Question

Express 2i+3j+k as a sum of two vectors out of which one vector is perpendicular to 2i-4j+k and another is parallel to 2i-4j+k

Answer 1

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

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

Let $\overrightarrow{a}=\overrightarrow{a_1}+\overrightarrow{a_2}$

$\overrightarrow{a_1}=$ vector parallel to $\overrightarrow{b}$

$\overrightarrow{a_2}=$ vector perpendicular to $\overrightarrow{b}$

Since, $\overrightarrow{a_1}$ is parallel to $\overrightarrow{b}$

by equation of line in vector form

$\overrightarrow{a_1}=\lambda \left(\overrightarrow{b}\right)=\lambda (2\hat{i}-4\hat{j}-\hat{k})$ scalar $\lambda -$ constant

$\overrightarrow{a_1}=2\lambda \hat{i}-4\lambda \hat{j}+\lambda \hat{k}$

Also,

$\overrightarrow{a_2}=\overrightarrow{a}-\overrightarrow{a_1}$

$\overrightarrow{a_2}=(2\hat{i}+3\hat{j}+\hat{k})-(2\lambda \hat{i}-4\lambda \hat{j}+\lambda \hat{k})$

$\overrightarrow{a_2}=\hat{i}(2-2\lambda )+\hat{j}(3+4\lambda )+\hat{k}(1-\lambda )$

Since, $\overrightarrow{a_2}$ and $\overrightarrow{b}$ are perpendicular.

$\overrightarrow{a_2}.\overrightarrow{b}=0$

$\left[\hat{i}\right(2-2\lambda )+\hat{j}(3+4\lambda )+\hat{k}(1-\lambda \left)\right].[2\hat{i}-4\hat{j}+\hat{k}]=0$

$2(2-2\lambda )+(3+4\lambda )(-4)+(1-\lambda ).1=0$

$4-4\lambda +(-12)-16\lambda +1-\lambda =0$

$-7=21\lambda$

$\lambda =\frac{-1}{3}$

$\therefore \overrightarrow{a_1}=\frac{-1}{3}(2i-4j+k)$

$\overrightarrow{a_2}=i(2-2(\frac{-1}{3}\left)\right)+j(3+4(\frac{-1}{3}\left)\right)+k(1-(\frac{-1}{3}\left)\right)$

and $\overrightarrow{a_2}=\frac{8\hat{i}}{3}+\hat{j}\frac{5}{3}+\frac{4\hat{k}}{3}$