Express the vector $\vec{a}=(6 \hat{i}-3 \hat{j}-6 \hat{k})$ as sum of two vectors such that one is parallel to the vector $\overrightarrow{\mathrm{b}}=(\hat{\mathrm{i}}+\hat{\mathrm{j}}+\hat{\mathrm{k}})$ and the other is perpendicular to $\overrightarrow{\mathrm{b}}$.

Express the vector $\vec{a}=(6 \hat{i}-3 \hat{j}-6 \hat{k})$ as sum of two vectors such that one is parallel to the vector $\overrightarrow{\mathrm{b}}=(\hat{\mathrm{i}}+\hat{\mathrm{j}}+\hat{\mathrm{k}})$ and the other is perpendicular to $\overrightarrow{\mathrm{b}}$ .

CBSE | Class 12 | Maths | RS Aggarwal | Scalar, or Dot, Product of Vectors

Solution:

$\begin{array}{l} \vec{a}=6 \hat{\imath}-3 \hat{\jmath}-6 \hat{k} \\ \vec{b}=\hat{\imath}+\hat{\jmath}+\hat{k} \end{array}$
Let $\vec{a}$ be written as sum of $\vec{c}+\vec{d}$ , where $\vec{c}$ is parallel to $\vec{b}$ and $\vec{d}$ is perpendicular to $\vec{b}$
Let $\vec{d}=\mathrm{d} \hat{\imath} \hat{\imath}+\mathrm{d}_{2} \hat{\jmath}+\mathrm{d}_{3} \hat{k}$
$\begin{array}{l} \mathrm{I} \vec{b} \mathrm{I}=\sqrt{1^{2}+1^{2}+1^{2}}=\sqrt{3} \\ \hat{b}=\frac{1}{\sqrt{3}} \hat{\imath}+\frac{1}{\sqrt{3}} \hat{\jmath}+\frac{1}{\sqrt{3}} \hat{k} \end{array}$
So according to the question
$\vec{c}=\mathrm{k}\left(\frac{1}{\sqrt{3}} \hat{\imath}+\frac{1}{\sqrt{3}} \hat{\jmath}+\frac{1}{\sqrt{3}} \hat{k}\right)$ for some constant $\mathrm{k}$
$\vec{a}=\vec{c}+\vec{d}$
$6 \hat{\imath}+3 \hat{\jmath}-6 \hat{k}=\mathrm{k}\left(\frac{1}{\sqrt{3}} \hat{\imath}+\frac{1}{\sqrt{3}} \hat{\jmath}+\frac{1}{\sqrt{3}} \hat{k}\right)+\left(\mathrm{d} \hat{\imath} \hat{\imath}+\mathrm{d}_{2} \hat{\jmath}+\mathrm{d}_{3} \hat{k}\right)---------I$
Also $\vec{d}$ is perpendicular to $\vec{b}$
$\vec{d} \cdot \vec{b}=0$
$\mathrm{d}_{1}+\mathrm{d}_{2}+\mathrm{d}_{3}=0---------II$
Comparing left hand side and right hand side in I
$\begin{array}{l} \frac{K}{\sqrt{3}}+\mathrm{d}_{1}=6--------III \\ \frac{K}{\sqrt{3}}+\mathrm{d}_{2}=-3------------IV \end{array}$
$\frac{K}{\sqrt{3}}+\mathrm{d}_{3}=-6---------V$
$\mathrm{III}+\mathrm{IV}+\mathrm{V}$
$\begin{array}{l} \mathrm{d}_{1}+\mathrm{d}_{2}+\mathrm{d}_{3}+\sqrt{3} \mathrm{~K}=-3 \\ \mathrm{~K}=-\sqrt{3} \end{array}$
Hence now we can calculate d1, d2, d3 from equations III, IV and V as
$\begin{array}{l} \mathrm{d}_{1}=7 \\ \mathrm{~d}_{2}=-2 \\ \mathrm{~d}_{3}=-5 \\ \vec{c}=-\hat{\imath}-\hat{\jmath}-\hat{k} \\ \vec{d}=7 \hat{\imath}-2 \hat{\jmath}-5 \hat{k} \end{array}$