Vector scalar product problem?

We are asked to calculate the angle theta, i.e. the angle between BC and BA.

In part (ii), the direction vector of line A'B turns out to be (-2,-1, 1). The direction vector of BC is therefore (-2, -1, 1) as we are told that A'BC is a straight line.

Additionally, we also know that that the direction vector of AB is (1,-1,2).

Since $\lambda = -1$, the direction vector (1,-1,2) must be pointing away from B. Hence the dot product calculated in the mark scheme is correct.

They tricked you a bit by giving you the "wrong" direction for the direction vector.

This might sound slightly stupid but what do you mean by lambda = -1 and how did you know that? I'm not the best with vectors!

1. They give you the vector equation of AB in which $\lambda$ appears. It turns out that AB intersects the plane at B when $\lambda=-1$.

Since $\lambda$ is the coefficient of the direction vector in the equation for the line, then if it is negative, the quantity $\lambda(1,-1,2)^T$ points in the opposite direction to the direction vector. However, it must point in the direction of $\vec{AB}$ else $(1,2,4)^T+\lambda(1,-1,2)^T$ wouldn't be the position vector for B.

Thus the direction vector $(1,-1,2)^T$ points in the opposite direction to the arrow that they've shown on the line AB, and hence in the correct direction for calculating the dot product with the other direction vector (as long as you calculate that as $\vec{A'B} = (0,3,2)^T-(2,4,1)^T$ - that's what they've used in the mark scheme)

I'm not sure if this is entirely clear - it would be much easier to explain it face-to-face. If it doesn't make any sense at all, however, you may need to revise vector equations of lines.

2. I knew that $\lambda=-1$ since I calculated it incorrectly to be -3/2, then looked at the mark scheme to get the right answer :-)