Please solve with a detailed explanation?

The velocity of a particle is vec v =6hati +2hatj-2hatkv=6ˆi+2ˆj2ˆk.The component of the velocity parallel to vector veca=hati +hatj +hatka=ˆi+ˆj+ˆk in vector form is

3 Answers
Feb 23, 2018

We can say , vec v.vec a = |vec v| |vec a| cos thetav.a=vacosθ (where, thetaθ is the angle between the two vector)

So, |vec v| cos theta = (vec v . vec a)/(|vec a|)vcosθ=v.aa

Now, |vec v| cos thetavcosθ is defined as the projection of vec vv on vec aa

So, the magnitude of the projection is (6+2-2)/sqrt(3) = 2sqrt(3)6+223=23 (given , vec v=6i+2j-2kv=6i+2j2k and vec a=i+j+ka=i+j+k)

So,the vector will be 2sqrt(3)(i+j+k)/(|(i+j+k)| )= 2(i+j+k)23i+j+k|(i+j+k)|=2(i+j+k) (as parallel to vec aa,so just multiplied the unit vector along vec aa with the magnitude of projection to get the vector)

Feb 23, 2018

2hat{i}+2hat{j}+2hat{k}2ˆi+2ˆj+2ˆk

Explanation:

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

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

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

|vec{p}| = vec{p} cdot hat {n} = vec{v} cdot hat{n}p=pˆn=vˆn

Thus,

vec{p} = |vec{p}| hat{n} = ( vec{v} cdot hat{n}) hat{n} p=pˆn=(vˆn)ˆn

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

(vec{v} cdot {vec{a}}/{|vec{a}|}){vec{a}}/{|vec{a}|}= {(vec{v} cdot vec{a})vec{a}}/{|vec{a}|^2vaaaa=(va)aa2

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}(6ˆi+2ˆj2ˆk)(ˆi+ˆj+ˆk)(ˆi+ˆj+ˆk)(ˆi+ˆj+ˆk)(ˆi+ˆj+ˆk)=2ˆi+2ˆj+2ˆk

Feb 23, 2018

Component of the velocity parallel to the vector
hati+hatj+hatkˆi+ˆj+ˆk
is

2(hati+hatj+hatk)2(ˆi+ˆj+ˆk)

Explanation:

Given:

vecv=6hati+2hatj-2hatkv=6ˆi+2ˆj2ˆk

Projection needed along

veca=hati+hatj+hatka=ˆi+ˆj+ˆk

((vecv*veca))/(|veca|^2)veca(va)a2a

vecv*veca=(6hati+2hatj-2hatk)*(hati+hatj+hatk)va=(6ˆi+2ˆj2ˆk)(ˆi+ˆj+ˆk)

=6xx1+2xx1-2xx1=6+2-2=6=6×1+2×12×1=6+22=6

vecv*veca=6va=6

|veca|^2=|hati+hatj+hatk|^2=1^2+1^2+1^2=1+1+1=3a2=ˆi+ˆj+ˆk2=12+12+12=1+1+1=3

|veca|^2=3a2=3

((vecv*veca))/(|veca|^2)veca=6/3(hati+hatj+hatk)=2(hati+hatj+hatk(va)a2a=63(ˆi+ˆj+ˆk)=2(ˆi+ˆj+ˆk

((vecv*veca))/(|veca|^2)veca=2(hati+hatj+hatk)(va)a2a=2(ˆi+ˆj+ˆk)

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

2(hati+hatj+hatk)2(ˆi+ˆj+ˆk)