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Date May 2011 Marks available 1 Reference code 11M.2.hl.TZ2.11
Level HL only Paper 2 Time zone TZ2
Command term Write down 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.

[2]
a.

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 .

[2]
b.

Hence calculate the area of parallelogram PQRS.

[2]
c.

Find the Cartesian equation of the plane, \({\prod _1}\) , containing the parallelogram PQRS.

[3]
d.

Write down the vector equation of the line through the origin (0, 0, 0) that is perpendicular to the plane \({\prod _1}\) .

[1]
e.

Hence find the point on the plane that is closest to the origin.

[3]
f.

A second plane, \({\prod _2}\) , has equation x − 2y + z = 3. Calculate the angle between the two planes.

[4]
g.

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]

a.

\(\overrightarrow {{\text{PQ}}} = \left( {\begin{array}{*{20}{c}}
  { - 1} \\
  { - 1} \\
  3
\end{array}} \right)\)\(\overrightarrow {{\text{PS}}} = \left( {\begin{array}{*{20}{c}}
  2 \\
  4 \\
  1
\end{array}} \right)\)     A1

\(\overrightarrow {{\text{PQ}}}  \times \overrightarrow {{\text{PS}}} = \left( {\begin{array}{*{20}{c}}
  { - 13} \\
  7 \\
  { - 2}
\end{array}} \right)\)

m = −2     A1

[2 marks]

b.

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]

c.

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]

d.

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]

e.

\(169\lambda + 49\lambda + 4\lambda = 33\)     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]

f.

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]

g.

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 = ...

a.

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 = ...

b.

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 = ...

c.

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 = ...

d.

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 \(\boldsymbol{r} = \) ...

e.

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 = ...

f.

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 = ...

g.

Syllabus sections

Topic 4 - Core: Vectors » 4.3 » Vector equation of a line in two and three dimensions: \(r = a + \lambda b\) .
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