Class 12th

(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.

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

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

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

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

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