Date | May 2011 | Marks available | 2 | Reference code | 11M.2.hl.TZ2.11 |
Level | HL only | Paper | 2 | Time zone | TZ2 |
Command term | Calculate and Hence | Question number | 11 | Adapted from | N/A |
Question
The points P(−1, 2, − 3), Q(−2, 1, 0), R(0, 5, 1) and S form a parallelogram, where S is diagonally opposite Q.
Find the coordinates of S.
The vector product \(\overrightarrow {{\text{PQ}}} \times \overrightarrow {{\text{PS}}} = \left( {\begin{array}{*{20}{c}}
{ - 13} \\
7 \\
m
\end{array}} \right)\). Find the value of m .
Hence calculate the area of parallelogram PQRS.
Find the Cartesian equation of the plane, \({\prod _1}\) , containing the parallelogram PQRS.
Write down the vector equation of the line through the origin (0, 0, 0) that is perpendicular to the plane \({\prod _1}\) .
Hence find the point on the plane that is closest to the origin.
A second plane, \({\prod _2}\) , has equation x − 2y + z = 3. Calculate the angle between the two planes.
Markscheme
\(\overrightarrow {{\text{PQ}}} = \left( {\begin{array}{*{20}{c}}
{ - 1} \\
{ - 1} \\
3
\end{array}} \right)\) , \(\overrightarrow {{\text{SR}}} = \left( {\begin{array}{*{20}{c}}
{0 - x} \\
{5 - y} \\
{1 - z}
\end{array}} \right)\) (M1)
point S = (1, 6, −2) A1
[2 marks]
\(\overrightarrow {{\text{PQ}}} = \left( {\begin{array}{*{20}{c}}
{ - 1} \\
{ - 1} \\
3
\end{array}} \right)\)\(\overrightarrow {{\text{PS}}} = \left( {\begin{array}{*{20}{c}}
2 \\
4 \\
1
\end{array}} \right)\) A1
\(\overrightarrow {{\text{PQ}}} \times \overrightarrow {{\text{PS}}} = \left( {\begin{array}{*{20}{c}}
{ - 13} \\
7 \\
{ - 2}
\end{array}} \right)\)
m = −2 A1
[2 marks]
area of parallelogram PQRS \( = \left| {\overrightarrow {{\text{PQ}}} \times \overrightarrow {{\text{PS}}} } \right| = \sqrt {{{( - 13)}^2} + {7^2} + {{( - 2)}^2}} \) M1
\( = \sqrt {222} = 14.9\) A1
[2 marks]
equation of plane is −13x + 7y − 2z = d M1A1
substituting any of the points given gives d = 33
−13x + 7y − 2z = 33 A1
[3 marks]
equation of line is \(\boldsymbol{r} = \left( {\begin{array}{*{20}{c}}
0 \\
0 \\
0
\end{array}} \right) + \lambda \left( {\begin{array}{*{20}{c}}
{ - 13} \\
7 \\
{ - 2}
\end{array}} \right)\) A1
Note: To get the A1 must have \(\boldsymbol{r} =\) or equivalent.
[1 mark]
\(169\lambda + 49\lambda + 4\lambda = 33\) M1
\(\lambda = \frac{{33}}{{222}}{\text{ }}( = 0.149…)\) A1
closest point is \(\left( { - \frac{{143}}{{74}},\frac{{77}}{{74}}, - \frac{{11}}{{37}}} \right){\text{ }}\left( { = ( - 1.93,{\text{ 1.04, - 0.297)}}} \right)\) A1
[3 marks]
angle between planes is the same as the angle between the normals (R1)
\(\cos \theta = \frac{{ - 13 \times 1 + 7 \times - 2 - 2 \times 1}}{{\sqrt {222} \times \sqrt 6 }}\) M1A1
\(\theta = 143^\circ \) (accept \(\theta = 37.4^\circ \) or 2.49 radians or 0.652 radians) A1
[4 marks]
Examiners report
This was a multi-part question that was well answered by many candidates. Wrong answers to part (a) were mainly the result of failing to draw a diagram. Follow through benefitted many candidates. A high proportion of candidates lost the mark in (e) by not writing their answer as an equation in the form r = ...
This was a multi-part question that was well answered by many candidates. Wrong answers to part (a) were mainly the result of failing to draw a diagram. Follow through benefitted many candidates. A high proportion of candidates lost the mark in (e) by not writing their answer as an equation in the form r = ...
This was a multi-part question that was well answered by many candidates. Wrong answers to part (a) were mainly the result of failing to draw a diagram. Follow through benefitted many candidates. A high proportion of candidates lost the mark in (e) by not writing their answer as an equation in the form r = ...
This was a multi-part question that was well answered by many candidates. Wrong answers to part (a) were mainly the result of failing to draw a diagram. Follow through benefitted many candidates. A high proportion of candidates lost the mark in (e) by not writing their answer as an equation in the form r = ...
This was a multi-part question that was well answered by many candidates. Wrong answers to part (a) were mainly the result of failing to draw a diagram. Follow through benefitted many candidates. A high proportion of candidates lost the mark in (e) by not writing their answer as an equation in the form \(\boldsymbol{r} = \) ...
This was a multi-part question that was well answered by many candidates. Wrong answers to part (a) were mainly the result of failing to draw a diagram. Follow through benefitted many candidates. A high proportion of candidates lost the mark in (e) by not writing their answer as an equation in the form r = ...
This was a multi-part question that was well answered by many candidates. Wrong answers to part (a) were mainly the result of failing to draw a diagram. Follow through benefitted many candidates. A high proportion of candidates lost the mark in (e) by not writing their answer as an equation in the form r = ...