Question

Please solve with a detailed explanation?

The velocity of a particle is `vec v =6hati +2hatj-2hatk`.The component of the velocity parallel to vector `veca=hati +hatj +hatk` in vector form is

Answer 1

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 = (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 `(6+2-2)/sqrt(3) = 2sqrt(3)` (given , `vec v=6i+2j-2k` and `vec a=i+j+k`)

So,the vector will be `2sqrt(3)(i+j+k)/(|(i+j+k)| )= 2(i+j+k)` (as parallel to `vec a`,so just multiplied the unit vector along `vec a` with the magnitude of projection to get the vector)

Answer 2

`2hat{i}+2hat{j}+2hat{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} = ( vec{v} cdot hat{n}) hat{n}`

Since the unit vector in the direction of `vec{a}` is given by `{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

`{(6 hat{i}+2 hat{j}-2 hat{k}) cdot (hat{i}+hat{j}+hat{k})} /{(hat{i}+hat{j}+hat{k})cdot (hat{i}+hat{j}+hat{k})} (hat{i}+hat{j}+hat{k}) = 2hat{i}+2hat{j}+2hat{k}`

Answer 3

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

`2(hati+hatj+hatk)`

Explanation:

Given:

`vecv=6hati+2hatj-2hatk`

Projection needed along

`veca=hati+hatj+hatk`

`((vecv*veca))/(|veca|^2)veca`

`vecv*veca=(6hati+2hatj-2hatk)*(hati+hatj+hatk)`

`=6xx1+2xx1-2xx1=6+2-2=6`

`vecv*veca=6`

`|veca|^2=|hati+hatj+hatk|^2=1^2+1^2+1^2=1+1+1=3`

`|veca|^2=3`

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

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

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

`2(hati+hatj+hatk)`