Question 1.2: Points P and Q are located at (0, 2, 4) and (-3, 1, 5). Calc...

(a) r_{P}=0a_{x}+2a_{y}+4a_{z}=2a_{y}+4a_{z}

(b) r_{PQ}=r_{Q}-r_{P}=\left(-3,1,5\right)-\left(0,2,4\right)=\left(-3,-1,1\right)

or r_{PQ}=-3a_{x}-a_{y}+a_{z}

(c) Since r_{PQ} is the distance vector from P to Q, the distance between P and Q is the magnitude of this vector; that is,

d=\left|r_{PQ}\right|=\sqrt{9+1+1}=3.317

Alternatively:

d=\sqrt{\left(x_{Q}-x_{P}\right)^{2}+\left(y_{Q}-y_{P}\right)^{2}+\left(z_{Q}-z_{P}\right)^{2}}

=\sqrt{9+1+1}=3.317

(d) Let the required vector be A, then

A=Aa_{A}

where A=10 is the magnitude of A. Since A is parallel to PQ, it must have the same unit vector as r_{PQ} or r_{QP}. Hence,

a_{A}=\pm\frac{r_{PQ}}{\left|r_{PQ}\right|}=\pm\frac{\left(-3,-1,1\right)}{3.317}

and

A=\pm \frac{10\left(-3,-1,1\right) }{3.317}=\pm\left(-9.045a_{x}-3.015a_{y}+3.015a_{z} \right)