# Please solve with a detailed explanation?

## The velocity of a particle is $\vec{v} = 6 \hat{i} + 2 \hat{j} - 2 \hat{k}$.The component of the velocity parallel to vector $\vec{a} = \hat{i} + \hat{j} + \hat{k}$ in vector form is

Feb 23, 2018

We can say , $\vec{v} . \vec{a} = | \vec{v} | | \vec{a} | \cos \theta$ (where, $\theta$ is the angle between the two vector)

So, $| \vec{v} | \cos \theta = \frac{\vec{v} . \vec{a}}{| \vec{a} |}$

Now, $| \vec{v} | \cos \theta$ is defined as the projection of $\vec{v}$ on $\vec{a}$

So, the magnitude of the projection is $\frac{6 + 2 - 2}{\sqrt{3}} = 2 \sqrt{3}$ (given , $\vec{v} = 6 i + 2 j - 2 k$ and $\vec{a} = i + j + k$)

So,the vector will be $2 \sqrt{3} \frac{i + j + k}{| \left(i + j + k\right) |} = 2 \left(i + j + k\right)$ (as parallel to $\vec{a}$,so just multiplied the unit vector along $\vec{a}$ with the magnitude of projection to get the vector)

Feb 23, 2018

$2 \hat{i} + 2 \hat{j} + 2 \hat{k}$

#### Explanation:

Given a vector $\vec{v}$ and a unit vector $\hat{n}$, you can divide the vector $\vec{v}$ into two parts

$\vec{v} = \vec{p} + \vec{q}$

where $\vec{p}$ and $\vec{q}$ are parallel and perpendicular to $\hat{n}$, respectively. Now, the magnitude of $\vec{p}$ is given by $\vec{p} \cdot \hat{n}$, and since $\vec{q} \cdot \hat{n} = 0$ we have

$| \vec{p} | = \vec{p} \cdot \hat{n} = \vec{v} \cdot \hat{n}$

Thus,

$\vec{p} = | \vec{p} | \hat{n} = \left(\vec{v} \cdot \hat{n}\right) \hat{n}$

Since the unit vector in the direction of $\vec{a}$ is given by $\frac{\vec{a}}{| \vec{a} |}$, the component of $\vec{v}$ parallel to $\vec{a}$ is given by

(vec{v} cdot {vec{a}}/{|vec{a}|}){vec{a}}/{|vec{a}|}= {(vec{v} cdot vec{a})vec{a}}/{|vec{a}|^2

So, for the given problem, the component that we seek is

$\frac{\left(6 \hat{i} + 2 \hat{j} - 2 \hat{k}\right) \cdot \left(\hat{i} + \hat{j} + \hat{k}\right)}{\left(\hat{i} + \hat{j} + \hat{k}\right) \cdot \left(\hat{i} + \hat{j} + \hat{k}\right)} \left(\hat{i} + \hat{j} + \hat{k}\right) = 2 \hat{i} + 2 \hat{j} + 2 \hat{k}$

Feb 23, 2018

Component of the velocity parallel to the vector
$\hat{i} + \hat{j} + \hat{k}$
is

$2 \left(\hat{i} + \hat{j} + \hat{k}\right)$

#### Explanation:

Given:

$\vec{v} = 6 \hat{i} + 2 \hat{j} - 2 \hat{k}$

Projection needed along

$\vec{a} = \hat{i} + \hat{j} + \hat{k}$

$\frac{\left(\vec{v} \cdot \vec{a}\right)}{| \vec{a} {|}^{2}} \vec{a}$

$\vec{v} \cdot \vec{a} = \left(6 \hat{i} + 2 \hat{j} - 2 \hat{k}\right) \cdot \left(\hat{i} + \hat{j} + \hat{k}\right)$

$= 6 \times 1 + 2 \times 1 - 2 \times 1 = 6 + 2 - 2 = 6$

$\vec{v} \cdot \vec{a} = 6$

$| \vec{a} {|}^{2} = | \hat{i} + \hat{j} + \hat{k} {|}^{2} = {1}^{2} + {1}^{2} + {1}^{2} = 1 + 1 + 1 = 3$

$| \vec{a} {|}^{2} = 3$

((vecv*veca))/(|veca|^2)veca=6/3(hati+hatj+hatk)=2(hati+hatj+hatk

$\frac{\left(\vec{v} \cdot \vec{a}\right)}{| \vec{a} {|}^{2}} \vec{a} = 2 \left(\hat{i} + \hat{j} + \hat{k}\right)$

Component of the velocity parallel to the vector
hati+hatj+hatk)
is

$2 \left(\hat{i} + \hat{j} + \hat{k}\right)$