Home / IB DP Math AA: Topic : AHL 3.12 :Concept of a vector: IB Style Questions HL Paper 2

# IB DP Math AA: Topic : AHL 3.12 :Concept of a vector: IB Style Questions HL Paper 2

### 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]

## 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]

a.

$$\overrightarrow {{\text{PQ}}} = \left( {\begin{array}{*{20}{c}} { – 1} \\ { – 1} \\ 3 \end{array}} \right)$$$$\overrightarrow {{\text{PS}}} = \left( {\begin{array}{*{20}{c}} 2 \\ 4 \\ 1 \end{array}} \right)$$     A1

$$\overrightarrow {{\text{PQ}}} \times \overrightarrow {{\text{PS}}} = \left( {\begin{array}{*{20}{c}} { – 13} \\ 7 \\ { – 2} \end{array}} \right)$$

m = −2     A1

[2 marks]

b.

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]

c.

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]

d.

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]

e.

$$169\lambda + 49\lambda + 4\lambda = 33$$     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]

f.

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

Two submarines A and B have their routes planned so that their positions at time t hours, 0 ≤ t < 20 , would be defined by the position vectors rA

$$\left( \begin{gathered} \,2 \hfill \\ \,4 \hfill \\ – 1 \hfill \\ \end{gathered} \right) + t\left( \begin{gathered} – 1 \hfill \\ \,1 \hfill \\ – 0.15 \hfill \\ \end{gathered} \right)$$

and rB $$= \left( \begin{gathered} \,0 \hfill \\ \,3.2 \hfill \\ – 2 \hfill \\ \end{gathered} \right) + t\left( \begin{gathered} – 0.5 \hfill \\ \,1.2 \hfill \\ \,0.1 \hfill \\ \end{gathered} \right)$$

relative to a fixed point on the surface of the ocean (all lengths are in kilometres).

To avoid the collision submarine B adjusts its velocity so that its position vector is now given by

rB $$= \left( \begin{gathered} \,0 \hfill \\ \,3.2 \hfill \\ – 2 \hfill \\ \end{gathered} \right) + t\left( \begin{gathered} – 0.45 \hfill \\ \,1.08 \hfill \\ \,0.09 \hfill \\ \end{gathered} \right)$$.

a.Show that the two submarines would collide at a point P and write down the coordinates of P.[4]

b.i.Show that submarine B travels in the same direction as originally planned.[1]

b.ii.Find the value of t when submarine B passes through P.[2]

c.i.Find an expression for the distance between the two submarines in terms of t.[5]

c.ii.Find the value of t when the two submarines are closest together.[2]

c.iii.Find the distance between the two submarines at this time.[1]

## Markscheme

rA rB        (M1)

2 − t = − 0.5t ⇒ t = 4       A1

checking t = 4 satisfies 4 + t = 3.2 + 1.2t and − 1 − 0.15t = − 2 + 0.1t      R1

P(−2, 8, −1.6)      A1

Note: Do not award final A1 if answer given as column vector.

[4 marks]

a.

$$0.9 \times \left( \begin{gathered} – 0.5 \hfill \\ \,1.2 \hfill \\ \,0.1 \hfill \\ \end{gathered} \right) = \left( \begin{gathered} – 0.45 \hfill \\ \,1.08 \hfill \\ \,0.09 \hfill \\ \end{gathered} \right)$$    A1

Note: Accept use of cross product equalling zero.

hence in the same direction      AG

[1 mark]

b.i.

$$\left( \begin{gathered} \, – 0.45t \hfill \\ 3.2 + 1.08t \hfill \\ – 2 + 0.09t \hfill \\ \end{gathered} \right) = \left( \begin{gathered} – 2 \hfill \\ \,8 \hfill \\ – 1.6 \hfill \\ \end{gathered} \right)$$     M1

Note: The M1 can be awarded for any one of the resultant equations.

$$\Rightarrow t = \frac{{40}}{9} = 4.44 \ldots$$     A1

[2 marks]

b.ii.

rA − rB = $$\left( \begin{gathered} \,2 – t \hfill \\ \,4 + t \hfill \\ – 1 – 0.15t \hfill \\ \end{gathered} \right) – \left( \begin{gathered} \, – 0.45t \hfill \\ 3.2 + 1.08t \hfill \\ – 2 + 0.09t \hfill \\ \end{gathered} \right)$$
(M1)(A1)

$$= \left( \begin{gathered} \,2 – 0.55t \hfill \\ \,0.8 – 0.08t \hfill \\ 1 – 0.24t \hfill \\ \end{gathered} \right)$$    (A1)

Note: Accept rA − rB.

distance $$D = \sqrt {{{\left( {2 – 0.55t} \right)}^2} + {{\left( {0.8 – 0.08t} \right)}^2} + {{\left( {1 – 0.24t} \right)}^2}}$$      M1A1

$$\left( { = \sqrt {8.64 – 2.688t + 0.317{t^2}} } \right)$$

[5 marks]

c.i.

minimum when $$\frac{{{\text{d}}D}}{{{\text{d}}t}} = 0$$      (M1)

t = 3.83      A1

[2 marks]

c.ii.

0.511 (km)      A1 [1 mark]

### Question

OABCDE is a regular hexagon and a , b denote respectively the position vectors of A, B with respect to O.

a.Show that OC = 2AB .[2]

b.Find the position vectors of C, D and E in terms of a and b .[7]

## Markscheme

$${\text{OC}} = {\text{AB}} + {\text{OA}}\cos 60 + {\text{BC}}\cos 60$$     M1

$$= {\text{AB}} + {\text{AB}} \times \frac{1}{2} + {\text{AB}} \times \frac{1}{2}$$     A1

$$= 2{\text{AB}}$$     AG

[2 marks]

a.

$$\overrightarrow {{\text{OC}}} = 2\overrightarrow {{\text{AB}}} =$$2(ba)     M1A1

$$\overrightarrow {{\text{OD}}} = \overrightarrow {{\text{OC}}} + \overrightarrow {{\text{CD}}}$$     M1

$$= \overrightarrow {{\text{OC}}} + \overrightarrow {{\text{AO}}}$$     A1

= 2b – 2aa = 2b – 3a     A1

$$\overrightarrow {{\text{OE}}} = \overrightarrow {{\text{BC}}}$$     M1

= 2b – 2ab = b – 2a     A1

[7 marks]

b.

### Question

OABCDE is a regular hexagon and a , b denote respectively the position vectors of A, B with respect to O.

a.Show that OC = 2AB .[2]

b.Find the position vectors of C, D and E in terms of a and b .[7]

## Markscheme

$${\text{OC}} = {\text{AB}} + {\text{OA}}\cos 60 + {\text{BC}}\cos 60$$     M1

$$= {\text{AB}} + {\text{AB}} \times \frac{1}{2} + {\text{AB}} \times \frac{1}{2}$$     A1

$$= 2{\text{AB}}$$     AG

[2 marks]

a.

$$\overrightarrow {{\text{OC}}} = 2\overrightarrow {{\text{AB}}} =$$2(ba)     M1A1

$$\overrightarrow {{\text{OD}}} = \overrightarrow {{\text{OC}}} + \overrightarrow {{\text{CD}}}$$     M1

$$= \overrightarrow {{\text{OC}}} + \overrightarrow {{\text{AO}}}$$     A1

= 2b – 2aa = 2b – 3a     A1

$$\overrightarrow {{\text{OE}}} = \overrightarrow {{\text{BC}}}$$     M1

= 2b – 2ab = b – 2a     A1

[7 marks]

b.
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