IB DP Maths Topic 4.1 Components of a vector; column representation SL Paper 2

 

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Question

Two points P and Q have coordinates (3, 2, 5) and (7, 4, 9) respectively.

Let \({\mathop {{\text{PR}}}\limits^ \to  }\) = 6i − j + 3k.

Find \(\mathop {{\text{PQ}}}\limits^ \to  \).

[2]
a.i.

Find \(\left| {\mathop {{\text{PQ}}}\limits^ \to  } \right|\).

[2]
a.ii.

Find the angle between PQ and PR.

[4]
b.

Find the area of triangle PQR.

[2]
c.

Hence or otherwise find the shortest distance from R to the line through P and Q.

[3]
d.
Answer/Explanation

Markscheme

valid approach      (M1)

eg   (7, 4, 9) − (3, 2, 5)  A − B

\(\mathop {{\text{PQ}}}\limits^ \to   = \) 4i + 2j + 4k \(\left( { = \left( \begin{gathered}
4 \hfill \\
2 \hfill \\
4 \hfill \\
\end{gathered} \right)} \right)\)     A1 N2

[2 marks]

a.i.

correct substitution into magnitude formula      (A1)
eg  \(\sqrt {{4^2} + {2^2} + {4^2}} \)

\(\left| {\mathop {{\text{PQ}}}\limits^ \to  } \right| = 6\)     A1 N2

[2 marks]

a.ii.

finding scalar product and magnitudes      (A1)(A1)

scalar product = (4 × 6) + (2 × (−1) + (4 × 3) (= 34)

magnitude of PR = \(\sqrt {36 + 1 + 9}  = \left( {6.782} \right)\)

correct substitution of their values to find cos \({\text{Q}}\mathop {\text{P}}\limits^ \wedge  {\text{R}}\)     M1

eg  cos \({\text{Q}}\mathop {\text{P}}\limits^ \wedge  {\text{R}}\,\,{\text{ = }}\frac{{24 – 2 + 12}}{{\left( 6 \right) \times \left( {\sqrt {46} } \right)}},\,\,0.8355\)

0.581746

\({\text{Q}}\mathop {\text{P}}\limits^ \wedge  {\text{R}}\) = 0.582 radians  or \({\text{Q}}\mathop {\text{P}}\limits^ \wedge  {\text{R}}\) = 33.3°     A1 N3

[4 marks]

b.

correct substitution (A1)
eg    \(\frac{1}{2} \times \left| {\mathop {{\text{PQ}}}\limits^ \to  } \right| \times \left| {\mathop {{\text{PR}}}\limits^ \to  } \right| \times \,\,{\text{sin}}\,P,\,\,\frac{1}{2} \times 6 \times \sqrt {46}  \times \,\,{\text{sin}}\,0.582\)

area is 11.2 (sq. units)      A1 N2

[2 marks]

c.

recognizing shortest distance is perpendicular distance from R to line through P and Q      (M1)

eg  sketch, height of triangle with base \(\left[ {{\text{PQ}}} \right],\,\,\frac{1}{2} \times 6 \times h,\,\,{\text{sin}}\,33.3^\circ  = \frac{h}{{\sqrt {46} }}\)

correct working      (A1)

eg  \(\frac{1}{2} \times 6 \times d = 11.2,\,\,\left| {\mathop {{\text{PR}}}\limits^ \to  } \right| \times \,\,{\text{sin}}\,P,\,\,\sqrt {46}  \times \,\,{\text{sin}}\,33.3^\circ \)

3.72677

distance = 3.73  (units)    A1 N2

[3 marks]

d.
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