1. ## c4 vector question

The point A with coordinates (0,-5,11) lies on the line l1, which has equation

r=6i+19j-k+(lamda)
(i+4-2k)

The point p lies on l1 and is such that OP is perindicular to l1, where O is the origin.

Find the position vectore of point P.

Given that B had coordinates (5,15,1)

Show that the points A, P and B are collinear and find ratio AP:PB

Thanks

Bryn

2. Have you found the values of $\lambda$ that corresponds to each point a , b and p ?

Once you find them you can consider differences of values of $\lambda$ as distances between points on the line, that should help with finding the ratio.

Finding the values of $\lambda$ for a should not be a problem

for p, your given it lies on $l_{1}$ so the position of p is $\left(\begin{array}{c}6 + \lambda \\19 + 4 \lambda \\-1 - 2 \lambda\end{array}\right)$ and you're given that OP it is perpendicular to the the line $l_{1}$ so $\left(\begin{array}{c}6 + \lambda \\19 + 4 \lambda \\-1 - 2 \lambda\end{array}\right) \cdot \left(\begin{array}{c}6 \\19 \\-1\end{array}\right) = 0$, So now you can solve for $\lambda$

Bobak