# If" "veca=3hati+4hatj+5hatk and vec b= 2hati+hatj-4hatk ;How will you find out the component of " "veca " ""perpendicular to" " " vecb?

Jun 3, 2016

${\vec{a}}_{T} = \frac{1}{21} \left\{83 \hat{i} , 94 \hat{j} , 65 \hat{k}\right\}$

#### Explanation:

Given two non null vectors $\vec{a}$ and $\vec{b}$ the first $\vec{a}$ allways can be decomposed as a sum of two components: one parallel to $\vec{b}$ and another perpendicular to $\vec{b}$.

The parallel component is the projection of $\vec{a}$ onto $\vec{b}$ or
${\vec{a}}_{P} = \left\langle\vec{a} , \frac{\vec{b}}{\left\lVert \vec{b} \right\rVert}\right\rangle \frac{\vec{b}}{\left\lVert \vec{b} \right\rVert} = \left\langle\vec{a} , \vec{b}\right\rangle \frac{\vec{b}}{\left\lVert \vec{b} \right\rVert} ^ 2$
and the perpendicular component given by
${\vec{a}}_{T} = \vec{a} - {\vec{a}}_{P} = \vec{a} - \left\langle\vec{a} , \vec{b}\right\rangle \frac{\vec{b}}{\left\lVert \vec{b} \right\rVert} ^ 2$

So index $P$ for parallel and $T$ for perpendicular. We can verify that

${\vec{a}}_{P} + {\vec{a}}_{T} = \vec{a}$
$\left\langle{\vec{a}}_{P} , {\vec{a}}_{T}\right\rangle = \left\langle\left\langle\vec{a} , \vec{b}\right\rangle \frac{\vec{b}}{\left\lVert \vec{b} \right\rVert} ^ 2 , \vec{a} - \left\langle\vec{a} , \vec{b}\right\rangle \frac{\vec{b}}{\left\lVert \vec{b} \right\rVert} ^ 2\right\rangle = 0$
$\left\langle{\vec{a}}_{T} , \vec{b}\right\rangle = \left\langle\vec{a} - \left\langle\vec{a} , \vec{b}\right\rangle \frac{\vec{b}}{\left\lVert \vec{b} \right\rVert} ^ 2 , \vec{b}\right\rangle = 0$

In our case

${\vec{a}}_{P} = \frac{10}{21} \left\{- 2 \hat{i} , - \hat{j} , 4 \hat{k}\right\}$
${\vec{a}}_{T} = \frac{1}{21} \left\{83 \hat{i} , 94 \hat{j} , 65 \hat{k}\right\}$