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

Mar 6, 2017

$\left(\frac{83}{21} , \frac{94}{21} , \frac{65}{21}\right)$

Explanation:

Let $\vec{a} = {\vec{a}}_{1} + {\vec{a}}_{2}$ with ${\vec{a}}_{1} \ne \vec{0} , {\vec{a}}_{2} \ne \vec{0}$such that $\left\langle{\vec{a}}_{1} , {\vec{a}}_{2}\right\rangle = 0$ and ${\vec{a}}_{2} = \lambda \vec{b}$ with $\lambda \in \mathbb{R}$

Making now $\left\langle\vec{a} , \vec{b}\right\rangle = \left\langle{\vec{a}}_{1} + {\vec{a}}_{2} , \vec{b}\right\rangle = \left\langle{\vec{a}}_{1} , \vec{b}\right\rangle + \left\langle{\vec{a}}_{2} , \vec{b}\right\rangle$ but by hypothesis ${\vec{a}}_{2} = \lambda \vec{b}$ so

$\left\langle\vec{a} , \vec{b}\right\rangle = \lambda \left\langle\vec{b} , \vec{b}\right\rangle$ so

$\lambda = \frac{\left\langle\vec{a} , \vec{b}\right\rangle}{\left\langle\vec{b} , \vec{b}\right\rangle}$ so

${\vec{a}}_{2} = \frac{\left\langle\vec{a} , \vec{b}\right\rangle}{\left\langle\vec{b} , \vec{b}\right\rangle} \vec{b}$ and

${\vec{a}}_{1} = \vec{a} - \frac{\left\langle\vec{a} , \vec{b}\right\rangle}{\left\langle\vec{b} , \vec{b}\right\rangle} \vec{b}$

or

${\vec{a}}_{1} = \left(3 , 4 , 5\right) - \frac{3 \times 2 + 4 \times 1 - 5 \times 4}{{2}^{2} + {1}^{2} + {4}^{2}} \left(2 , 1 , - 4\right) = \left(\frac{83}{21} , \frac{94}{21} , \frac{65}{21}\right)$

Mar 6, 2017

The component of $\vec{a}$ perpendicular to $\vec{b}$

$= \vec{a} - \text{projection of "veca" " on " } \vec{b}$

$= \vec{a} - \frac{\vec{a} \cdot \vec{b}}{\left\mid \vec{b} \right\mid} \times \frac{\vec{b}}{\left\mid \vec{b} \right\mid}$

$= \vec{a} - \frac{\vec{a} \cdot \vec{b}}{\left\mid \vec{b} \right\mid} ^ 2 \times \vec{b}$

$= \left(3 \hat{i} + 4 \hat{j} + 5 \hat{k}\right) - \frac{\left(3 \hat{i} + 4 \hat{j} + 5 \hat{k}\right) . \left(2 \hat{i} + \hat{j} - 4 \hat{k}\right)}{\left\mid 2 \hat{i} + \hat{j} - 4 \hat{k} \right\mid} ^ 2 \times \left(2 \hat{i} + \hat{j} - 4 \hat{k}\right)$

$= \left(3 \hat{i} + 4 \hat{j} + 5 \hat{k}\right) - \frac{2 \cdot 3 + 4 \cdot 1 + 5 \cdot - 4}{{2}^{2} + {1}^{2} + {4}^{2}} \times \left(2 \hat{i} + \hat{j} - 4 \hat{k}\right)$

$= \left(3 \hat{i} + 4 \hat{j} + 5 \hat{k}\right) + \frac{10}{21} \times \left(2 \hat{i} + \hat{j} - 4 \hat{k}\right)$

$= \frac{1}{21} \left[21 \left(3 \hat{i} + 4 \hat{j} + 5 \hat{k}\right) + 10 \left(2 \hat{i} + \hat{j} - 4 \hat{k}\right)\right]$

$= \frac{1}{21} \left[83 \hat{i} + 94 \hat{j} + 65 \hat{k}\right]$