Question

If ¯a=i+2j , ¯b=2i+j , ¯c=4i+3j , find x and y such that ¯c=x¯a+y¯b​

Answer 1

Step-by-step explanation:

here is your answer...hope it will help u

Answer 2

$\begin{gathered}\begin{gathered}\bf\: Given-\begin{cases} &\sf{\vec{a} = \hat{i} + 2\hat{j}} \\ &\sf{\vec{b} = 2\hat{i} + \hat{j}}\\ &\sf{\vec{c} = 4\hat{i} + 3\hat{j}}\\ &\sf{\vec{c} = x\vec{a} + y\vec{b}} \end{cases}\end{gathered}\end{gathered}$

$\begin{gathered}\begin{gathered}\bf \: To\:Find-\begin{cases} &\sf{value \: of \: x \: and \: y}\end{cases}\end{gathered}\end{gathered}$

$\large\underline{\sf{Solution-}}$

Given that

$\rm :\longmapsto\:\vec{a} = \hat{i} + 2\hat{j}$

$\rm :\longmapsto\:\vec{b} = 2\hat{i} + \hat{j}$

$\rm :\longmapsto\:\vec{c} = 4\hat{i} + 3\hat{j}$

and

$\rm :\longmapsto\:\vec{c} = x\vec{a} + y\vec{b}$

On substituting the values, we have

$\rm :\longmapsto\:4\hat{i} + 3\hat{j} = x(\hat{i} + 2\hat{j}) + y(2\hat{i} + \hat{j})$

$\rm :\longmapsto\:4\hat{i} + 3\hat{j} = x\hat{i} + 2x\hat{j} + 2y\hat{i} + y\hat{j}$

$\rm :\longmapsto\:4\hat{i} + 3\hat{j} = (x + 2y)\hat{i} + (2x + y)\hat{j}$

On comparing we get,

$\rm :\longmapsto\:x + 2y = 4 \implies \: x = 4 - 2y - - (1)$

and

$\rm :\longmapsto\:2x + y = 3$

$\rm :\longmapsto\:2(4 - 2y) + y = 3$

$\rm :\longmapsto\:8 - 4y + y = 3$

$\rm :\longmapsto\:8 - 3y = 3$

$\rm :\longmapsto\: - 3y = 3 - 8$

$\rm :\longmapsto\: - 3y = - 5$

$\rm :\implies\:y = \dfrac{5}{3} - - - (2)$

Put the value of y in equation (1), we get

$\rm :\longmapsto\:x = 4 - 2 \times \dfrac{5}{3}$

$\rm :\longmapsto\:x = 4 - \dfrac{10}{3}$

$\rm :\longmapsto\:x = \dfrac{12 - 10}{3}$

$\rm :\implies\:x = \dfrac{2}{3}$

$\red{\begin{gathered}\begin{gathered}\bf\: Hence-\begin{cases} &\sf{x = \dfrac{2}{3} } \\ &\sf{y = \dfrac{5}{3} } \end{cases}\end{gathered}\end{gathered}}$

$\boxed{\pink{\bf\:Additional \: Information}}$

$\sf \: Unit \: vector \: along \: \vec{a} = \dfrac{\vec{a}}{ |\vec{a}| }$

$\sf \: If \: \vec{a} = x\hat{i} + y\hat{j} + z\hat{k} \: then \: direction \: ratios \: are \: (x, y, z)$

$\sf \: If \: \vec{a} \parallel \: \vec{b} \: then \: \vec{a} = k \: \vec{b} \: and \: vice \: versa$

$\sf \: If \: \vec{a} \perp\vec{b} \: then \: \vec{a}.\vec{b} = 0 \: and \: vice - versa$