Question 1043241
The position vector of OP, OQ and OR is <0,0,0>, <1,2,3> and <2,1,-1> respectively. Find
i) the area of the triangle PQR
<pre>
The cross product is the area of the parallelogram,
and so the area of the triangle is half that.

{{{drawing(400,400,-4,4,-5,3,

line(-12,0,12,0),line(0,-12,0,12),line(-12,-18,12,18),

green(line(-2,-3,-.5,-3), line(-.5,-3,-.5,-4.5), line(-1,-1.5,2,-1.5),
line(2,-1.5,2,1.5)),

locate(2,1.8,"Q(1,2,3)"), locate(-2,-4.5,"R(2,1,-1)"), locate(-1.6,.3,"O(0,0,0)"),
locate(-3.5,-4.7,x),locate(3.9,0,y),locate(0.1,2.97,z),
red(triangle(0,0,2,1.5,-.5,-4.5),
line(0,0,2,1.5), line(2,1.5,-.5,-4.5), line(0,0,-.5,-4.5)),
red(arc(0,0,1,-1.6,260,383))


  )}}} 

The area is &#8741;OQ×OR&#8741;/2 = {{{expr(1/2)abs(abs(matrix(3,3,i,j,k,1,2,3,2,1,-1)))}}} = &#8741;-5i+7j-3k&#8741;/2 = {{{sqrt((-5)^2+(7)^2+(-3)^2)/2=sqrt(11025)/2=105/2 = 52.5}}} 

We find the magnitude of vectors OQ and OR

&#8741;OQ&#8741; = {{{sqrt(1^2+2^2+3^2)=sqrt(14)}}}

&#8741;OR&#8741; = {{{sqrt(2^2+1^2+(-1)^2)=sqrt(6)}}}

-----------------------

ii) the projection of QR and OR

I think you meant "onto", not "and".

the projection of QR onto OR is given by

            OQ•OR
proj<sub>OR</sub>OQ  = ————— OR 
            &#8741;OR&#8741;²

                          
OQ•OR = <1,2,3>•<2,1,-1> = (1)(2)+(2)(1)+(3)(-1) = 1

&#8741;OR&#8741;² = 2²+1²+(-1)² = 4+1+1 = 6

So,

proj<sub>OR</sub>OQ = {{{expr(1/6)*"<2,1,-1>"}}}

Edwin</pre>