Analyzing Three-Dimensional Line Intersections

00:00 10 marks

Given three lines in 3D space with vector equations:

Line 1: \( \mathbf{r} = \begin{pmatrix} 1 \\ 2 \\ 3 \end{pmatrix} + t \begin{pmatrix} 2 \\ -1 \\ 4 \end{pmatrix} \)
Line 2: \( \mathbf{r} = \begin{pmatrix} 3 \\ 0 \\ 5 \end{pmatrix} + s \begin{pmatrix} 0 \\ 1 \\ 2 \end{pmatrix} \)
Line 3: \( \mathbf{r} = \begin{pmatrix} a \\ b \\ c \end{pmatrix} + u \begin{pmatrix} p \\ q \\ r \end{pmatrix} \)

(a) Determine the intersection point of lines 1 and 2. Find the values of the parameters \(t\) and \(s\) at the point of intersection and state the coordinates of this point.

(b) State the condition on the parameters \(a, b, c, p, q, r\) such that the intersection point found in part (a) lies on line 3. Explain what this condition means in terms of the concurrency of the three lines. Consider cases where \(p, q,\) or \(r\) could be zero.

Part a

6 minutes 5 marks

Question

Find the intersection point of lines 1 and 2 by equating their parametric equations. Determine the values of \(t\) and \(s\) and compute the coordinates of the intersection.

Hint Hint

Hint

Set up a system of linear equations by equating the x, y, and z components of Line 1 and Line 2. Solve for \(t\) and \(s\) using the first two equations and then verify in the third.

Answer Answer
Mark Scheme Mark Scheme
Explanation Explanation
Formula Booklet

Answer Answer

Complete Answer

The intersection occurs at \(t=1\) and \(s=1\), giving the intersection point \((3, 1, 7)\).

Mark Scheme Mark Scheme

Answer Explanation Answer Explanation

Formula Booklet

Part b

6 minutes 5 marks

Question

Determine the condition on \(a, b, c, p, q, r\) such that the intersection point \((3, 1, 7)\) from part (a) lies on line 3. Express your answer in terms of \(a, b, c, p, q, r\) and consider cases where \(p, q,\) or \(r\) might be zero. Explain the meaning of this condition in the context of line concurrency.

Hint Hint

Hint

Substitute the coordinates \((3, 1, 7)\) into the vector equation of Line 3. Consider what happens if any of the direction components \(p, q, r\) are zero and how this affects the conditions for \(a,b,c\).

Answer Answer
Mark Scheme Mark Scheme
Explanation Explanation
Formula Booklet

Answer Answer

Complete Answer

For the point \((3, 1, 7)\) to lie on Line 3, there must exist a scalar \(u\) such that \(3 = a + up\), \(1 = b + uq\), and \(7 = c + ur\).

Case 1: If \(p, q, r
eq 0\), the condition is \(\frac{3-a}{p} = \frac{1-b}{q} = \frac{7-c}{r}\).

Case 2: If \(p = 0\), then we must have \(a = 3\). The other components give \(u = \frac{1-b}{q}\) and \(u = \frac{7-c}{r}\) (assuming \(q, r
eq 0\)), so the condition is \(a = 3\) and \(\frac{1-b}{q} = \frac{7-c}{r}\).

Similar conditions apply if \(q = 0\) or \(r = 0\). For example, if \(q = 0\), then \(b = 1\) and \(\frac{3-a}{p} = \frac{7-c}{r}\) (assuming \(p, r
eq 0\)). If \(p=q=0\), then \(a=3, b=1\) and \(u = \frac{7-c}{r}\) (assuming \(r
eq 0\)). If \(p=q=r=0\), then we must have \(a=3, b=1, c=7\), meaning Line 3’s position vector is exactly the intersection point.

This condition ensures that all three lines intersect at a single common point, which is the definition of concurrency for lines in 3D space.

Mark Scheme Mark Scheme

Answer Explanation Answer Explanation

Formula Booklet