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Date November 2011 Marks available 4 Reference code 11N.2.hl.TZ0.13
Level HL only Paper 2 Time zone TZ0
Command term Show that Question number 13 Adapted from N/A

Question

Two planes \({\Pi _1}\) and \({\Pi _2}\) have equations \(2x + y + z = 1\) and \(3x + y - z = 2\) respectively.

Find the vector equation of L, the line of intersection of \({\Pi _1}\) and \({\Pi _2}\).

[6]
a.

Show that the plane \({\Pi _3}\) which is perpendicular to \({\Pi _1}\) and contains L, has equation \(x - 2z = 1\).

[4]
b.

The point P has coordinates (−2, 4, 1) , the point Q lies on \({\Pi _3}\) and PQ is perpendicular to \({\Pi _2}\). Find the coordinates of Q.

[6]
c.

Markscheme

(a)     METHOD 1

solving simultaneously (gdc)     (M1)

\(x = 1 + 2z;{\text{ }}y = - 1 - 5z\)     A1A1

\(L:\boldsymbol{r} = \left( {\begin{array}{*{20}{c}}
  1 \\
  { - 1} \\
  0
\end{array}} \right) + \lambda \left( {\begin{array}{*{20}{c}}
  2 \\
  { - 5} \\
  1
\end{array}} \right)\)     A1A1A1

Note: \({1^{{\text{st}}}}\) A1 is for r =.

 

[6 marks]

METHOD 2

direction of line \( = \left| {\begin{array}{*{20}{c}}
  \boldsymbol{i}&\boldsymbol{j}&\boldsymbol{k} \\
  3&1&{ - 1} \\
  2&1&1
\end{array}} \right|\) (last two rows swapped)     M1

= 2i − 5j + k     A1

putting z = 0, a point on the line satisfies \(2x + y = 1,{\text{ }}3x + y = 2\)     M1

i.e. (1, −1, 0)     A1

the equation of the line is

\(\left( {\begin{array}{*{20}{c}}
  x \\
  y \\
  z
\end{array}} \right) = \left( {\begin{array}{*{20}{c}}
  1 \\
  { - 1} \\
  0
\end{array}} \right) + \lambda \left( {\begin{array}{*{20}{c}}
  2 \\
  { - 5} \\
  1
\end{array}} \right)\)     A1A1

Note: Award A0A1 if \(\left( {\begin{array}{*{20}{c}}
  x \\
  y \\
  z
\end{array}} \right)\) is missing.

 

[6 marks]

a.

\(\left( {\begin{array}{*{20}{c}}
  2 \\
  1 \\
  1
\end{array}} \right) \times \left( {\begin{array}{*{20}{c}}
  2 \\
  { - 5} \\
  1
\end{array}} \right)\)     M1

= 6i − 12k     A1

hence, n = i − 2k

\({\boldsymbol{n}} \cdot {\boldsymbol{a}} = \left( {\begin{array}{*{20}{c}}
  1 \\
  0 \\
  { - 2}
\end{array}} \right) \cdot \left( {\begin{array}{*{20}{c}}
  1 \\
  { - 1} \\
  0
\end{array}} \right) = 1\)     M1A1

therefore \( \cdot \) n = \( \cdot \) n \( \Rightarrow x - 2z = 1\)     AG

[4 marks]

b.

METHOD 1

P = (−2, 4, 1), Q = \((x,{\text{ }}y,{\text{ }}z)\)

\(\overrightarrow {{\text{PQ}}}  = \left( {\begin{array}{*{20}{c}}
  {x + 2} \\
  {y - 4} \\
  {z - 1}
\end{array}} \right)\)     A1

\(\overrightarrow {{\text{PQ}}} \) is perpendicular to \(3x + y - z = 2\)

\( \Rightarrow \overrightarrow {{\text{PQ}}} \) is parallel to 3i + jk     R1

\( \Rightarrow x + 2 = 3t;{\text{ }}y - 4 = t;{\text{ }}z - 1 = - t\)     A1

\(1 - z = t \Rightarrow x + 2 = 3 - 3z \Rightarrow x + 3z = 1\)     A1

solving simultaneously \(x + 3z = 1;{\text{ }}x - 2z = 1\)     M1

\(5z = 0 \Rightarrow z = 0;{\text{ }}x = 1,{\text{ }}y = 5\)     A1

hence, Q = (1, 5, 0)

[6 marks]

 

METHOD 2

Line passing through PQ has equation

\({\mathbf{r}} = \begin{array}{*{20}{c}}
  { - 2} \\
  4 \\
  1
\end{array} + t\begin{array}{*{20}{c}}
  3 \\
  1 \\
  { - 1}
\end{array}\)     M1A1

Meets \({\pi _3}\) when:

\( - 2 + 3t - 2(1 - t) = 1\)     M1A1

t = 1     A1

Q has coordinates (1, 5, 0)     A1

[6 marks]

c.

Examiners report

Candidates generally attempted this question but with varying degrees of success. Although (a) was answered best of all the parts, quite a few did not use correct notation to designate the vector equation of a line, i.e., r =, or its equivalent. In (b) some candidates incorrectly assumed the result and worked the question from there. In (c) some candidates did not understand the necessary relationships to make a meaningful attempt.

a.

Candidates generally attempted this question but with varying degrees of success. Although (a) was answered best of all the parts, quite a few did not use correct notation to designate the vector equation of a line, i.e., r =, or its equivalent. In (b) some candidates incorrectly assumed the result and worked the question from there. In (c) some candidates did not understand the necessary relationships to make a meaningful attempt.

b.

Candidates generally attempted this question but with varying degrees of success. Although (a) was answered best of all the parts, quite a few did not use correct notation to designate the vector equation of a line, i.e., r =, or its equivalent. In (b) some candidates incorrectly assumed the result and worked the question from there. In (c) some candidates did not understand the necessary relationships to make a meaningful attempt.

c.

Syllabus sections

Topic 4 - Core: Vectors » 4.6 » Vector equation of a plane \(r = a + \lambda b + \mu c\) .

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