Question

If $\underset{a}{\to }=3\hat{i}+4\hat{j}+5\hat{k}$ and $\underset{\beta }{\to }=2\hat{i}+\hat{j}-4\hat{k}$ , then express $\beta$ as the sum of two vectors such that one is parallel to $\underset{a}{\to }$ and other is perpendicular to $\underset{a}{\to }$

Answer 1

$\overrightarrow{a}=3\hat{i}+4\hat{j}+5\hat{k}$

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

Let $\beta ={\beta }_1+{\beta }_2$

So, ${\beta }_1=\lambda \overrightarrow{a}$

And, ${\beta }_2.\overrightarrow{a}=0$

Now,

${\beta }_1=\lambda (3\hat{i}+4\hat{j}+5\hat{k})$

${\beta }_1=3\lambda \hat{i}+4\lambda \hat{j}+5\lambda \hat{k}$

Now,

${\beta }_2=\beta -{\beta }_1$

${\beta }_2=(2-3\lambda )\hat{i}+(1-4\lambda )\hat{j}+(-4-5\lambda )\hat{k}$

Since, ${\beta }_2.\overrightarrow{a}=0$

So, $3(2-3\lambda )+4(1-4\lambda )+5(-4-5\lambda )=0$

$6-9\lambda +4-16\lambda -20-25\lambda =0$

$-10-50\lambda =0$

$50\lambda =-10$

$\lambda =\frac{-1}{5}$

So, ${\beta }_1=\frac{-1}{5}(3\hat{i}+4\hat{j}+5\hat{k})=\frac{-3}{5}\hat{i}-\frac{4}{5}\hat{j}-\hat{k}$

And,

${\beta }_2=\frac{13}{5}\hat{i}+\frac{9}{5}\hat{j}-3\hat{k}$

Hence,

$\beta =(\frac{-3}{5}\hat{i}-\frac{4}{5}\hat{j}-\hat{k})+(\frac{13}{5}\hat{i}+\frac{9}{5}\hat{j}-3\hat{k})$