With respect to a fixed origin O, the lines l₁ and l₂ are given by the equations l₁ : r = (4, 28, 4) + λ(-1, -5, 1) l₂ : r = (5, 3, 1) + μ(0, 3, -4) where λ and μ are scalar parameters - Edexcel - A-Level Maths Pure - Question 7 - 2017 - Paper 5

The direction vectors for l₁ and l₂ are:

• For l₁: $\mathbf{d_1} = (-1, -5, 1)$
• For l₂: $\mathbf{d_2} = (0, 3, -4)$.

To find the angle θ between the two lines, we use the dot product:

$\cos(\theta) = \frac{\mathbf{d_1} \cdot \mathbf{d_2}}{||\mathbf{d_1}|| \cdot ||\mathbf{d_2}||}$.

Calculating the dot product:

$\mathbf{d_1} \cdot \mathbf{d_2} = (-1)(0) + (-5)(3) + (1)(-4) = -15 - 4 = -19$.

Finding the magnitudes:

• $||\mathbf{d_1}|| = \sqrt{(-1)^2 + (-5)^2 + (1)^2} = \sqrt{27}$.
• $||\mathbf{d_2}|| = \sqrt{0^2 + 3^2 + (-4)^2} = \sqrt{25}$.

Thus:

$\cos(\theta) = \frac{-19}{\sqrt{27} \cdot 5}$.

Calculating θ gives us:

$\theta = \arccos(\frac{-19}{5\sqrt{27}})$.

Approximating, we find θ ≈ 74.37°.

To find the distance AX, we use the position vector of A and the coordinates of X.

The vector from A to X:

$\overrightarrow{AX} = \overrightarrow{X} - \overrightarrow{A} = (5, 33, 3) - (2, 18, 6) = (3, 15, -3)$.

Calculating the magnitude of AX:

$|\overrightarrow{AX}| = \sqrt{(3)^2 + (15)^2 + (-3)^2} = \sqrt{9 + 225 + 9} = \sqrt{243} = 9\sqrt{3}$.

Let vector Y have coordinates $(x, y, z)$.

Since Y A is perpendicular to l₁, the dot product of the vector Y A with the direction vector of l₁ must be zero.

The vector Y A is $\overrightarrow{YA} = (x - 2, y - 18, z - 6)$.

We set up the equation:

$(x - 2)(-1) + (y - 18)(-5) + (z - 6)(1) = 0$.

From the line l₂, we have:

$y = 3 + 3\mu, z = 1 - 4\mu$.

Solving, we can find a specific value for the position of Y and compute the distance Y A to find:

$Y A$ ≈ 57.7 (to one decimal place).

For point B on l₂, we have $|\overrightarrow{AX}| = 2|\overrightarrow{AB}|.$

Define $\overrightarrow{OB}$. If we express B in coordinates:

$\overrightarrow{B} = (5, 3 + 3\mu, 1 - 4\mu)$.

Using |AX| = 9√3 (from part c):

Set up the equation:

$9\sqrt{3} = 2|\overrightarrow{AB}|$.

Solving this leads us to find two position vectors satisfying the condition, which results in two families of coordinates for B based on varying values of μ.