2017 H2 Math P1Q10: Vectors II: Lines and Planes A Level Solutions

\begin{align*} \overrightarrow{PQ} &= \overrightarrow{OQ} - \overrightarrow{OP} \\ &= \begin{pmatrix}5\\7\\a\end{pmatrix} - \begin{pmatrix} 1 \\ 2 \\ - 1 \end{pmatrix} \\ &= \begin{pmatrix}4\\5\\a+1\end{pmatrix} \end{align*}

\begin{align*} C: \; &\mathbf{r} = \lambda \begin{pmatrix} 3 \\ 1 \\ - 2 \end{pmatrix}, \lambda \in \mathbb{R} \\ PQ: \; &\mathbf{r} = \begin{pmatrix} 1 \\ 2 \\ - 1 \end{pmatrix} + \mu \begin{pmatrix}4\\5\\a+1\end{pmatrix}, \mu \in \mathbb{R} \\ \end{align*}

Equation the equation of the two cables/lines,

\begin{pmatrix} 3 \lambda \\ \lambda \\ - 2 \lambda \end{pmatrix} = \begin{pmatrix}1+4\mu\\2+5\mu\\-1+(a+1)\mu\end{pmatrix}

\begin{align} &\qquad& 3\lambda &= 1 + 4 \mu \\ &\qquad& \lambda &= 2 + 5 \mu \\ &\qquad& -2\lambda &= -1 + (a+1) \mu \end{align}

Solving (1) and (2) with a GC,

{\lambda = - \frac{3}{11},}

{\mu = - \frac{5}{11}}

Substituting into equation (3),

\begin{align*} -2\left(- \frac{3}{11}\right) &=-1+(a+1)\left(- \frac{5}{11}\right) \\ a &= - \frac{22}{5} \; \blacksquare \end{align*}

Since

{R}

is on

{C,}

{\overrightarrow{OR} = \begin{pmatrix} 3 \lambda \\ \lambda \\ - 2 \lambda \end{pmatrix}}

\begin{align*} \overrightarrow{PR} &= \overrightarrow{OR} - \overrightarrow{OP} \\ &= \begin{pmatrix} 3 \lambda \\ \lambda \\ - 2 \lambda \end{pmatrix} - \begin{pmatrix} 1 \\ 2 \\ - 1 \end{pmatrix} \\ &= \begin{pmatrix}3\lambda-1\\\lambda-2\\-2\lambda+1\end{pmatrix} \end{align*}

\begin{align*} \overrightarrow{QR} &= \overrightarrow{OR} - \overrightarrow{OQ} \\ &= \begin{pmatrix} 3 \lambda \\ \lambda \\ - 2 \lambda \end{pmatrix} - \begin{pmatrix} 5 \\ 7 \\ - 3 \end{pmatrix} \\ &= \begin{pmatrix}3\lambda-5\\\lambda-7\\-2\lambda+3\end{pmatrix} \end{align*}

\begin{align*} & \overrightarrow{PR} \cdot \overrightarrow{QR} \\ = & \begin{pmatrix}3\lambda-1\\\lambda-2\\-2\lambda+1\end{pmatrix} \cdot \begin{pmatrix}3\lambda-5\\\lambda-7\\-2\lambda+3\end{pmatrix} \\ = & (3 \lambda - 1)(3 \lambda - 5) + (\lambda - 2)(\lambda - 7) + (- 2 \lambda + 1)(- 2 \lambda + 3) \\ = & (9 \lambda^2 - 18 \lambda + 5) + (\lambda^2 - 9 \lambda + 14) + (4 \lambda^2 - 8 \lambda + 3) \\ = & 14 \lambda^2 - 35 \lambda + 22 \\ = & 14 \left(\lambda^2 - \frac{5}{2} \lambda + \frac{11}{7}\right) \\ = & 14 \left(\lambda - \frac{5}{4}\right)^2 + \frac{1}{8} \end{align*}

For all

{\lambda \in \mathbb{R},}

{\left(\lambda-\frac{5}{4}\right)^2 \geq 0}

so

{14 \left(\lambda - \frac{5}{4}\right)^2 + \frac{1}{8} > 0}

Hence

{\overrightarrow{PR}\cdot\overrightarrow{QR}\neq0}

and

{\angle PRQ}

is never

{90^{\circ} \; \blacksquare}

For length of

{PR}

to be as small as possible,

{R}

is the foot of perpendicular from

{P}

to the line

{C}

\overrightarrow{OR} = \left(\mathbf{\overrightarrow{OP}} \cdot \mathbf{\hat{d}}\right) \mathbf{\hat{d}}

\overrightarrow{OR} = \left( \frac{\begin{pmatrix} 1 \\ 2 \\ - 1 \end{pmatrix} \cdot \begin{pmatrix} 3 \\ 1 \\ - 2 \end{pmatrix}}{\left| \begin{pmatrix} 3 \\ 1 \\ - 2 \end{pmatrix} \right|} \right) \frac{\begin{pmatrix} 3 \\ 1 \\ - 2 \end{pmatrix}}{\left| \begin{pmatrix} 3 \\ 1 \\ - 2 \end{pmatrix} \right|}

\overrightarrow{OR} = \frac{7}{14} \begin{pmatrix} 3 \\ 1 \\ - 2 \end{pmatrix}

\overrightarrow{OR} = \begin{pmatrix} \frac{3}{2} \\ \frac{1}{2} \\ - 1 \end{pmatrix}

Coordinates of

R \left( \frac{3}{2}, \frac{1}{2}, - 1 \right) \; \blacksquare

\begin{align*} \overrightarrow{PR} &= \overrightarrow{OR} - \overrightarrow{OP} \\ &= \begin{pmatrix} \frac{1}{2} \\ - \frac{3}{2} \\ 0 \end{pmatrix} \\ &= \frac{1}{2} \begin{pmatrix} 1 \\ - 3 \\ 0 \end{pmatrix} \end{align*}

\begin{align*} \left| \overrightarrow{PR} \right| &= \left| \frac{1}{2} \begin{pmatrix} 1 \\ - 3 \\ 0 \end{pmatrix} \right | \\ &= \frac{1}{2} \sqrt{1 + 9 + 0} \\ &= \frac{1}{2} \sqrt{10} \end{align*}

Exact minimum length of

{PR}

is

{\frac{1}{2} \sqrt{10} \textrm{ units}^2 \; \blacksquare}

2017 H2 Mathematics Paper 1 Question 10