Question

9.3 Resolve the force vector F = (990 N)i - (330 N) j + (660 N)k into two components, one parallel to vector a = i + 3k , the other perpendicular to a .

Answer 1

@Arnab09 would you please explain how you got it?

Answer 2

yeah, obviously..

Answer 3

Your perpendicular component's dot product with a is not zero!!!

Answer 4

lemme check again

Answer 5

No worries:)

Answer 6

oh sorry, my method was wrong.. :(
trying it afresh

Answer 7

yeah, answer is coming:
297Ni+891Nj (parallel)
693Ni-330Nj-231Nk (perpendicular)

Answer 8

@ash2326 , check it :)

Answer 9

Isn't the sum of these two vectors be F?

Answer 10

yes, of course..

Answer 11

F=990i-330j+660k
a=i+3k
Let's find the angle between F and a
let A be unit vector parallel to a
\[A= (a)/|a|=\frac{i+3k}{\sqrt {10}}\]
let Pa be F's component parallel to a
\[Pa= (F.A)A\]
\[Pa=((990i-330j+660k).\frac{i+3k}{\sqrt {10}})\frac{i+3k}{\sqrt {10}}\]
\[Pa=(990+1980)\frac{i+3k}{ 10}\]
\[Pa=297i+891j\]
Perpendicular component Pe
\[Pe=F-Pa=990i-330j+660k-(297i+891k)\]
\[Pe=693i-330j-231k\]
So parallel Component=297i+891j
Perpendicular Component=693i-330j-231k

Answer 12

@ash2326 , it can be solved in easier way, i guess

Answer 13

@Arnab09 great work but next time explain your answer. Just giving answers won't help the user who has asked the question.

Answer 14

yep..

Answer 15

i solved it this way,
say, the perpendicular vector is xi+yj+zk
so, dot product of this vector and a will be zero
so, after dot product, we get a condition: x+3z=0
also, the other vector is (990-x)i-330j+(660-z)k
but this is parallel to a
so, we get another condition: 3(990-x)=660-z
solving these two, we can get x and z

Answer 16

also, the parallel vector should not have any j component, as it is parallel to a. so, y=-330