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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.

[5]
a.

(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.

[7]
b.

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]

a.

(i)     direction of line = (3i + j + 2k) – (i + 2j + 3k) = 2ijk     M1A1

equation of line is

r = (i + 2j + 3k) + \(\lambda \)(2ijk)     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]

b.

Examiners report

[N/A]
a.
[N/A]
b.

Syllabus sections

Topic 4 - Core: Vectors » 4.5 » The definition of the vector product of two vectors.

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