Vector Algebra

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

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

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

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.

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 $x=2\&y=3$