Question
The points P(−1, 2, − 3), Q(−2, 1, 0), R(0, 5, 1) and S form a parallelogram, where S is diagonally opposite Q.
a.Find the coordinates of S.[2]
b.The vector product \(\overrightarrow {{\text{PQ}}} \times \overrightarrow {{\text{PS}}} = \left( {\begin{array}{*{20}{c}}
{ – 13} \\
7 \\
m
\end{array}} \right)\). Find the value of m .[2]
c.Hence calculate the area of parallelogram PQRS.[2]
d.Find the Cartesian equation of the plane, \({\prod _1}\) , containing the parallelogram PQRS.[3]
e.Write down the vector equation of the line through the origin (0, 0, 0) that is perpendicular to the plane \({\prod _1}\) .[1]
f.Hence find the point on the plane that is closest to the origin.[3]
g.A second plane, \({\prod _2}\) , has equation x − 2y + z = 3. Calculate the angle between the two planes.[4]
▶️Answer/Explanation
Markscheme
\(\overrightarrow {{\text{PQ}}} = \left( {\begin{array}{*{20}{c}}
{ – 1} \\
{ – 1} \\
3
\end{array}} \right)\) , \(\overrightarrow {{\text{SR}}} = \left( {\begin{array}{*{20}{c}}
{0 – x} \\
{5 – y} \\
{1 – z}
\end{array}} \right)\) (M1)
point S = (1, 6, −2) A1
[2 marks]
\(\overrightarrow {{\text{PQ}}} = \left( {\begin{array}{*{20}{c}}
{ – 1} \\
{ – 1} \\
3
\end{array}} \right)\)\(\overrightarrow {{\text{PS}}} = \left( {\begin{array}{*{20}{c}}
2 \\
4 \\
1
\end{array}} \right)\) A1
\(\overrightarrow {{\text{PQ}}} \times \overrightarrow {{\text{PS}}} = \left( {\begin{array}{*{20}{c}}
{ – 13} \\
7 \\
{ – 2}
\end{array}} \right)\)
m = −2 A1
[2 marks]
area of parallelogram PQRS \( = \left| {\overrightarrow {{\text{PQ}}} \times \overrightarrow {{\text{PS}}} } \right| = \sqrt {{{( – 13)}^2} + {7^2} + {{( – 2)}^2}} \) M1
\( = \sqrt {222} = 14.9\) A1
[2 marks]
equation of plane is −13x + 7y − 2z = d M1A1
substituting any of the points given gives d = 33
−13x + 7y − 2z = 33 A1
[3 marks]
equation of line is \(\boldsymbol{r} = \left( {\begin{array}{*{20}{c}}
0 \\
0 \\
0
\end{array}} \right) + \lambda \left( {\begin{array}{*{20}{c}}
{ – 13} \\
7 \\
{ – 2}
\end{array}} \right)\) A1
Note: To get the A1 must have \(\boldsymbol{r} =\) or equivalent.
[1 mark]
\(169\lambda + 49\lambda + 4\lambda = 33\) M1
\(\lambda = \frac{{33}}{{222}}{\text{ }}( = 0.149…)\) A1
closest point is \(\left( { – \frac{{143}}{{74}},\frac{{77}}{{74}}, – \frac{{11}}{{37}}} \right){\text{ }}\left( { = ( – 1.93,{\text{ 1.04, – 0.297)}}} \right)\) A1
[3 marks]
angle between planes is the same as the angle between the normals (R1)
\(\cos \theta = \frac{{ – 13 \times 1 + 7 \times – 2 – 2 \times 1}}{{\sqrt {222} \times \sqrt 6 }}\) M1A1
\(\theta = 143^\circ \) (accept \(\theta = 37.4^\circ \) or 2.49 radians or 0.652 radians) A1
[4 marks]
Question
The points A and B have coordinates (1, 2, 3) and (3, 1, 2) relative to an origin O.
(i) Find \(\overrightarrow {{\text{OA}}} \times \overrightarrow {{\text{OB}}} \) .
(ii) Determine the area of the triangle OAB.
(iii) Find the Cartesian equation of the plane OAB.
(i) Find the vector equation of the line \({L_1}\) containing the points A and B.
(ii) The line \({L_2}\) has vector equation \(\left( {\begin{array}{*{20}{c}}
x \\
y \\
z
\end{array}} \right) = \left( {\begin{array}{*{20}{c}}
2 \\
4 \\
3
\end{array}} \right) + \mu \left( {\begin{array}{*{20}{c}}
1 \\
3 \\
2
\end{array}} \right)\).
Determine whether or not \({L_1}\) and \({L_2}\) are skew.
▶️Answer/Explanation
Markscheme
(i) \(\overrightarrow {{\text{OA}}} \times \overrightarrow {{\text{OB}}} = \) i + 7j – 5k A1
(ii) area \( = \frac{1}{2}|\)i + 7j – 5k\(| = \frac{{5\sqrt 3 }}{2}{\text{(4.33)}}\) M1A1
(iii) equation of plane is \(x + 7y – 5z = k\) M1
\(x + 7y – 5z = 0\) A1
[5 marks]
(i) direction of line = (3i + j + 2k) – (i + 2j + 3k) = 2i – j – k M1A1
equation of line is
r = (i + 2j + 3k) + \(\lambda \)(2i – j – k) A1
(ii) at a point of intersection,
\(1 + 2\lambda = 2 + \mu \)
\(2 – \lambda = 4 + 3\mu \) M1A1
\(3 – \lambda = 3 + 2\mu \)
solving the \({2^{{\text{nd}}}}\) and \({3^{{\text{rd}}}}\) equations, \(\lambda = 4{\text{, }}\mu = – 2\) A1
these values do not satisfy the \({1^{{\text{st}}}}\) equation so the lines are skew R1
[7 marks]