IBDP Maths analysis and approaches Topic: AHL 3.18 Angle between: a line and a plane; two planes. HL Paper 1

 Question: [Maximum mark: 15]

Consider the three planes

1 : 2x – y + z = 4
2 : x – 2y + 3z = 5
3 : -9x + 3y – 2z = 32

(a) Show that the three planes do not intersect.
(b) (i) Verify that the point P(1 , -2 , 0) lies on both ∏1 and ∏2 .
      (ii) Find a vector equation of L, the line of intersection of ∏1 and ∏2 .
(c) Find the distance between L and ∏3 .

▶️Answer/Explanation

Ans:

(a) METHOD 1
attempt to eliminate a variable
obtain a pair of equations in two variables

EITHER

-3x + z = -3 and
-3x + z = 44

OR

-5x + y = -7 and
-5x + y = 40

OR

3x – z = 3 and
3x – z = \(\frac{79}{5}\)

THEN

the two lines are parallel ( – 3 ≠ 44 or – 7 ≠ 40 or 3 ≠ -\(\frac{79}{5}\)

Note: There are other possible pairs of equations in two variables.
To obtain the final R1, at least the initial M1 must have been awarded.

hence the three planes do not intersect

METHOD 2

vector product of the two normals = \(\begin{pmatrix}-1\\-5\\-3\end{pmatrix}\) (or equivalent)

\(r = \begin{pmatrix}1\\-2 \\0 \end{pmatrix} + \lambda \begin{pmatrix}1\\5 \\3 \end{pmatrix}\)      (or equivalent)

Note: Award A0 if “r =” is missing. Subsequent marks may still be awarded.

Attempt to substitute (1 + λ, -2 + 5λ, 3λ) in ∏3

-9(1 + λ) + 3(-2+5λ) – 2(3λ) = 32

− 15 = 32, a contradiction
hence the three planes do not intersect

(b) (i)     ∏1 : 2 + 2 + 0 = 4 and ∏2 : 1 + 4 + 0 = 5

      (ii) METHOD 1
attempt to find the vector product of the two normals

\(\begin{pmatrix}2\\-1 \\1 \end{pmatrix} \times \begin{pmatrix}1\\-2 \\3 \end{pmatrix}\)

\(=\begin{pmatrix}-1\\-5 \\-3 \end{pmatrix}\)

\(r = \begin{pmatrix}1\\-2 \\0 \end{pmatrix} + \lambda \begin{pmatrix}1\\5 \\3 \end{pmatrix}\) 

Note: Award A1A0 if “r =” is missing.
Accept any multiple of the direction vector.
Working for (b)(ii) may be seen in part (a) Method 2. In this case penalize
lack of “r =” only once.

METHOD 2
attempt to eliminate a variable from ∏1 and ∏2

3x – z = 3 OR 3y – 5z = -6  OR 5x – y = 7
Let x = t
substituting x = t in 3x – z = 3 to obtain
z = -3 + 3t and y = 5t – 7 (for all three variables in parametric form)

\(r = \begin{pmatrix}1\\-2 \\0 \end{pmatrix} + \lambda \begin{pmatrix}1\\5 \\3 \end{pmatrix}\) 

Note: Award A1A0 if “r =” is missing.
Accept any multiple of the direction vector. Accept other position vectors
which satisfy both the planes ∏1 and ∏2 .

(c) METHOD 1
the line connecting L and ∏3 is given by L1
attempt to substitute position and direction vector to form L1

\(s = \begin{pmatrix}1\\-2 \\0 \end{pmatrix} + t \begin{pmatrix}-9\\3 \\-2 \end{pmatrix}\)
substitute (1 – 9t, – 2 + 3t, – 2t) in ∏3
-9(1-9t) + 3(-2+3t) – 2(-2t) = 32
\(94t = 47\Rightarrow t=\frac{1}{2}\)

attempt to find distance between (1, -2,0) and their point \(\begin{pmatrix}-\frac{7}{2}, & -\frac{1}{2}, & -1 \end{pmatrix}\)

\(=\left | \begin{pmatrix}1\\-2 \\0 \end{pmatrix} +\frac{1}{2}\begin{pmatrix}-9\\3 \\-2 \end{pmatrix}-\begin{pmatrix}1\\-2 \\0 \end{pmatrix}\right | = \frac{1}{2}\sqrt{(-9)^{2}+3^{2}+(-2)^{2}}\)

\(=\frac{\sqrt{94}}{2}\)

METHOD 2

unit normal vector equation of ∏3 is given by

\(=\frac{32}{\sqrt{94}}\)

Question

The points A, B, C have position vectors i + j + 2k , i + 2j + 3k , 3i + k respectively and lie in the plane \(\pi \) .

(a)     Find

(i)     the area of the triangle ABC;

(ii)     the shortest distance from C to the line AB;

(iii)     the cartesian equation of the plane \(\pi \) .

The line L passes through the origin and is normal to the plane \(\pi \) , it intersects \(\pi \) at the

point D.

(b)     Find

(i)     the coordinates of the point D;

(ii)     the distance of \(\pi \) from the origin.

▶️Answer/Explanation

Markscheme

(a)     (i)     METHOD 1

\(\overrightarrow {{\text{AB}}}  = \boldsymbol{b} – \boldsymbol{a} = \left( {\begin{array}{*{20}{c}}
  1 \\
  2 \\
  3
\end{array}} \right) – \left( {\begin{array}{*{20}{c}}
  1 \\
  1 \\
  2
\end{array}} \right) = \left( {\begin{array}{*{20}{c}}
  0 \\
  1 \\
  1
\end{array}} \right)\)     (A1)

\(\overrightarrow {{\text{AC}}}  = \boldsymbol{c} – \boldsymbol{a} = \left( {\begin{array}{*{20}{c}}
  3 \\
  0 \\
  1
\end{array}} \right) – \left( {\begin{array}{*{20}{c}}
  1 \\
  1 \\
  2
\end{array}} \right) = \left( {\begin{array}{*{20}{c}}
  2 \\
  { – 1} \\
  { – 1}
\end{array}} \right)\)     (A1)

\(\overrightarrow {{\text{AB}}}  \times \overrightarrow {{\text{AC}}}  = \left| {\begin{array}{*{20}{c}}
  \boldsymbol{i}&\boldsymbol{j}&\boldsymbol{k} \\
  0&1&1 \\
  2&{ – 1}&{ – 1}
\end{array}} \right|\)     M1

i(−1 + 1) − j(0 − 2) + k (0 − 2)     (A1)

= 2j − 2k     A1

Area of triangle ABC}} \( = \frac{1}{2}\left| {2{\mathbf{j}} – 2{\mathbf{k}}} \right| = {\text{ }}\frac{1}{2}\sqrt 8 \)   \(( = \sqrt 2 )\)   sq. units     M1A1

Note: Allow FT on final A1.

METHOD 2

\(\left| {{\text{AB}}} \right| = \sqrt 2 ,{\text{ }}\left| {{\text{BC}}} \right| = \sqrt {12} ,{\text{ }}\left| {{\text{AC}}} \right| = \sqrt 6 \)     A1A1A1

Using cosine rule, e.g. on \({\hat C}\)     M1

\(\cos C = \frac{{6 + 12 – 2}}{{2\sqrt {72} }} = \frac{{2\sqrt 2 }}{3}\)     A1

\(\therefore {\text{Area }}\Delta {\text{ABC}} = \frac{1}{2}ab\sin C\)     M1

\( = \frac{1}{2}\sqrt {12} \sqrt 6 \sin \left( {\arccos \frac{{2\sqrt 2 }}{3}} \right)\)

\( = 3\sqrt 2 \sin \left( {\arccos \frac{{2\sqrt 2 }}{3}} \right){\text{ }}\left( { = \sqrt 2 } \right)\)     A1

Note: Allow FT on final A1.

(ii)     \({\text{AB}} = \sqrt 2 \)     A1

\(\sqrt 2  = \frac{1}{2}{\text{AB}} \times h = \frac{1}{2}\sqrt 2  \times h{\text{ , }}h\) equals the shortest distance     (M1)

\( \Rightarrow h = 2\)     A1

(iii)     METHOD 1

(\pi \) has form \(r \cdot \left( {\begin{array}{*{20}{c}}
  0 \\
  2 \\
  { – 2}
\end{array}} \right) = d\)     (M1)

Since (1, 1, 2) is on the plane

\(d = \left( {\begin{array}{*{20}{c}}
  1 \\
  1 \\
  2
\end{array}} \right) \cdot \left( {\begin{array}{*{20}{c}}
  0 \\
  2 \\
  { – 2}
\end{array}} \right) = 2 – 4 = -2\)     M1A1

Hence \(r \cdot \left( {\begin{array}{*{20}{c}}
  0 \\
  2 \\
  { – 2}
\end{array}} \right) = – 2\)

\(2y – 2z = – 2{\text{ (or }}y – z = – 1)\)     A1

METHOD 2

\(r = \left( {\begin{array}{*{20}{c}}
  1 \\
  1 \\
  2
\end{array}} \right) + \lambda \left( {\begin{array}{*{20}{c}}
  0 \\
  1 \\
  1
\end{array}} \right) + \mu \left( {\begin{array}{*{20}{c}}
  2 \\
  { – 1} \\
  { – 1}
\end{array}} \right)\)     (M1)

\(x = 1 + 2\mu \)     (i)

\(y = 1 + \lambda – \mu \)     (ii)

\(z = 2 + \lambda – \mu \)     (iii)     A1

Note: Award A1 for all three correct, A0 otherwise.

From (i) \(\mu = \frac{{x – 1}}{2}\)

substitute in (ii) \(y = 1 + \lambda – \left( {\frac{{x – 1}}{2}} \right)\)

\( \Rightarrow \lambda = y – 1 + \left( {\frac{{x – 1}}{2}} \right)\)

substitute \(\lambda \) and \(\mu \) in (iii)     M1

\( \Rightarrow z = 2 + y – 1 + \left( {\frac{{x – 1}}{2}} \right) – \left( {\frac{{x – 1}}{2}} \right)\)

\( \Rightarrow y – z = – 1\)     A1

[14 marks]

(b)     (i)     The equation of OD is

\(r = \lambda \left( {\begin{array}{*{20}{c}}
  0 \\
  2 \\
  { – 2}
\end{array}} \right)\), \(\left( {{\text{or }}r = \lambda \left( {\begin{array}{*{20}{c}}
  0 \\
  1 \\
  { – 1}
\end{array}} \right)} \right)\)     M1

This meets \(\pi \) where

\(2\lambda + 2\lambda = – 1\)     (M1)

\(\lambda = – \frac{1}{4}\)     A1

Coordinates of D are \(\left( {0, – \frac{1}{2},\frac{1}{2}} \right)\)     A1

(ii)     \(\left| {\overrightarrow {{\text{OD}}} } \right| = \sqrt {0 + {{\left( { – \frac{1}{2}} \right)}^2} + {{\left( {\frac{1}{2}} \right)}^2}} = \frac{1}{{\sqrt 2 }}\)     (M1)A1

[6 marks]

Total [20 marks]

Examiners report

It was disappointing to see that a number of candidates did not appear to be well prepared for this question and made no progress at all. There were a number of schools where no candidate made any appreciable progress with the question. A good number of students, however, were successful with part (a) (i). A good number of candidates were also successful with part a (ii) but few realised that the shortest distance was the height of the triangle. Candidates used a variety of methods to answer (a) (iii) but again a reasonable number of correct answers were seen. Candidates also had a reasonable degree of success with part (b), with a respectable number of correct answers seen.

Question

The following figure shows a square based pyramid with vertices at O(0, 0, 0), A(1, 0, 0), B(1, 1, 0), C(0, 1, 0) and D(0, 0, 1).

The Cartesian equation of the plane \({\Pi _2}\), passing through the points B , C and D , is \(y + z = 1\).

The plane \({\Pi _3}\) passes through O and is normal to the line BD.

\({\Pi _3}\) cuts AD and BD at the points P and Q respectively.

a.Find the Cartesian equation of the plane \({\Pi _1}\), passing through the points A , B and D.[3]

b.Find the angle between the faces ABD and BCD.[4]

c.Find the Cartesian equation of \({\Pi _3}\).[3]

d.Show that P is the midpoint of AD.[4]

e.Find the area of the triangle OPQ.[5]

 
▶️Answer/Explanation

Markscheme

recognising normal to plane or attempting to find cross product of two vectors lying in the plane      (M1)

for example, \(\mathop {{\text{AB}}}\limits^ \to \,\, \times \mathop {{\text{AD}}}\limits^ \to = \left( \begin{gathered}
0 \hfill \\
1 \hfill \\
0 \hfill \\
\end{gathered} \right) \times \left( \begin{gathered}
– 1 \hfill \\
\,0 \hfill \\
\,1 \hfill \\
\end{gathered} \right) = \left( \begin{gathered}
1 \hfill \\
0 \hfill \\
1 \hfill \\
\end{gathered} \right)\)     (A1)

\({\Pi _1}\,{\text{:}}\,\,x + z = 1\)     A1

[3 marks]

a.

EITHER

\(\left( \begin{gathered}
1 \hfill \\
0 \hfill \\
1 \hfill \\
\end{gathered} \right) \bullet \left( \begin{gathered}
0 \hfill \\
1 \hfill \\
1 \hfill \\
\end{gathered} \right) = 1 = \sqrt 2 \sqrt 2 \,{\text{cos}}\,\theta \)     M1A1

OR

\(\left| {\left( \begin{gathered}
1 \hfill \\
0 \hfill \\
1 \hfill \\
\end{gathered} \right) \times \left( \begin{gathered}
0 \hfill \\
1 \hfill \\
1 \hfill \\
\end{gathered} \right)} \right| = \sqrt 3 = \sqrt 2 \sqrt 2 \,{\text{sin}}\,\theta \)     M1A1

Note: M1 is for an attempt to find the scalar or vector product of the two normal vectors.

\( \Rightarrow \theta  = 60^\circ \left( { = \frac{\pi }{3}} \right)\)     A1

angle between faces is \(20^\circ \left( { = \frac{{2\pi }}{3}} \right)\)     A1

[4 marks]

b.

\(\mathop {{\text{DB}}}\limits^ \to = \left( \begin{gathered}
\,1 \hfill \\
\,1 \hfill \\
– 1 \hfill \\
\end{gathered} \right)\) or \(\mathop {{\text{BD}}}\limits^ \to = \left( \begin{gathered}
– 1 \hfill \\
– 1 \hfill \\
\,1 \hfill \\
\end{gathered} \right)\)     (A1)

\({\Pi _3}\,{\text{:}}\,\,x + y – z = k\)     (M1)

\({\Pi _3}\,{\text{:}}\,\,x + y – z = 0\)     A1

[3 marks]

c.

METHOD 1

line AD : (r =)\(\left( \begin{gathered}
0 \hfill \\
0 \hfill \\
1 \hfill \\
\end{gathered} \right) + \lambda \left( \begin{gathered}
\,1 \hfill \\
\,0 \hfill \\
– 1 \hfill \\
\end{gathered} \right)\)     M1A1

intersects \({\Pi _3}\) when \(\lambda  – \left( {1 – \lambda } \right) = 0\)     M1

so \(\lambda  = \frac{1}{2}\)     A1

hence P is the midpoint of AD      AG

METHOD 2

midpoint of AD is (0.5, 0, 0.5)      (M1)A1

substitute into \(x + y – z = 0\)     M1

0.5 + 0.5 − 0.5 = 0     A1

hence P is the midpoint of AD     AG

[4 marks]

d.

METHOD 1

\({\text{OP}} = \frac{1}{{\sqrt 2 }},\,\,{\text{O}}\mathop {\text{P}}\limits^ \wedge  {\text{Q}} = 90^\circ ,\,\,{\text{O}}\mathop {\text{Q}}\limits^ \wedge  {\text{P}} = 60^\circ \)      A1A1A1

\({\text{PQ}} = \frac{1}{{\sqrt 6 }}\)     A1

area \( = \frac{1}{{2\sqrt {12} }} = \frac{1}{{4\sqrt 3 }} = \frac{{\sqrt 3 }}{{12}}\)     A1

METHOD 2

line BD : ( =)\(\left( \begin{gathered}
1 \hfill \\
1 \hfill \\
0 \hfill \\
\end{gathered} \right) + \lambda \left( \begin{gathered}
– 1 \hfill \\
– 1 \hfill \\
\,1 \hfill \\
\end{gathered} \right)\)

\( \Rightarrow \lambda  = \frac{2}{3}\)     (A1)

\(\mathop {{\text{OQ}}}\limits^ \to = \left( \begin{gathered}
\frac{1}{3} \hfill \\
\frac{1}{3} \hfill \\
\frac{2}{3} \hfill \\
\end{gathered} \right)\)    A1

area = \(\frac{1}{2}\left| {\mathop {{\text{OP}}}\limits^ \to  \, \times \mathop {{\text{OQ}}}\limits^ \to  } \right|\)     M1

\(\mathop {{\text{OP}}}\limits^ \to = \left( \begin{gathered}
\frac{1}{2} \hfill \\
0 \hfill \\
\frac{1}{2} \hfill \\
\end{gathered} \right)\)    A1

Note: This A1 is dependent on M1.

area = \(\frac{{\sqrt 3 }}{{12}}\)     A1

[5 marks]

e.

Examiners report

[N/A]

a.

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b.

[N/A]

c.

[N/A]

d.

[N/A]

e.
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