Why Does the Cosine of Angle OAB Calculate to Negative Two Thirds?

[nokia8650]

Relative to a fixed origin O, the point A has position vector 4i + 8j – k, and the point B has position vector 7i + 14j + 5k.

(a) Find the vector A to B.

(b) Calculate the cosine of OAB.

(c) Show that, for all values of t, the point P with position vector ti + 2tj + (2t - 9)k
lies on the line through A and B.

I am having problem with parts b and c.

For a) - A to B is (3i + 6j + 6k)

for b), I get the cosine of the angle using the dot product formula to be positive 2/3 - however the markscheme says negative 2/3.

I am really stumped with c - I don't know how to tackle this problem. I can find the equation of the line AB to be r= (4i + 8j - k) + t(i + 2j + 2k). However, I don't understand how this ties into the question. I can see that the "t" coefficients are the same, but where does the -9 in the question come from?

Thanks

 

[Dick]

For computing the angle OAB I think they want you to find the vectors relative to the origin A (i.e. two arrows going outward from A). One is the AB that you found, but the vector from A to O is actually -A. For c) there are lots of different ways of writing the parametric equation of corresponding to changes in the parameter t. You line equation is (t+4)i+(2t+8)j+(2t-1)k. Try substituting t->t-4. Same line, different equation.