#### Prove that the line through points A (0, -1, -1) and B (4, 5, 1) intersects the line through C (3, 9, 4 ) and D (-4, 4, 4).

Given: A (0, -1, -1), B (4, 5, 1), C (3, 9, 4), D (-4, 4, 4).

To prove: The line passing through A and B intersects the line passing through C and D.

Proof: We know, equation of a line passing through two points (x1 , y1 , z1) and (x2 , y2 , z2) is:

$\frac{x-x_{1}}{x_{2}-x_{1}}=\frac{y-y_{1}}{y_{2}-y_{1}}=\frac{z-z_{1}}{z_{2}-z_{1}}$

Hence, the equation of the line passing through A (0, -1, -1) and B (4, 5,1) is:

$\frac{x-0}{4-0}=\frac{y-(-1)}{5-(-1)}=\frac{z-(-1)}{1-(-1)}$

, where x1 = 0, y1 = -1, z1 = -1; and x2 = 4, y2 = 5, z2 = 1

$\Rightarrow \frac{x-0}{4}=\frac{y+1}{6}=\frac{z+1}{2}\\ \Rightarrow \frac{x}{4}=\frac{y+1}{6}=\frac{z+1}{2}$

Let

$L_{1}: \frac{x}{4}=\frac{y+1}{6}=\frac{z+1}{2}=\lambda\\ \\ \frac{x}{4}=\lambda, \frac{y+1}{6}=\lambda,\frac{z+1}{2}=\lambda\\$

We must find the values of x, y, and z. Therefore,

$Take \frac{x}{4}=\lambda\\ \Rightarrow x=4\lambda\\ \\ Take \frac{y+1}{6}=\lambda\\ \Rightarrow y+1=6\lambda\\ \Rightarrow y=6\lambda-1\\ \\ Take \frac{z+1}{2}=\lambda\\ \Rightarrow z+1=2\lambda\\ \Rightarrow z=2\lambda-1\\ \\$

This implies that any point on the line L1 is (4λ, 6λ – 1, 2λ – 1).

The equation of the line passing through points C (3, 9, 4) and D (-4, 4, 4)           is:

$\frac{x-3}{-4-3}=\frac{y-9}{4-9}=\frac{z-4}{4-4}$

, where x1 = 3, y1 = 9, z1 = 4; and x2 = -4, y2 = 4, z2 = 4

$\frac{x-3}{-7}=\frac{y-9}{-5}=\frac{z-4}{0}$

Let

$L_{2}:\frac{x-3}{-7}=\frac{\left (y-9 \right )}{-5}=\frac{z-4}{0}=\mu$

$\Rightarrow \frac{x-3}{-7}=\mu,\frac{\left (y-9 \right )}{-5}=\mu,\frac{z-4}{0}=\mu$

We must find the values of x, y, and z. Therefore,

$Take \, \frac{x-3}{-7}=\mu\\ \Rightarrow x-3=-7\mu\\ \Rightarrow x=-7\mu+3\\ \\ Take \,\frac{\left (y-9 \right )}{-5}=\mu \\ \Rightarrow y-9=-5\mu\\ \Rightarrow y=-5\mu+9\\ \\ Take\frac{z-4}{0}=\mu \\ \Rightarrow z-4=0\\ \Rightarrow z=4$

This implies that any point on line L2 is (-7μ +3, -5μ + 9, 4).

If the lines intersect, then there must exist a value of λ and for μ, for which

$\left (4\lambda, 6\lambda - 1, 2\lambda -1\right) \equiv \left(-7\mu + 3, -5\mu + 9, 4\right ) \\ \Rightarrow 4\lambda = -7\mu + 3...(i)\\ 6\lambda - 1 = -5\mu + 9..(ii)\\ 2\lambda - 1 = 4 %u2026(iii)\\$

From equation (iii), we get

$2\lambda - 1 = 4 \\ \Rightarrow 2\lambda=4+1\\ \Rightarrow 2\lambda=5\\ \Rightarrow \lambda=\frac{5}{2}$

Substituting the value of λ in equation (i),

$4\left ( \frac{5}{2} \right )=-7\mu+3\\ \Rightarrow 2 \times 5=-7\mu+3\\ \Rightarrow 10=-7\mu+3\\ \Rightarrow 7\mu=3-10\\ \Rightarrow 7\mu=-7 \\ \Rightarrow -\frac{7}{7}\\ \Rightarrow \mu=-1$

Substituting these values of λ and μ in equation (ii),

$6\left ( \frac{5}{2} \right )-1=-5\left ( -1 \right )+9\\ \Rightarrow 3 \times 5 - 1 = 5 + 9\\ \Rightarrow 15 - 1 = 14\\ \Rightarrow 14 = 14$

Since the values of λ and μ satisfy eq (ii), the lines intersect.

Hence, proved that the line through A and B intersects the line through C and D.