Question

I'm taking Calc 3 right now, and we're learning about vectors. "A constant force of magnitude 20 N has the same direction as the vector T = 2i + j - 2k. Distance is measured in meters, find the work done if the point of application moves along the straight line path from the point P(2,3,1) to the point Q(5,7,3)."

I've got so far as finding that the Distance = PQ = <3,4,2> meters and found that force is 20/3<2i+j-2k> Newtons. I also know that Work = Force x Distance. I just don't know if I leave the "i,j,k" as is, or combine them with the distance?

Answer 1

What i'm asking is, does \[\frac{ 20 }{ 3}<2i+j-2k>N \times <3,4,2>m = \frac{20}{3}<6i+4j-4k>Nm?\]or \[\frac{20}{3}<6+4-4>Nm=40Nm?\]

Answer 2

The work will be the magnitude of the force, times , the magnitude of the displacement, times, cosine of the angle between the two

Answer 3

So you got Vector PQ, and the direction of the force <T>. Find the angle between those two vectors.

Answer 4

They gave you the magnitude of the force 20 N, and the magnitude of the displacement is the magnitude of PQ.

Just find the angle between <PQ> and <T> using the dot product. THen multiply ..

Work = 20 * (magnitude of <PQ>) * COS( angle btw PQ and T)

Answer 5

So would it be,\[20(\sqrt{29})(\cos(68.2))=7.43\sqrt{29}?\] @DanJS

Answer 6

Do i just get rid of the i j k?

Answer 7

Is <2i+j-2k> = <2,1,-2>??

Answer 8

yes,

i = <1,0,0>
j = <0,1,0>
k= <0,0,1>

Answer 9

Magnitude of a vector,
\[\left| \left| <a,b,c> \right| \right|=\sqrt{a^2+b^2+c^2}\]

Answer 10

Angle between 2 vectors, from the dot product.
\[\cos(\theta)=\frac{ a*b }{ \left| \left| a \right| \right|\left| \left| b \right| \right| }\]

Answer 11

the i,j,k is just a different way to represent a vector.

For example
\[A = 4i-6j+8k\]is the same as \[A=4*<1,0,0>-6*<0,1,0>+8*<0,0,1> = <4,-6,8>\]

Answer 12

ahh okay i've got it! Thank you for the reply!