2020 H2 Mathematics Paper 1 Question 1

Vectors II: Lines and Planes

Answers

(i)

(−11−3){\begin{pmatrix} - 1 \\ 1 \\ - 3 \end{pmatrix}}

(ii)

49.2∘{49.2^\circ}

Full solutions

(i)

Let d1{\mathbf{d_1}} and d2{\mathbf{d_2}} denote the two vectors provided
n1′=d1×d2=(110)×(1−5−2)=(−22−6)=2(−11−3)\begin{align*} \mathbf{n'_1} &= \mathbf{d}_1 \times \mathbf{d}_2 \\ &= \begin{pmatrix} 1 \\ 1 \\ 0 \end{pmatrix} \times \begin{pmatrix} 1 \\ - 5 \\ - 2 \end{pmatrix} \\ &= \begin{pmatrix} - 2 \\ 2 \\ - 6 \end{pmatrix} \\ &= 2 \begin{pmatrix} - 1 \\ 1 \\ - 3 \end{pmatrix} \end{align*}
Hence a vector normal to π1{\pi_1} is (−11−3){\begin{pmatrix} - 1 \\ 1 \\ - 3 \end{pmatrix}}

(ii)

∣n1⋅n2∣=∣n1∣∣n2∣cos⁡θ\left|\mathbf{n_1} \cdot \mathbf{n_2}\right| = \left| \mathbf{n_1} \right| \left| \mathbf{n_2} \right| \cos \theta
∣(−11−3)⋅(45−6)∣=∣(−11−3)∣∣(45−6)∣cos⁡θ∣19∣=(11)(77)cos⁡θ\begin{align*} \left|\begin{pmatrix} - 1 \\ 1 \\ - 3 \end{pmatrix} \cdot \begin{pmatrix} 4 \\ 5 \\ - 6 \end{pmatrix} \right| &= \left| \begin{pmatrix} - 1 \\ 1 \\ - 3 \end{pmatrix} \right| \left| \begin{pmatrix} 4 \\ 5 \\ - 6 \end{pmatrix} \right| \cos \theta \\ \left|19 \right| &= (\sqrt{11}) (\sqrt{77}) \cos \theta \end{align*}
cos⁡θ=19(11)(77)θ=49.2∘  ■\begin{align*} \cos \theta &= \frac{19}{(\sqrt{11})(\sqrt{77})} \\ \theta &= 49.2^\circ \; \blacksquare \end{align*}