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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

3 Answers
Jane
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 = (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)

Ananda Dasgupta
Feb 23, 2018

#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}#

Shiva Prakash M V
Feb 23, 2018

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)#

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