{a}=2\hat{i}-\hat{j}+2\hat{k}\&\overrightarrow{b}=-\hat{i}+\hat{j}-\hat{k}$

The sum of given vectors is given by

44. The given system of inequality is

x+y≤ 9- (1)

y>x- (2)

x≥ 0 - (3)

The corresponding equation of (1) is x+y=9 and (2) is y=x

Substituting (x, y)= (0,0) in (1),

0+0 ≤ 9  => 0 ≤ 9 which is true.

And putting (1,0) in (2)

0> 1which is false.

So, solution region of inequality (1) includes origin (0,0) and solution region of inequality (2) excludes plane having (1,0).

? Solution of region of given system of inequality is the shaded region.

Given,  $P\left(1,2,3\right)\&Q\left(4,5,6\right)$

So,

$\begin{array}{l}\overrightarrow{PQ}=(4-1)\hat{i}+(5-2)\hat{j}+(6-3)\hat{k}\\ =3\hat{i}+3\hat{j}+3\hat{k}\end{array}$

Kindly go through the solution

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The given vectors are

$\overrightarrow{a}=\hat{i}-2\hat{j}+\hat{k}$

$\overrightarrow{b}=\widehat{-2i}+4\hat{j}+5\hat{k}$

$\overrightarrow{c}=\hat{i}-6\hat{j}-7\hat{k}$

The sum of the vector is

$\overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c}=\left(a_1+a_2+a_3\right)\hat{i}+\left(b_1+b_2+b_3\right)\widehat{\hat{j}+\left(c_1+c_2+c_3\right)\hat{k}}$

$\begin{array}{l}=(1-2+1)\hat{i}+(-2+4-6)\hat{j}+(1+5-7)\hat{k}\\ =0.\hat{i}+(-4)\hat{j}+(-1)\hat{k}\\ =-4\hat{j}-\stackrel{}{k}\end{array}$

Let the vector with initial point P (2,1) and terminal point Q. (-5,7) can be shown as,

$\begin{array}{l}\overrightarrow{PQ}=(-5,-2)\hat{i}+(7,-1)\hat{j}\\ \overrightarrow{PQ}=-7\hat{i}+6\hat{j}\end{array}$

The scalar components are -7 and 6.

The vector components are -7i and 6j.

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43. Given system of inequality is

2x+y≥ 8- (1)

x+2y≥ 10- (2)

The corresponding equations are

2x + y = 8

and x + 2y = 10

Now, putting (x, y)= (0,0) in inequality (1) and (2),

2 × 0+8 ≥ 8

0 ≥ 8 which is not true.

and 0+2 × 0 ≥ 10

0 ≥ 10 which is not true.

So, solution of plane of inequality (1) and (2) does not include the origin (0,0)

? The required solution of the given system of inequality is the shaded region.

Note that two vector are equal only if their corresponding components are equal.

Thus, the given vectors $\overrightarrow{a}$ and $\overrightarrow{b}$ will be equal if and only if $x=2\&y=3$