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]
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]
\(\mathop {{\text{DB}}}\limits^ \to = \left( \begin{gathered}
\,1 \hfill \\
\,1 \hfill \\
– 1 \hfill \\
\end{gathered} \right)\) 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]
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]
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 : (r =)\(\left( \begin{gathered}
1 \hfill \\
1 \hfill \\
0 \hfill \\
\end{gathered} \right) + \lambda \left( \begin{gathered}
– 1 \hfill \\
– 1 \hfill \\
\,1 \hfill \\
\end{gathered} \right)\)
\( \Rightarrow \lambda = \frac{2}{3}\) (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]
Examiners report
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