Answer to Question #139092 in Electricity and Magnetism for Arsalan

Question

If A=10ax-4ay+6az and B=2ax+ay, find a the component of A along ay b the magnitude of 3A-B, c a unit vector along A+2B

Answer 1

$A=10a_x-4a_y+6a_z$ and $B=2a_x+a_y$

a) If $A=A_xa_x+A_ya_y+A_za_z$ , then the component of $A$ along $a_y$ is $A_y$ .

For $A=10a_x-4a_y+6a_z$ we have that $A_y=-4$ is the component of $A$ along $a_y$.

b) $3A-B=3 (10a_x-4a_y+6a_z )-(2a_x+a_y)=(3\times 10-2)a_x+(3\times(-4)-1)a_y+3\times6a_z$

$3A-B=28a_x-13a_y+18a_z$

$|3A-B|=\sqrt{28^2+(-13)^2+18^2}=\sqrt{1277}=35.74$

c) $A+2B= 10a_x-4a_y+6a_z+2(2a_x+a_y)=(10+2\times2)a_x+(-4+2)a_y+6a_z$

$A+2B=14a_x-2a_y+6a_z$

A unit vector along $A+2B$ is vector $\frac{14a_x-2a_y+6a_z}{\sqrt{14^2+(-2)^2+6^2}}= \frac{14a_x-2a_y+6a_z}{\sqrt{236}}\approx0.91a_x-0.13a_y+0.39a_z$

Answer: a) $-4$ ; b) $35.74$ ; c) $0.91a_x-0.13a_y+0.39a_z$