Answer:
magnitude =3 unit vector [tex]\frac{2i-2j+k}{3}[/tex]
Step-by-step explanation:
We have given source point [tex](x_1,y_1,z_1)=(2,8,7)[/tex]
And the field point [tex](x_2,y_2,z_2)=(4,6,8)[/tex]
Resultant vector [tex]=(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[/tex]
The resultant vector [tex]=2i-2j+k[/tex]
Magnitude of the vector [tex]=\sqrt{2^2+2^2+1^2}=3[/tex]
Now we have to find the unit vector
So unit vector [tex]=\frac{2i-2j+k}{3}[/tex]
