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]
Question
Two planes \({\Pi _1}\) and \({\Pi _2}\) have equations \(2x + y + z = 1\) and \(3x + y – z = 2\) respectively.
a.Find the vector equation of L, the line of intersection of \({\Pi _1}\) and \({\Pi _2}\).[6]
b.Show that the plane \({\Pi _3}\) which is perpendicular to \({\Pi _1}\) and contains L, has equation \(x – 2z = 1\).[4]
c.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]
▶️Answer/Explanation
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]
\(\left( {\begin{array}{*{20}{c}}
2 \\
1 \\
1
\end{array}} \right) \times \left( {\begin{array}{*{20}{c}}
2 \\
{ – 5} \\
1
\end{array}} \right)\) 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 r \( \cdot \) n = a \( \cdot \) n \( \Rightarrow x – 2z = 1\) AG
[4 marks]
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 + j − k 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]