# IB DP Maths Topic 4.1 :Algebraic and geometric approaches to the zero vector HL Paper 2

## Question

OACB is a parallelogram with $$\overrightarrow {{\text{OA}}} =$$ a and $$\overrightarrow {{\text{OB}}} =$$ b, where a and b are non-zero vectors.

Show that

(i)     $${\left| {\overrightarrow {{\text{OC}}} } \right|^2} = |$$a$${|^2} + 2$$a $$\bullet$$ b $$+ |$$b$${|^2}$$;

(ii)     $${\left| {\overrightarrow {{\text{AB}}} } \right|^2} = |$$a$${|^2} – 2$$a $$\bullet$$ b $$+ |$$b$${|^2}$$.

[4]
a.

Given that $$\left| {\overrightarrow {{\text{OC}}} } \right| = \left| {\overrightarrow {{\text{AB}}} } \right|$$, prove that OACB is a rectangle.

[4]
b.

## Markscheme

METHOD 1

$${\left| {\overrightarrow {{\text{OC}}} } \right|^2} = \overrightarrow {{\text{OC}}} \bullet \overrightarrow {{\text{OC}}}$$

= (+ b) $$\bullet$$ (b)     A1

a $$\bullet$$ a $$\bullet$$ b $$\bullet$$ b $$\bullet$$ b     A1

= $$|$$a$${|^2}$$ + 2a $$\bullet$$ + $$|$$b$${|^2}$$     AG

METHOD 2

$${\left| {\overrightarrow {{\text{OC}}} } \right|^2} = {\left| {\overrightarrow {{\text{OA}}} } \right|^2} + {\left| {\overrightarrow {{\text{OB}}} } \right|^2} – 2\left| {\overrightarrow {{\text{OA}}} } \right|\left| {\overrightarrow {{\text{OB}}} } \right|\cos ({\rm{O\hat AC}})$$    A1

$$\left| {\overrightarrow {{\text{OA}}} } \right|\left| {\overrightarrow {{\text{OB}}} } \right|\cos ({\rm{O\hat AC}}) = –$$(a $$\bullet$$ b)     A1

$${\left| {\overrightarrow {{\text{OC}}} } \right|^2} = |$$a$${|^2}$$ + 2a $$\bullet$$ + $$|$$b$${|^2}$$     AG

(ii)     METHOD 1

$${\left| {\overrightarrow {{\text{AB}}} } \right|^2} = \overrightarrow {{\text{AB}}} \bullet \overrightarrow {{\text{AB}}}$$

= (b a) $$\bullet$$ (b a)     A1

= $$\bullet$$ b $$\bullet$$ a a $$\bullet$$ b + a $$\bullet$$ a     A1

= $$|$$a$${|^2}$$ – 2a $$\bullet$$ b + $$|$$b$${|^2}$$     AG

METHOD 2

$${\left| {\overrightarrow {{\text{AB}}} } \right|^2} = {\left| {\overrightarrow {{\text{AC}}} } \right|^2} + {\left| {\overrightarrow {{\text{BC}}} } \right|^2} – 2\left| {\overrightarrow {{\text{AC}}} } \right|\left| {\overrightarrow {{\text{BC}}} } \right|\cos ({\rm{A\hat CB}})$$    A1

$$\left| {\overrightarrow {{\text{AC}}} } \right|\left| {\overrightarrow {{\text{BC}}} } \right|\cos ({\rm{A\hat CB}}) =$$ a $$\bullet$$ b     A1

$${\left| {\overrightarrow {{\text{AB}}} } \right|^2} = |$$a $${|^2} –$$ 2a $$\bullet$$ b + $$|$$b$${|^2}$$     AG

[4 marks]

a.

$$\left| {\overrightarrow {{\text{OC}}} } \right| = \left| {\overrightarrow {{\text{AB}}} } \right| \Rightarrow {\left| {\overrightarrow {{\text{OC}}} } \right|^2} = {\left| {\overrightarrow {{\text{AB}}} } \right|^2} \Rightarrow |$$a$${|^2} + {\text{2}}$$a $$\bullet$$ b $$+ |$$b$${|^2} = |$$a$${|^2} – 2$$a $$\bullet$$ b $$+ |$$b$${|^2}$$     R1(M1)

Note:     Award R1 for $$\left| {\overrightarrow {{\text{OC}}} } \right| = \left| {\overrightarrow {{\text{AB}}} } \right| \Rightarrow {\left| {\overrightarrow {{\text{OC}}} } \right|^2} = {\left| {\overrightarrow {{\text{AB}}} } \right|^2}$$ and (M1) for $$|$$a$${|^2} + {\text{2}}$$a $$\bullet$$ b $$+ |$$b$${|^2} = |$$a$${|^2} – 2$$a $$\bullet$$ b $$+ |$$b$${|^2}$$.

a $$\bullet$$ b $$= 0$$     A1

hence OACB is a rectangle (a and b both non-zero)

with adjacent sides at right angles     R1

Note:     Award R1(M1)A0R1 if the dot product has not been used.

[4 marks]

b.

## Examiners report

In part (a), a significant number of candidates either did not use correct vector notation or simply did not use vector notation at all. A large number of candidates who appeared to adopt a scalar product approach, did not use the scalar product ‘dot’ and represented a $$\bullet$$ b as ab. A few candidates successfully used the cosine rule with correct vector notation. A small number of candidates expressed a and b in general component form. In part (a) (ii), quite a number of candidates expressed $${\overrightarrow {{\text{AB}}} }$$ as a – b rather than as b – a.

a.

In part (b), some very well structured proofs were offered by a small number of candidates.

b.

## Question

OACB is a parallelogram with $$\overrightarrow {{\text{OA}}} =$$ a and $$\overrightarrow {{\text{OB}}} =$$ b, where a and b are non-zero vectors.

Show that

(i)     $${\left| {\overrightarrow {{\text{OC}}} } \right|^2} = |$$a$${|^2} + 2$$a $$\bullet$$ b $$+ |$$b$${|^2}$$;

(ii)     $${\left| {\overrightarrow {{\text{AB}}} } \right|^2} = |$$a$${|^2} – 2$$a $$\bullet$$ b $$+ |$$b$${|^2}$$.

[4]
a.

Given that $$\left| {\overrightarrow {{\text{OC}}} } \right| = \left| {\overrightarrow {{\text{AB}}} } \right|$$, prove that OACB is a rectangle.

[4]
b.

## Markscheme

METHOD 1

$${\left| {\overrightarrow {{\text{OC}}} } \right|^2} = \overrightarrow {{\text{OC}}} \bullet \overrightarrow {{\text{OC}}}$$

= (+ b) $$\bullet$$ (b)     A1

a $$\bullet$$ a $$\bullet$$ b $$\bullet$$ b $$\bullet$$ b     A1

= $$|$$a$${|^2}$$ + 2a $$\bullet$$ + $$|$$b$${|^2}$$     AG

METHOD 2

$${\left| {\overrightarrow {{\text{OC}}} } \right|^2} = {\left| {\overrightarrow {{\text{OA}}} } \right|^2} + {\left| {\overrightarrow {{\text{OB}}} } \right|^2} – 2\left| {\overrightarrow {{\text{OA}}} } \right|\left| {\overrightarrow {{\text{OB}}} } \right|\cos ({\rm{O\hat AC}})$$    A1

$$\left| {\overrightarrow {{\text{OA}}} } \right|\left| {\overrightarrow {{\text{OB}}} } \right|\cos ({\rm{O\hat AC}}) = –$$(a $$\bullet$$ b)     A1

$${\left| {\overrightarrow {{\text{OC}}} } \right|^2} = |$$a$${|^2}$$ + 2a $$\bullet$$ + $$|$$b$${|^2}$$     AG

(ii)     METHOD 1

$${\left| {\overrightarrow {{\text{AB}}} } \right|^2} = \overrightarrow {{\text{AB}}} \bullet \overrightarrow {{\text{AB}}}$$

= (b a) $$\bullet$$ (b a)     A1

= $$\bullet$$ b $$\bullet$$ a a $$\bullet$$ b + a $$\bullet$$ a     A1

= $$|$$a$${|^2}$$ – 2a $$\bullet$$ b + $$|$$b$${|^2}$$     AG

METHOD 2

$${\left| {\overrightarrow {{\text{AB}}} } \right|^2} = {\left| {\overrightarrow {{\text{AC}}} } \right|^2} + {\left| {\overrightarrow {{\text{BC}}} } \right|^2} – 2\left| {\overrightarrow {{\text{AC}}} } \right|\left| {\overrightarrow {{\text{BC}}} } \right|\cos ({\rm{A\hat CB}})$$    A1

$$\left| {\overrightarrow {{\text{AC}}} } \right|\left| {\overrightarrow {{\text{BC}}} } \right|\cos ({\rm{A\hat CB}}) =$$ a $$\bullet$$ b     A1

$${\left| {\overrightarrow {{\text{AB}}} } \right|^2} = |$$a $${|^2} –$$ 2a $$\bullet$$ b + $$|$$b$${|^2}$$     AG

[4 marks]

a.

$$\left| {\overrightarrow {{\text{OC}}} } \right| = \left| {\overrightarrow {{\text{AB}}} } \right| \Rightarrow {\left| {\overrightarrow {{\text{OC}}} } \right|^2} = {\left| {\overrightarrow {{\text{AB}}} } \right|^2} \Rightarrow |$$a$${|^2} + {\text{2}}$$a $$\bullet$$ b $$+ |$$b$${|^2} = |$$a$${|^2} – 2$$a $$\bullet$$ b $$+ |$$b$${|^2}$$     R1(M1)

Note:     Award R1 for $$\left| {\overrightarrow {{\text{OC}}} } \right| = \left| {\overrightarrow {{\text{AB}}} } \right| \Rightarrow {\left| {\overrightarrow {{\text{OC}}} } \right|^2} = {\left| {\overrightarrow {{\text{AB}}} } \right|^2}$$ and (M1) for $$|$$a$${|^2} + {\text{2}}$$a $$\bullet$$ b $$+ |$$b$${|^2} = |$$a$${|^2} – 2$$a $$\bullet$$ b $$+ |$$b$${|^2}$$.

a $$\bullet$$ b $$= 0$$     A1

hence OACB is a rectangle (a and b both non-zero)

with adjacent sides at right angles     R1

Note:     Award R1(M1)A0R1 if the dot product has not been used.

[4 marks]

b.

## Examiners report

In part (a), a significant number of candidates either did not use correct vector notation or simply did not use vector notation at all. A large number of candidates who appeared to adopt a scalar product approach, did not use the scalar product ‘dot’ and represented a $$\bullet$$ b as ab. A few candidates successfully used the cosine rule with correct vector notation. A small number of candidates expressed a and b in general component form. In part (a) (ii), quite a number of candidates expressed $${\overrightarrow {{\text{AB}}} }$$ as a – b rather than as b – a.

a.

In part (b), some very well structured proofs were offered by a small number of candidates.

b.
Scroll to Top