IBDP Maths AA: Topic : AHL 3.18: Intersections of: a line with a plane: IB style Questions HL Paper 2

Question

Consider the planes Π1 and Π2 with the following equations.

Π1 : 3x + 2y + z = 6
Π2 : x – 2y + z = 4

    1. Find a Cartesian equation of the plane Π3 which is perpendicular to Π1 and Π2 and passes through the origin (0, 0, 0). [3]

    2. Find the coordinates of the point where Π1 , Π2 and Π3 intersect. [2]

▶️Answer/Explanation

Ans:

(a)

attempt to find a vector perpendicular to Π1 , Π2 using a cross product

(2-(-2))i+(1-3)j+(-6-2)k

equation is 4x-2y-8z=0 (⇒ 2x -y- 4z= 0)

(b)

attempt to solve 3 simultaneous equations in 3 variables

\((\frac{41}{21},\frac{-10}{21},\frac{23}{21})\) = 1.95, -0.476,1.10

Question

The plane Π1 has equation 3x − y + z = −13 and the line L has vector equation

  1. Given that L meets Π1at the point P, find the coordinates of P. [4]

  2. Find the shortest distance from the point O( 0, 0, 0) to Π1 . [4] The plane Π2 contains the point O and the line L .

  3. Find the equation of Π2 , giving your answer in the form r.n = d . [3]

  4. Determine the acute angle between Π1 and Π2 . [5]

▶️Answer/Explanation

Ans:

choice of any point on the plane, eg ( – 8 , -1, 10 ) to use in distance formula

c) EITHER

identify two vectors

Question

Find the vector equation of the line of intersection of the three planes represented by the following system of equations.

\[2x – 7y + 5z = 1\]

\[6x + 3y – z = – 1\]

\[ – 14x – 23y + 13z = 5\]

▶️Answer/Explanation

Markscheme

METHOD 1

(from GDC)

\(\left( {\begin{array}{*{20}{ccc|c}}
  1&0&{\frac{1}{6}}&{ – \frac{1}{{12}}} \\
  0&1&{ – \frac{2}{3}}&{ – \frac{1}{6}} \\
  0&0&0&0
\end{array}} \right)\)     (M1)

\(x + \frac{1}{6}\lambda = – \frac{1}{{12}}\)     A1

\(y – \frac{2}{3}\lambda = – \frac{1}{6}\)     A1

\(\boldsymbol{r} = \left( { – \frac{1}{{12}}\boldsymbol{i} – \frac{1}{6}\boldsymbol{j}} \right) + \lambda \left( { – \frac{1}{6}\boldsymbol{i} + \frac{2}{3}\boldsymbol{j} + \boldsymbol{k}} \right)\)     A1A1A1     N3

[6 marks] 

METHOD 2

(Elimination method either for equations or row reduction of matrix)

Eliminating one of the variables     M1A1

Finding a point on the line     (M1)A1

Finding the direction of the line     M1

The vector equation of the line     A1     N3

[6 marks]

Examiners report

A large number of candidates did not use their GDC in this question. Some candidates who attempted analytical solutions looked for a point solution although the question specifically states that the planes intersect in a line. Other candidates eliminated one variable and then had no clear strategy for proceeding with the solution.

Some candidates failed to write ‘r =’, and others did not give the equation in vector form.

Question

(a)     If \(a = 4\) find the coordinates of the point of intersection of the three planes.

(b)     (i)     Find the value of \(a\) for which the planes do not meet at a unique point.

  (ii)     For this value of \(a\) show that the three planes do not have any common point.

▶️Answer/Explanation

Markscheme

(a)     let \({\boldsymbol{A}} = \left( {\begin{array}{*{20}{c}}
  1&1&2 \\
  2&{ – 1}&3 \\
  5&{ – 1}&4
\end{array}} \right)\) , \({\boldsymbol{X}} = \left( {\begin{array}{*{20}{c}}
  x \\
  y \\
  z
\end{array}} \right)\) , and \({\boldsymbol{B}} = \left( {\begin{array}{*{20}{c}}
  2 \\
  2 \\
  5
\end{array}} \right)\)     (M1)

point of intersection is \(\left( {\frac{{11}}{{12}},\frac{7}{{12}},\frac{1}{4}} \right)\)   or \(\left( {{\text{or }}\left( {{\text{0}}{\text{.917, 0}}{\text{.583, 0}}{\text{.25}}} \right)} \right)\)     A1

(b)     METHOD 1

(i)     \(\det \left( {\begin{array}{*{20}{c}}
  1&1&2 \\
  2&{ – 1}&3 \\
  5&{ – 1}&4
\end{array}} \right) = 0\)     M1

\( – 3a + 24 = 0\)     (A1)

\(a = 8\)     A1     N1

(ii)     consider the augmented matrix \(\left( {\begin{array}{*{20}{ccc|c}}
  1&1&2&2 \\
  2&{ – 1}&3&2 \\
  5&{ – 1}&4&5
\end{array}} \right)\)     M1

use row reduction to obtain \(\left( {\begin{array}{*{20}{ccc|c}}
  1&1&2&2 \\
  0&{ – 3}&{ – 1}&{ – 2} \\
  0&0&0&{ – 1}
\end{array}} \right)\) or \(\left( {\begin{array}{*{20}{ccc|c}}
  1&0&{\frac{5}{3}}&0 \\
  0&1&{\frac{1}{3}}&0 \\
  0&0&0&1
\end{array}} \right)\) (or equivalent)     A1

any valid reason     R1

(e.g. as the last row is not all zeros, the planes do not meet)     N0

METHOD 2

use of row reduction (or equivalent manipulation of equations)     M1

e.g. \(\left( {\begin{array}{*{20}{c}}
  1&1&2&2 \\
  2&{ – 1}&3&2 \\
  5&{ – 1}&a&5
\end{array}} \right) \Rightarrow \left( {\begin{array}{*{20}{c}}
  1&1&2&2 \\
  0&{ – 3}&{ – 1}&{ – 2} \\
  0&{ – 6}&{a – 10}&{ – 5}
\end{array}} \right)\)     A1A1

Note: Award an A1 for each correctly reduced row.

 

(i)     \(a -10 = -2 \Rightarrow a = 8\)     M1A1     N1

(ii)     when \(a = 8\) , row 3 \( \ne \) 2 \( \times \) row 2     R1     N0

[8 marks]

Examiners report

Few students were able to do this question efficiently. Many students were able to do part (a) by manipulating equations, whereas calculator methods would yield the solution quickly and easily. Part (b) was poorly attempted and it was apparent that many students used a lot of time manipulating equations without real understanding of what they were looking for.

Question

The diagram shows a cube OABCDEFG.

 

Let O be the origin, (OA) the x-axis, (OC) the y-axis and (OD) the z-axis.

Let M, N and P be the midpoints of [FG], [DG] and [CG], respectively.

The coordinates of F are (2, 2, 2).

(a)     Find the position vectors \(\overrightarrow {{\text{OM}}} \), \(\overrightarrow {{\text{ON}}} \) and \(\overrightarrow {{\text{OP}}} \) in component form.

(b)     Find \(\overrightarrow {{\text{MP}}}  \times \overrightarrow {{\text{MN}}} \).

(c)     Hence,

  (i)     calculate the area of the triangle MNP;

  (ii)     show that the line (AG) is perpendicular to the plane MNP;

  (iii)     find the equation of the plane MNP.

(d)     Determine the coordinates of the point where the line (AG) meets the plane MNP.

▶️Answer/Explanation

Markscheme

(a)     \(\overrightarrow {{\text{OM}}}  = \left( {\begin{array}{*{20}{c}}
  1 \\
  2 \\
  2
\end{array}} \right)\), \(\overrightarrow {{\text{ON}}}  = \left( {\begin{array}{*{20}{c}}
  0 \\
  1 \\
  2
\end{array}} \right)\) and \(\overrightarrow {{\text{OP}}}  = \left( {\begin{array}{*{20}{c}}
  0 \\
  2 \\
  1
\end{array}} \right)\)     A1A1A1

[3 marks]

 

(b)     \(\overrightarrow {{\text{MP}}}  = \left( {\begin{array}{*{20}{c}}
  { – 1} \\
  0 \\
  { – 1}
\end{array}} \right)\) and \(\overrightarrow {{\text{MN}}}  = \left( {\begin{array}{*{20}{c}}
  { – 1} \\
  { – 1} \\
  0
\end{array}} \right)\)     A1A1

\(\overrightarrow {{\text{MP}}}  \times \overrightarrow {{\text{MN}}}  = \left( {\begin{array}{*{20}{c}}
  i&j&k \\
  { – 1}&0&{ – 1} \\
  { – 1}&{ – 1}&0
\end{array}} \right) = \left( {\begin{array}{*{20}{c}}
  { – 1} \\
  1 \\
  1
\end{array}} \right)\)     (M1)A1

[4 marks]

 

(c)     (i)     area of MNP \( = \frac{1}{2}\left| {\overrightarrow {{\text{MP}}}  \times \overrightarrow {{\text{MN}}} } \right|\)     M1

\( = \frac{1}{2}\left| {\left( {\begin{array}{*{20}{c}}
  { – 1} \\
  1 \\
  1
\end{array}} \right)} \right|\)

\( = \frac{{\sqrt 3 }}{2}\)     A1

 

(ii)     \(\overrightarrow {{\text{OA}}}  = \left( {\begin{array}{*{20}{c}}
  2 \\
  0 \\
  0
\end{array}} \right)\), \(\overrightarrow {{\text{OG}}}  = \left( {\begin{array}{*{20}{c}}
  0 \\
  2 \\
  2
\end{array}} \right)\)

\(\overrightarrow {{\text{AG}}}  = \left( {\begin{array}{*{20}{c}}
  { – 2} \\
  2 \\
  2
\end{array}} \right)\)     A1

since \(\overrightarrow {{\text{AG}}} = 2(\overrightarrow {{\text{MP}}}  \times \overrightarrow {{\text{MN}}} )\) AG is perpendicular to MNP     R1

 

(iii)     \(r \cdot \left( {\begin{array}{*{20}{c}}
  { – 1} \\
  1 \\
  1
\end{array}} \right) = \left( {\begin{array}{*{20}{c}}
  1 \\
  2 \\
  2
\end{array}} \right) \cdot \left( {\begin{array}{*{20}{c}}
  { – 1} \\
  1 \\
  1
\end{array}} \right)\)     M1A1

\(r \cdot \left( {\begin{array}{*{20}{c}}
  { – 1} \\
  1 \\
  1
\end{array}} \right) = 3\) (accept \( – x + y + z = 3\))     A1

[7 marks]

 

(d)     \(r = \left( {\begin{array}{*{20}{c}}
  2 \\
  0 \\
  0
\end{array}} \right) + \lambda \left( {\begin{array}{*{20}{c}}
  { – 2} \\
  2 \\
  2
\end{array}} \right)\)     A1

\(\left( {\begin{array}{*{20}{c}}
  {2 – 2\lambda } \\
  {2\lambda } \\
  {2\lambda }
\end{array}} \right) \cdot \left( {\begin{array}{*{20}{c}}
  { – 1} \\
  1 \\
  1
\end{array}} \right) = 3\)     M1A1

\( – 2 + 2\lambda + 2\lambda + 2\lambda = 3\)

\(\lambda = \frac{5}{6}\)     A1

\(r = \left( {\begin{array}{*{20}{c}}
  2 \\
  0 \\
  0
\end{array}} \right) + \frac{5}{6}\left( {\begin{array}{*{20}{c}}
  { – 2} \\
  2 \\
  2
\end{array}} \right)\)     M1

coordinates of point \(\left( {\frac{1}{3},\frac{5}{3},\frac{5}{3}} \right)\)     A1

[6 marks]

Total [20 marks]

Examiners report

This was the most successfully answered question in part B, with many candidates achieving full marks. There were a few candidates who misread the question and treated the cube as a unit cube. The most common errors were either algebraic or arithmetic mistakes. A variety of notation forms were seen but in general were used consistently. In a few cases, candidates failed to show all the work or set it properly.

Question

The planes \(2x + 3y – z = 5\) and \(x – y + 2z = k\) intersect in the line \(5x + 1 = 9 – 5y = – 5z\) .

Find the value of k .

▶️Answer/Explanation

Markscheme

point on line is \(x = \frac{{ – 1 – 5\lambda }}{5}{\text{, }}y = \frac{{9 + 5\lambda }}{5}{\text{, }}z = \lambda \) or similar     M1A1

 Note: Accept use of point on the line or elimination of one of the variables using the equations of the planes

\(\frac{{ – 1 – 5\lambda }}{5} – \frac{{9 + 5\lambda }}{5} + 2\lambda  = k\)     M1A1

 Note: Award M1A1 if coordinates of point and equation of a plane is used to obtain linear equation in k or equations of the line are used in combination with equation obtained by elimination to get linear equation in k.

 \(k = – 2\)     A1

[5 marks]

Examiners report

Many different attempts were seen, sometimes with success. Unfortunately many candidates wasted time with aimless substitutions showing little understanding of the problem.

Question

A ray of light coming from the point (−1, 3, 2) is travelling in the direction of vector and meets the plane (π : x + 3y + 2z -24=0) .

Find the angle that the ray of light makes with the plane.

▶️Answer/Explanation

Markscheme

The normal vector to the plane is \(\left( {\begin{array}{*{20}{c}}
  1 \\
  3 \\
  2
\end{array}} \right)\) .     (A1)

EITHER

\(\theta \) is the angle between the line and the normal to the plane.

\(\cos \theta = \frac{{\left( {\begin{array}{*{20}{c}}
  4 \\
  1 \\
  { – 2}
\end{array}} \right) \cdot \left( {\begin{array}{*{20}{c}}
  1 \\
  3 \\
  2
\end{array}} \right)}}{{\sqrt {14} \sqrt {21} }} = \frac{3}{{\sqrt {14} \sqrt {21} }} = \left( {\frac{3}{{7\sqrt 6 }}} \right)\)     (M1)A1A1

\( \Rightarrow \theta = 79.9^\circ {\text{ }}( = 1.394…)\)     A1

The required angle is 10.1° (= 0.176)     A1

OR

\(\phi \) is the angle between the line and the plane.

\(\sin \phi = \frac{{\left( {\begin{array}{*{20}{c}}
  4 \\
  1 \\
  { – 2}
\end{array}} \right) \cdot \left( {\begin{array}{*{20}{c}}
  1 \\
  3 \\
  2
\end{array}} \right)}}{{\sqrt {14} \sqrt {21} }} = \frac{3}{{\sqrt {14} \sqrt {21} }}\)     (M1)A1A1

\(\phi \) = 10.1° (= 0.176)     A2

[6 marks]

 

Question

A plane \(\pi \) has vector equation r = (−2i + 3j − 2k) + \(\lambda \)(2i + 3j + 2k) + \(\mu \)(6i − 3j + 2k).

(a)     Show that the Cartesian equation of the plane \(\pi \) is 3x + 2y − 6z = 12.

(b)     The plane \(\pi \) meets the x, y and z axes at A, B and C respectively. Find the coordinates of A, B and C.

(c)     Find the volume of the pyramid OABC.

(d)     Find the angle between the plane \(\pi \) and the x-axis.

(e)     Hence, or otherwise, find the distance from the origin to the plane \(\pi \).

(f)     Using your answers from (c) and (e), find the area of the triangle ABC.

▶️Answer/Explanation

Markscheme

(a)     EITHER

normal to plane given by

\(\left| {\begin{array}{*{20}{c}}
  i&j&k \\
  2&3&2 \\
  6&{ – 3}&2
\end{array}} \right|\)     M1A1

= 12i + 8j – 24k     A1

equation of \(\pi \) is \(3x + 2y – 6z = d\)     (M1)

as goes through (–2, 3, –2) so d = 12     M1A1

\(\pi :3x + 2y – 6z = 12\)     AG

OR

\(x = – 2 + 2\lambda + 6\mu \)

\(y = 3 + 3\lambda – 3\mu \)

\(z = – 2 + 2\lambda + 2\mu \)

eliminating \(\mu \)

\(x + 2y = 4 + 8\lambda \)

\(2y + 3z = 12\lambda \)     M1A1A1

eliminating \(\lambda \)

\(3(x + 2y) – 2(2y + 3z) = 12\)     M1A1A1

\(\pi :3x + 2y – 6z = 12\)     AG

[6 marks]

 

(b)     therefore A(4, 0, 0), B(0, 6, 0) and C(0, 0, 2)     A1A1A1

Note: Award A1A1A0 if position vectors given instead of coordinates.

 

[3 marks]

 

(c)     area of base \({\text{OAB}} = \frac{1}{2} \times 4 \times 6 = 12\)     M1

\(V = \frac{1}{3} \times 12 \times 2 = 8\)     M1A1

[3 marks]

 

(d)     \(\left( {\begin{array}{*{20}{c}}
  3 \\
  2 \\
  { – 6}
\end{array}} \right) \cdot \left( {\begin{array}{*{20}{c}}
  1 \\
  0 \\
  0
\end{array}} \right) = 3 = 7 \times 1 \times \cos \phi \)     M1A1

\(\phi = \arccos \frac{3}{7}\)

so \(\theta = 90 – \arccos \frac{3}{7} = 25.4^\circ \,\,\,\,\,\)(accept 0.443 radians)     M1A1

[4 marks]

 

(e)     \(d = 4\sin \theta = \frac{{12}}{7}\,\,\,\,\,( = 1.71)\)     (M1)A1

[2 marks]

 

(f)     \(8 = \frac{1}{3} \times \frac{{12}}{7} \times {\text{area}} \Rightarrow {\text{area}} = 14\)     M1A1

Note: If answer to part (f) is found in an earlier part, award M1A1, regardless of the fact that it has not come from their answers to part (c) and part (e).

 

[2 marks]

Total [20 marks]

Examiners report

The question was generally well answered, although there were many students who failed to recognise that the volume was most logically found using a base as one of the coordinate planes.

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