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c4 maths

http://mei.org.uk/files/papers/c4_june_2012.pdf
question 8 can someone run me through he solutions and give me questions like these so i can practice vectors c4.
Original post by robert150
http://mei.org.uk/files/papers/c4_june_2012.pdf
question 8 can someone run me through he solutions and give me questions like these so i can practice vectors c4.


8i) To show A'A is a parallel vector, all you need to do is first find what it is. Then you need to argue that it is a multiple of the vector (12−3)\begin{pmatrix} 1 \\ 2 \\ -3 \end{pmatrix} because the vector equation of that plane is r⋅(12−3)=0\mathbf{r} \cdot \begin{pmatrix} 1 \\ 2 \\ -3 \end{pmatrix} = 0, and from the theory you should know that the eq. of the plane is in the form r⋅n=d\mathbf{r} \cdot \mathbf{n} = d where n\mathbf{n} is a perpendicular vector to the plane.

8ii) The position vector of BB is precisely the point where the line r=(124)+λ(1−12)\mathbf{r} = \begin{pmatrix} 1 \\ 2 \\ 4 \end{pmatrix} + \lambda \begin{pmatrix} 1 \\ -1 \\ 2 \end{pmatrix} intersects the plane r⋅(12−3)=0\mathbf{r} \cdot \begin{pmatrix} 1 \\ 2 \\ -3 \end{pmatrix} = 0 so sub it its equation and solve for λ\lambda and sub that back into the line's equation to determine the actual point.

8iii) To find θ\theta, you get it from using cos⁡θ=a⋅b∣a∣∣b∣\cos \theta = \dfrac{\mathbf{a} \cdot \mathbf{b}}{|\mathbf{a}| |\mathbf{b}|} equation. Clearly, this angle is between the vectors AB and CB. But you don't know C, and in fact you don't need it, because the direction of the vector CB is exactly that of BA'.
So apply that formula with a=AB and b=BA' to get the angle.

8iv) You can determine the eq. of the line BC, and hence set y=0y=0, solve for that parameter, and sub it back in to get your x and z coords.

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