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

Concept: Collinearity Condition – Two lines intersect if there exist parameters $\lambda$ and $\mu$ such that the coordinates of a point on one line equal those on the other.

Step 1: Parametric equations of line $AB$:

$A(0,-1,-1)$, direction $\vec{AB} = (4,6,2)$.

So $P(\lambda) = (4\lambda,\ -1+6\lambda,\ -1+2\lambda)$.

Step 2: Parametric equations of line $CD$:

$C(3,9,4)$, direction $\vec{CD} = (-7,-5,0)$.

So $Q(\mu) = (3-7\mu,\ 9-5\mu,\ 4)$.

Step 3: Equate coordinates for intersection:

$4\lambda = 3 - 7\mu$

$-1 + 6\lambda = 9 - 5\mu$

$-1 + 2\lambda = 4$ …

Concept

Two lines in space intersect exactly when they are coplanar and not parallel. Coplanarity is tested by the scalar triple product of the two direction vectors and a vector joining a point on each line; if it is zero and the directions are not proportional, a unique intersection exists.

Solution

1. Direction vectors.

2. Not parallel. $(4,6,2)=k(-7,-5,0)$ is impossible (the $z$-component forces $2=0$). So they are not parallel.

3. Coplanarity. With $\vec{AC}=C-A=(3,10,5)$,

Expand along the first row:

$=4(-25)-6(-35)+2(-55)=-100+210-110=0.$ …

Method: Proving two lines (each given by two points) intersect

Use this when the lines are specified by pairs of points and you must show they meet.

Steps

Step 1: Get a point and a direction for each line.

The direction of line $AB$ is $\vec{AB}=B-A$; of line $CD$ it is $\vec{CD}=D-C$. Confirm these are NOT proportional (the lines are not parallel).

Step 2: Test coplanarity with the scalar triple product.

The lines are coplanar (a prerequisite for intersecting) when

$[\vec{AB}\;\;\vec{CD}\;\;\vec{AC}]=0,$ …

Common Mistakes

Mistake 1: Declaring the lines skew just because they are not parallel.

Why it's wrong: non-parallel lines in 3D can either intersect or be skew, so coplanarity must be tested. Correct approach: check the scalar triple product $[\vec{AB},\vec{CD},\vec{AC}]=0$, then solve for the point.

Mistake 2: Using one parameter for both lines.

Why it's wrong: the meeting point corresponds to different parameter values on each line. Correct approach: parametrise $AB$ and $CD$ separately and solve the resulting system. …