The points A and B have position vectors $2i + 6j - k$ and $3i + 4j + k$ respectively - Edexcel - A-Level Maths Pure - Question 7 - 2008 - Paper 8

The vector equation for the line $l_1$ can be expressed in terms of a point and a direction vector. Using point A and the direction vector $\vec{AB}$, we can write:

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

Where $t$ is a scalar. Substituting in:

$\vec{r} = (2i + 6j - k) + t(i - 2j + 2k)$

Thus, the equation becomes:

$\vec{r} = (2 + t)i + (6 - 2t)j + (-1 + 2t)k$

The direction vector of line $l_1$ is $\vec{AB} = (1, -2, 2)$ and the direction vector of line $l_2$ is $\vec{d} = (1, 0, 1)$. The cosine of the angle $\theta$ between the two lines can be found using the dot product formula:

$\cos(\theta) = \frac{\vec{AB} \cdot \vec{d}}{\|\vec{AB}\| \|\vec{d}\|}$

Calculating the dot product:

$\vec{AB} \cdot \vec{d} = 1 \cdot 1 + (-2) \cdot 0 + 2 \cdot 1 = 3$

Finding the magnitudes:

$\|\vec{AB}\| = \sqrt{1^2 + (-2)^2 + 2^2} = 3$ $\|\vec{d}\| = \sqrt{1^2 + 0^2 + 1^2} = \sqrt{2}$

Thus:

$\cos(\theta) = \frac{3}{3 \cdot \sqrt{2}} = \frac{1}{\sqrt{2}}$

This gives $\theta = 45^\circ$, which is the acute angle.

To find the point C where line $l_1$ intersects line $l_2$, we express line $l_2$ as:

$\vec{r} = \lambda(1i + 0j + 1k)$

Equating this with the parametric equation of $l_1$ determined earlier:

From $l_1$: $\vec{r} = (2 + t)i + (6 - 2t)j + (-1 + 2t)k$

Setting the two equations equal:

1. $2 + t = \lambda$
2. $6 - 2t = 0$
3. $-1 + 2t = \lambda$

From equation (2): $t = 3$. Substituting $t = 3$ into the other equations:

From (1): $\lambda = 2 + 3 = 5$.
From (3): $-1 + 2(3) = 5$.

Thus, both equations are consistent; hence, the intersection point is:

$\vec{C} = 5i + 0j + 5k$

So, the position vector of point C is $5i + 0j + 5k$.