Mathematics Applications & Interpretation (AI) International Baccalaureate (IB) Practice Question: Two lines L1 and L2 are given by the following equations, where p R. L_1: r = pmatrix 2 \\ p+9 \\ -3 pmatrix + pmatrix p...

• Set the scalar product of the direction vectors equal to zero: $\begin{pmatrix} p \\ 2p \\ 4 \end{pmatrix} \cdot \begin{pmatrix} p+4 \\ 4 \\ -7 \end{pmatrix} = 0$ M1

• Expand the expression: $p(p+4) + 8p - 28 = 0$

• Simplify to a quadratic equation: $p^2 + 12p - 28 = 0$ A1

• Solve for $p$: $(p+14)(p-2) = 0$

• State the possible values: $p = -14, p = 2$ A1

3 marks total

Method 1

• Substitute $p = -14$ to find the equations of the lines: $L_1: \mathbf{r} = \begin{pmatrix} 2 \\ -5 \\ -3 \end{pmatrix} + \lambda \begin{pmatrix} -14 \\ -28 \\ 4 \end{pmatrix}, \quad L_2: \mathbf{r} = \begin{pmatrix} 14 \\ 7 \\ -2 \end{pmatrix} + \mu \begin{pmatrix} -10 \\ 4 \\ -7 \end{pmatrix}$

• State the condition for a common point by equating components: M1 $\begin{aligned} 2 - 14\lambda &= 14 - 10\mu \\ -5 - 28\lambda &= 7 + 4\mu \\ -3 + 4\lambda &= -2 - 7\mu \end{aligned}$

• Attempt to solve the system using the first two equations: $2 - 14\lambda = 14 - 10\mu$ and $-5 - 28\lambda = 7 + 4\mu$

• Obtain the parameter values: $\lambda = -\frac{1}{2}, \mu = \frac{1}{2}$ A1

• Substitute these values into the third equation to verify: M1 LHS: $-3 + 4(-\frac{1}{2}) = -5$ RHS: $-2 - 7(\frac{1}{2}) = -5.5$

• Conclude that the lines do not intersect since $-5 \neq -5.5$ A1

Method 2

• Attempt to solve the system of equations using a GDC M1

• Identify inconsistent system or no solution found A1

• Verify using parameters or geometric check M1

• Conclude that the lines do not intersect A1

4 marks total

NoteAccept equivalent methods based on the order in which the equations are considered.