Maths

The Position vector of mid-point R of the vector joining point P (2,3,4) and Q (4,1, -2) is given by;

$\begin{array}{l}\overrightarrow{OR}=\frac{\left(2\hat{i}+3\hat{j}+4\hat{k}\right)+\left(4\hat{i}+\hat{j}-\widehat{2k}\right)}{2}\\ =\frac{(2+4)\hat{i}+(3+1)\hat{j}+(4-2)\hat{k}}{2}\\ =\frac{6\hat{i}+4\hat{j}+2\hat{k}}{2}=\frac{6\hat{i}}{2}+\frac{4\hat{j}}{2}+\frac{2\hat{k}}{2}\\ =3\hat{i}+2\hat{j}+\stackrel{}{k}\end{array}$

(i) The position vector of point R dividing the join of P and Q. internally in

the ratio 2:1 is,

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

(ii) The position vector of the point k dividing the join of P and Q. externally in the ratio 2:1

A15. (ii)

$\begin{array}{l}\overrightarrow{OR}=\frac{2\left(-\hat{i}+\hat{j}+\hat{k}\right)-1\left(\hat{i}+\widehat{2j}-\hat{k}\right)}{2-1}\\ =-2\hat{i}+\widehat{2j}+2\hat{k}-\hat{i}-\widehat{2j}+\hat{k}\\ =-3\hat{i}+\hat{k}\end{array}$

Here,

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

Then,

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 $\left(\frac{-1}{3},\frac{-2}{3},\frac{2}{3}\right)$

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.