Relative to a fixed origin O, the point A has position vector (10i + 2j + 3k), and the point B has position vector (8i + 3j + 4k) - Edexcel - A-Level Maths Pure - Question 1 - 2012 - Paper 7

A vector equation for the line l can be expressed as:

$$\vec{r} = \vec{A} + t \vec{AB}$$

where ( t ) is a parameter.

Substituting the values:

$$\vec{r} = (10i + 2j + 3k) + t(-2i + j + k) = (10 - 2t)i + (2 + t)j + (3 + t)k$$

Therefore, the vector equation of the line l is:

$$\vec{r} = (10 - 2t)i + (2 + t)j + (3 + t)k.$$

Let the position vector of point P be denoted as ( \vec{P} = (x, y, z) ).

To find the coordinates, we use that ( \vec{CP} ) is perpendicular to the direction vector of line l, which can be represented as ( \vec{AB} = (-2i + j + k) ).

First, we need to express ( \vec{CP} ):

$$\vec{CP} = (x - 3)i + (y - 12)j + (z - 3)k$$

Since ( \vec{CP} ) is perpendicular to ( \vec{AB} ), their dot product must be zero:

$$\vec{CP} \cdot \vec{AB} = 0$$

Substituting:

$$(x - 3)(-2) + (y - 12)(1) + (z - 3)(1) = 0$$

Expanding gives:

$$-2x + 6 + y - 12 + z - 3 = 0$$

Thus,

$$-2x + y + z - 9 = 0$$

We also know from the vector equation of line l that:

$$(x, y, z) = (10 - 2t, 2 + t, 3 + t)$$

Now substituting these into our equation:

$$-2(10 - 2t) + (2 + t) + (3 + t) - 9 = 0$$

Simplifying:

$$-20 + 4t + 2 + t + 3 + t - 9 = 0$$ $$6t - 24 = 0 \implies t = 4$$

Substituting ( t = 4 ) back into the vector equation gives us:

$$\vec{P} = (10 - 2(4), 2 + 4, 3 + 4) = (2, 6, 7)$$

Therefore, the position vector of point P is:\n $\vec{P} = (2i + 6j + 7k)$.