Date | None Specimen | Marks available | 5 | Reference code | SPNone.2.hl.TZ0.10 |
Level | HL only | Paper | 2 | Time zone | TZ0 |
Command term | Determine and Find | Question number | 10 | Adapted from | N/A |
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.
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]