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

The vector equation of a line can be expressed in the form:

ar{r} = ar{a} + tar{d}

where ar{a} is a point on the line (we can use point A) and ar{d} is the direction vector (which is ar{AB}):

Substituting in: ar{r} = (2i + 6j - k) + t(1i - 2j + 2k) This simplifies to:

ar{r} = (2 + t)i + (6 - 2t)j + (-1 + 2t)k

To find the acute angle $heta$ between lines $l_1$ and $l_2$, we can use the direction vectors of these lines:

1. Direction vector of line $l_1$: ar{AB} = (1, -2, 2)
2. Direction vector of line $l_2$: $(1, 0, 1)$

We use the dot product formula:

ar{a} ullet ar{b} = |ar{a}| |ar{b}| ext{cos}( heta)

Calculating the dot product:

ar{AB} ullet (1, 0, 1) = 1(1) + (-2)(0) + 2(1) = 3

Calculating the magnitudes:

|ar{AB}| = \\sqrt{(1^2 + (-2)^2 + 2^2)} = \\sqrt{9} = 3 $|l_2| = \\sqrt{(1^2 + 0^2 + 1^2)} = \\sqrt{2}$

So we have:

ightarrow ext{cos}( heta) = 1 ightarrow heta = 0^ ext{o}$$ The acute angle between the two lines is therefore $0^ ext{o}$.

Knowing that the line $l_2$ passes through the origin and has a direction vector of $(1,0,1)$, we can express $l_2$ as:

ar{r}_2 = s(1i + 0j + 1k)

To find the intersection, we set: ar{r}_1 = ar{r}_2 From our earlier equation: $2 + t = s$ $6 - 2t = 0$ $-1 + 2t = s$

Solving gives: From the second equation: $6 - 2t = 0 ightarrow t = 3$. Substituting $t = 3$ into the first equation:

ightarrow s = 5$$ Thus: Now substituting $s = 5$ back into $ar{r}_2$ to find C: $$ar{r}_2 = 5(1i + 0j + 1k) = 5i + 0j + 5k$$ Therefore, the position vector of point C is: $$ar{C} = 5i + 0j + 5k$$