Solved: Consider the points P(-3, 4, 2) and Q(8, -1, 5). A line M has vector equation... [IB Mathematics Analysis and Approaches (AA)]

Question

HLPaper 2

Consider the points P(-3, 4, 2) and Q(8, -1, 5).

A line M has vector equation

${r} = \begin{pmatrix} 2 \\ 0 \\ - 5 \end{pmatrix} + t\begin{pmatrix} 1 \\ -2 \\ 2 \end{pmatrix}$ .

The point R (5, y, 1) lies on line M.

Find $\overrightarrow {\text{ PQ}}$ .

[2]

Verified

Solution

Note: There may be slight differences in answers, depending on which combination of unrounded values and previous correct 3 sf values the candidates carry through in subsequent parts. Accept answers that are consistent with their working.

valid approach (M1) eg Q− P, PO + OQ, $\left( {\begin{pmatrix} 8 \\ { - 1} \\ 5 \end{pmatrix}} \right) - \left( {\begin{pmatrix} { - 3} \\ 4 \\ 2 \end{pmatrix}} \right)$ $\overrightarrow {{\text{PQ}}} = \left( {\begin{pmatrix} {11} \\ { - 5} \\ 3 \end{pmatrix}} \right)$ A1 N2

[2 marks]

Find $|\overrightarrow{PQ}|$ .

[2]

Verified

Solution

correct substitution into formula (A1) eg ${\sqrt{11^2 + (-5)^2 + 3^2}}$ 12.4498 ${|\overrightarrow{PQ}| = \sqrt{155}}$ (exact), 12.4 A1 N2

[2 marks]

Find the value of $y$ .

[3]

Verified

Solution

valid approach to find $t$ (M1) eg

$\begin{pmatrix} 5 \\ y \\ 1 \end{pmatrix} = \begin{pmatrix} 2 \\ 0 \\ - 5 \end{pmatrix} + t\begin{pmatrix} 1 \\ - 2 \\ 2 \end{pmatrix}$

$5 = 2 + t$

$1 = - 5 + 2t$ $t = 3$ (seen anywhere) (A1) attempt to substitute their parameter into the vector equation (M1) eg
$\begin{pmatrix} 5 \\ y \\ 1 \end{pmatrix} =\begin{pmatrix} 2 \\ 0 \\ { - 5} \end{pmatrix} + 3\begin{pmatrix} 1 \\ { - 2} \\ 2 \end{pmatrix}$

$3 \cdot \left( { - 2} \right)$

$y = - 6$ A1 N2

[3 marks]

Show that $\overrightarrow {{\text{PR}}} = \begin{pmatrix} 8 \\ - 10 \\ - 1 \end{pmatrix}$ .

[2]

Verified

Solution

correct approach A1 eg $\begin{pmatrix} 5 \\ - 6 \\ 1 \end{pmatrix} - \begin{pmatrix} - 3 \\ 4 \\ 2 \end{pmatrix}$ , PO + OR, $c - a$

$\overrightarrow{\text{PR}} = \begin{pmatrix} 8 \\ { - 10} \\ { - 1} \end{pmatrix}$ AG N0 Note: Do not award A1 in part (ii) unless answer in part (i) is correct and does not result from working backwards.

[2 marks]

Find the angle between ${\vec{PQ}}$ and ${\vec{PR}}$ .

[5]

Verified

Solution

finding scalar product and magnitude (A1)(A1) scalar product = 11 × 8 + −5 × −10 + 3 × −1 (=135) $\overrightarrow {\left| {{\text{PR}}} \right|} = \sqrt {{8^2} + {{\left( { - 10} \right)}^2} + {{\left( { - 1} \right)}^2}} \,\,\left( { = \sqrt {165} ,\,\,12.8452} \right)$ evidence of substitution into formula (M1) eg ${\text{cos}}\theta = \frac{{11 \times 8 + - 5 \times - 10 + 3 \times - 1}}{{\left| {\overrightarrow {{\text{PQ}}} } \right| \times \sqrt {{8^2} + {{\left( { - 10} \right)}^2} + {{\left( { - 1} \right)}^2}} }}{\text{, cos}}\theta = \frac{{\overrightarrow {{\text{PQ}}} \bullet \overrightarrow {{\text{PR}}} }}{{\sqrt {155} \times \sqrt {{8^2} + {{\left( { - 10} \right)}^2} + {{\left( { - 1} \right)}^2}} }}$ correct substitution (A1) eg ${\text{cos}}\theta = \frac{{11 \times 8 + - 5 \times - 10 + 3 \times - 1}}{{\sqrt {155} \times \sqrt {{8^2} + {{\left( { - 10} \right)}^2} + {{\left( { - 1} \right)}^2}} }}$ , ${\text{cos}}\theta = \frac{{135}}{{159.921 \ldots }}$ , ${\text{cos}}\theta = 0.844162 \ldots$ 0.565795, 32.4177° $\hat A = 0.566,32.4° \emph{A1 N3}$

[5marks]

Find the area of triangle $PQR$ .

[2]

Verified

Solution

correct substitution into area formula (A1) eg $\frac{1}{2} \times \sqrt {155} \times \sqrt {165} \times {\text{sin}}\left( {0.566 \ldots } \right)$ , $\frac{1}{2} \times \sqrt {155 \times 165} \times {\text{sin}}\left( {32.4} \right)$ 42.8660 area = 42.9 A1 N2

[2 marks]

Related topics