Calculating QP, QR, and the Angle PQR

[lemon]

1. Three points P, Q and R have position vectors p, q and r respectively, where:
p=8i+11j, q=7i-5j and r=2i+4j.
Write down the vectors QP and QR and show that they are not perpendicular. Hence determine the angle PQR.



2. |QP|.|QR|cos(theta)




3. QP= QO+OP=i+16j
QR=QO+OR=-5i+9j
|QP|=root257
|QR|=root106

If the vectors are not perpendicular then a.b=0
a.b= QP=i+16j x QR=-5i+9j = (1)(-5)+(16)(9)=-5+144=139 - not perpendicular

a.b=|QP|.|QR|cos(theta)
cos(theta)=a.b/|QP|.|QR|=139/root257 x root106 = 32.6

Could anybody check to see how pathetic this attempt is, please?

 

[tiny-tim]

Hi lemon!

Yes, that looks fine, except for the very last bit (the 32.6).

 

[lemon]

errr. can't see it
thats the same figure i keep getting out.

?

 

[tiny-tim]

cos(theta)=a.b/|QP|.|QR|=139/root257 x root106 = 32.6

i don't know how you get that …

139/√257√106 is approximately 139/16*10 < 1

 

[lemon]

yeah but that is cos(theta), right?
so to find theta i need to inverse cos - 32.6

 

[tiny-tim]

yeah but that is cos(theta), right?
so to find theta i need to inverse cos - 32.6

But it isn't 32.6!

 

[lemon]

it is on my calculator. I get 0.8421613497
inverse cos = 32.63093847

 

[lemon]

errr. can't see it
thats the same figure i keep getting out.

?

 

[tiny-tim]

it is on my calculator. I get 0.8421613497
inverse cos = 32.63093847

ohhh! you wrote cos(theta) = 32.6 …

cos(theta)=a.b/|QP|.|QR|=139/root257 x root106 = 32.6

yes, theta = 32.6º is fine.

 

[lemon]

thanks tT