IBDP Maths AHL 1.16 Solutions of systems of linear equations AA HL Paper 2- Exam Style Questions- New Syllabus

(a)
1. Identify two vectors lying in the plane. The first is the direction vector of \( L_1 \): \( \mathbf{v}_1 = \begin{pmatrix} 1 \\ 0 \\ 1 \end{pmatrix} \).

The second is a vector from a point on the line \( (0,0,2) \) to the given point \( (2,1,5) \): \( \mathbf{v}_2 = \begin{pmatrix} 2-0 \\ 1-0 \\ 5-2 \end{pmatrix} = \begin{pmatrix} 2 \\ 1 \\ 3 \end{pmatrix} \).
2. Find the normal vector \( \mathbf{n} \) using the vector product:
\( \mathbf{n} = \mathbf{v}_1 \times \mathbf{v}_2 = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 1 & 0 & 1 \\ 2 & 1 & 3 \end{vmatrix} = \mathbf{i}(0-1) – \mathbf{j}(3-2) + \mathbf{k}(1-0) = \begin{pmatrix} -1 \\ -1 \\ 1 \end{pmatrix} \).
3. Use the point \( (0,0,2) \) to form the equation: \( -1(x-0) – 1(y-0) + 1(z-2) = 0 \).
Simplifying: \( -x – y + z – 2 = 0 \implies x + y – z = -2 \).
Shown.

(b)
For three planes to intersect in a line, the system must have infinite solutions with one free variable. This occurs when the equations are linearly dependent.
System: \[ \begin{cases} x + y – z = -2 & (1)\\ 2x + by – z = 3 & (2)\\ x – y + 2z = d & (3) \end{cases} \] Using elimination:
From (1): \( z = x + y + 2 \).
Substitute into (2): \( 2x + by – (x + y + 2) = 3 \implies x + (b-1)y = 5 \quad (4) \).
Substitute into (3): \( x – y + 2(x + y + 2) = d \implies 3x + y = d – 4 \quad (5) \).

To intersect in a line, (4) and (5) must represent the same linear relationship. Multiply (4) by 3:
\( 3x + 3(b-1)y = 15 \).
Compare coefficients with (5):
\( 3(b-1) = 1 \implies 3b – 3 = 1 \implies b = \frac{4}{3} \).
\( 15 = d – 4 \implies d = 19 \).
Answer: \( \boxed{b = \frac{4}{3}, \ d = 19} \)