Home / IBDP Maths AA: Topic : AHL 3.12: Components of a vector: IB style Questions HL Paper 2

# IBDP Maths AA: Topic : AHL 3.12: Components of a vector: IB style Questions HL Paper 2

### 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.

a.i.Find $$\mathop {{\text{PQ}}}\limits^ \to$$.[2]

a.ii.Find $$\left| {\mathop {{\text{PQ}}}\limits^ \to } \right|$$.[2]

b.Find the angle between PQ and PR.[4]

c.Find the area of triangle PQR.[2]

d.Hence or otherwise find the shortest distance from R to the line through P and Q.[3]

▶️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]

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