Question

Find the separation vector r from the source point (2,8,7) to the field point (4,6,8). Determine its magnitude (r), and construct the unit vector ˆr.

Answer 1

Answer:

magnitude =3 unit vector $\frac{2i-2j+k}{3}$

Step-by-step explanation:

We have given source point $(x_1,y_1,z_1)=(2,8,7)$

And the field point $(x_2,y_2,z_2)=(4,6,8)$

Resultant vector $=(x_2-x_1)i+(y_2-y_1)j+(z_2-z_1)k=(4-2)i+(6-8)j+(8-7)k=2i-2j+k$

The resultant vector $=2i-2j+k$

Magnitude of the vector $=\sqrt{2^2+2^2+1^2}=3$

Now we have to find the unit vector

So unit vector $=\frac{2i-2j+k}{3}$