IB DP Maths Topic 4.3 Simple applications to kinematics HL Paper 2

(a)     \(\overrightarrow {{\text{OP}}}  = {\boldsymbol{i}} + 2{\boldsymbol{j}} – {\boldsymbol{k}}\)     (M1)

the coordinates of P are (1, 2, –1)     A1

[2 marks]

(b)     EITHER

\(x = 1 + t\) , \(y = 2 – 2t\) , \(z = 3t – 1\)     M1

\(x – 1 = t\) , \(\frac{{y – 2}}{{ – 2}} = t\) , \(\frac{{z + 1}}{3} = t\)     A1

\(x – 1 = \frac{{y – 2}}{{ – 2}} = \frac{{z + 1}}{3}\)     AG     N0

OR

\(\left( {\begin{array}{*{20}{c}}
  x \\
  y \\
  z
\end{array}} \right) = \left( {\begin{array}{*{20}{c}}
  1 \\
  2 \\
  { – 1}
\end{array}} \right) + t\left( {\begin{array}{*{20}{c}}
  1 \\
  { – 2} \\
  3
\end{array}} \right)\)     M1A1

\(x – 1 = \frac{{y – 2}}{{ – 2}} = \frac{{z + 1}}{3}\)     AG

[2 marks]

(c)     (i)     \(2\left( {1 + t} \right) + \left( {2 – 2t} \right) + \left( {3t – 1} \right) = 6 \Rightarrow t = 1\)     M1A1     N1

(ii)     coordinates are (2, 0, 2)     A1

Note: Award A0 for position vector.

(iii)     distance travelled is the distance between the two points     (M1)

\(\sqrt {{{\left( {2 – 1} \right)}^2} + {{\left( {0 – 2} \right)}^2} + {{\left( {2 + 1} \right)}^2}}  = \sqrt {14} \)   (\( = 3.74\))     (M1)A1

[6 marks]

(d)     (i)     distance from Q to the origin is given by

\(d(t) = \sqrt {{t^4} + {{\left( {1 – t} \right)}^2} + {{\left( {1 – {t^2}} \right)}^2}} \)   (or equivalent)     M1A1

e.g. for labelled sketch of graph of \(d\) or \({d^2}\)     (M1)(A1)

     or    

the minimum value is obtained for \(t = 0.761\)     A1     N3

(ii)     the coordinates are (0.579, 0.239, 0.421)     A1

Note: Accept answers given as a position vector.

[6 marks]

(e)     (i)     \({\boldsymbol{a}} = \left( {\begin{array}{*{20}{c}}
  0 \\
  1 \\
  1
\end{array}} \right)\), \({\boldsymbol{b}} = \left( {\begin{array}{*{20}{c}}
  1 \\
  0 \\
  0
\end{array}} \right)\) and \({\boldsymbol{c}} = \left( {\begin{array}{*{20}{c}}
  4 \\
  { – 1} \\
  { – 3}
\end{array}} \right)\)     (M1)A1

substituting in the equation \({\boldsymbol{a}} – {\boldsymbol{b}} = k\left( {{\boldsymbol{b}} – {\boldsymbol{c}}} \right)\) , we have     (M1)

\(\left( {\begin{array}{*{20}{c}}
  0 \\
  1 \\
  1
\end{array}} \right) – \left( {\begin{array}{*{20}{c}}
  1 \\
  0 \\
  0
\end{array}} \right) = k\left( {\left( {\begin{array}{*{20}{c}}
  1 \\
  0 \\
  0
\end{array}} \right) – \left( {\begin{array}{*{20}{c}}
  4 \\
  { – 1} \\
  { – 3}
\end{array}} \right)} \right) \Leftrightarrow \left( {\begin{array}{*{20}{c}}
  { – 1} \\
  1 \\
  1
\end{array}} \right) = k\left( {\begin{array}{*{20}{c}}
  { – 3} \\
  1 \\
  3
\end{array}} \right)\)     A1

\( \Rightarrow k = 1\) and \(k = \frac{1}{3}\) which is impossible

so there is no solution for \(k\)     R1

(ii)     \({{\text{BA}}}\)and \(\overrightarrow {{\text{CB}}} \)are not parallel     R2

(hence A, B, and C cannot be collinear)

Note: Only accept answers that follow from part (i).

[7 marks]

Total [23 marks]

Generally this question was answered well by those students who attempted it. It was common to see confusion between coordinates and position vectors. Part (d) was most easily answered with the use of a GDC, but fewer candidates took advantage of this. In part (e) many students had difficulties expressing their reasoning well to obtain the marks.