Hard Vectors question - Please help!!!

LouN1997

Find the coordinates of P if OP=3 (distance from origin) and the vector OP is in the direction of i - 2j + 3k

Reply 1

kennz

Original post by LouN1997

Find the coordinates of P if OP=3 (distance from origin) and the vector OP is in the direction of i - 2j + 3k

is the 3k part of the j component?

Reply 2

Ishan_2000

Need some kind of picture or diagram of the question to answer it.

Posted from TSR Mobile

Reply 3

LouN1997

Original post by kennz

is the 3k part of the j component?

no, separate component, 3d vector

Reply 4

LouN1997

Original post by Ishan_2000

Need some kind of picture or diagram of the question to answer it.

Posted from TSR Mobile

that is all the information I was given

Reply 5

Infraspecies

Find the magnitude of the direction vector, find the proportion between the magnitude of the direction vector and the magnitude of OP

Reply 6

LouN1997

Original post by Infraspecies

Find the magnitude of the direction vector, find the proportion between the magnitude of the direction vector and the magnitude of OP

thank you very much

Reply 7

kennz

Original post by LouN1997

Find the coordinates of P if OP=3 (distance from origin) and the vector OP is in the direction of i - 2j + 3k

I'm not too sure but I did OP^2=1^2+2^2+(3k)^2 so 9=9k^2+5 so rearrange to get k=2/3.

Reply 8

Infraspecies

Original post by LouN1997

thank you very much

Essentially what you're doing is finding a unit vector from the direction vector, then multiplying it by |OP|.

Reply 9

Infraspecies

Original post by kennz

I'm not too sure but I did OP^2=1^2+2^2+(3k)^2 so 9=9k^2+5 so rearrange to get k=2/3.

k is a basis vector, not a constant part of the magnitude of the j component.

Reply 10

Luke Kostanjsek

Original post by LouN1997

Find the coordinates of P if OP=3 (distance from origin) and the vector OP is in the direction of i - 2j + 3k

The vector OP is in the direction of the given vector, and the magnitude of our given vector is sqrt(1+4+9) which is sqrt(14). We want this vector to have magnitude 3, so we simply multiply each term by 3/sqrt(14). This gives you the answer:

(3/sqrt{14})i -(6/sqrt{14})j +(9/sqrt{14})k

Hope that helps :smile:

Reply 11

kennz

Original post by Infraspecies

k is a basis vector, not a constant part of the magnitude of the j component.

Yeah I just looked this up and its c4 vectors I think haha😊

Reply 12

Luke Kostanjsek

Strictly, I've taken a shortcut there cause I've done these sort of questions before. If I wanted to give my full working, it'd be more like this:

The vector OP is in the direction of our given vector. Therefore, OP is of the form X(i-2j+3k)= Xi-2Xj+3Xk, where X is a variable scalar. Therefore, the magnitude OP is:

sqrt(X^2 + 4X^2 +9X^2)=sqrt(14X^2)=Xsqrt(14)

We want to have magnitude 3, so therefore Xsqrt(14)=3

rearranging gives X=3/sqrt(14)

We now know what value our variable scalar must take to give our vector magnitude 3, so we plug it back in to our equation for OP to give the aforementioned answer