Question 1.1: If A=10ax-4ay+6az and B=2ax+ay, find (a) the component of A ...

(a) The component of A along a_{y} is A_{y}=-4.

(b) 3A – B = 3(10, -4, 6) – (2, 1, 0)
= (30, -12, 18) – (2, 1, 0)
= (28, -13, 18)

Hence,

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

(c) Let C = A + 2B = (10, -4, 6) + (4, 2, 0) = (14, -2, 6). A unit vector along C is

a_{c}=\frac{C}{\left|C\right| }=\frac{\left(14,-2,6\right) }{\sqrt{14^{2}+\left(-2\right)^{2}+6^{2}  } }

or

a_{c}=0.9113a_{x}-0.1302a_{y}+0.3906a_{z}

Note that \left|a_{c}\right|=1 as expected.