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2015
P1 Q7
Topical
Vectors I
15 P1 Q7
2015 H2 Mathematics Paper 1 Question 7
Vectors I: Basics, Dot and Cross Products
Answers
(i)
O
C
→
=
3
5
a
{\overrightarrow{OC} = \frac{3}{5} \mathbf{a}}
OC
=
5
3
a
O
D
→
=
5
11
b
{\overrightarrow{OD} = \frac{5}{11} \mathbf{b}}
O
D
=
11
5
b
(ii)
l
B
C
:
r
=
3
5
λ
a
+
(
1
−
λ
)
b
{l_{BC}: \mathbf{r} = \frac{3}{5}\lambda\mathbf{a} + (1-\lambda)\mathbf{b}}
l
BC
:
r
=
5
3
λ
a
+
(
1
−
λ
)
b
l
A
D
:
r
=
(
1
−
μ
)
a
+
5
11
μ
b
{l_{AD}: \mathbf{r} = (1-\mu)\mathbf{a} + \frac{5}{11}\mu\mathbf{b}}
l
A
D
:
r
=
(
1
−
μ
)
a
+
11
5
μ
b
(iii)
O
E
→
=
9
20
a
+
1
4
b
{\overrightarrow{OE} = \frac{9}{20}\mathbf{a}+\frac{1}{4}\mathbf{b}}
OE
=
20
9
a
+
4
1
b
A
E
:
E
D
=
11
:
9
{AE:ED = 11:9}
A
E
:
E
D
=
11
:
9
Full solutions
(i)
O
C
→
=
3
5
a
■
{\overrightarrow{OC} = \frac{3}{5} \mathbf{a} \; \blacksquare}
OC
=
5
3
a
■
O
D
→
=
5
11
b
■
{\overrightarrow{OD} = \frac{5}{11} \mathbf{b} \; \blacksquare}
O
D
=
11
5
b
■
(ii)
B
C
→
=
O
C
→
−
O
B
→
=
3
5
a
−
b
\begin{align*} \overrightarrow{BC} &= \overrightarrow{OC} - \overrightarrow{OB} \\ &= \frac{3}{5} \mathbf{a} - \mathbf{b} \end{align*}
BC
=
OC
−
OB
=
5
3
a
−
b
Equation of line
B
C
:
{BC:}
BC
:
r
=
O
B
→
+
λ
B
C
→
,
λ
∈
R
=
b
+
λ
(
3
5
a
−
b
)
l
B
C
:
r
=
3
5
λ
a
+
(
1
−
λ
)
b
■
\begin{align*} \mathbf{r} &= \overrightarrow{OB} + \lambda \overrightarrow{BC}, \; \lambda \in \mathbb{R} \\ &= \mathbf{b} + \lambda (\frac{3}{5} \mathbf{a} - \mathbf{b}) \\ l_{BC}: \mathbf{r} &= \frac{3}{5}\lambda\mathbf{a} + (1-\lambda)\mathbf{b} \; \blacksquare \end{align*}
r
l
BC
:
r
=
OB
+
λ
BC
,
λ
∈
R
=
b
+
λ
(
5
3
a
−
b
)
=
5
3
λ
a
+
(
1
−
λ
)
b
■
A
D
→
=
O
D
→
−
O
A
→
=
5
11
b
−
a
\begin{align*} \overrightarrow{AD} &= \overrightarrow{OD} - \overrightarrow{OA} \\ &= \frac{5}{11} \mathbf{b} - \mathbf{a} \end{align*}
A
D
=
O
D
−
O
A
=
11
5
b
−
a
Equation of line
A
D
:
{AD:}
A
D
:
r
=
O
A
→
+
μ
A
D
→
,
μ
∈
R
=
a
+
μ
(
5
11
b
−
a
)
l
A
D
:
r
=
(
1
−
μ
)
a
+
5
11
μ
b
■
\begin{align*} \mathbf{r} &= \overrightarrow{OA} + \mu \overrightarrow{AD}, \; \mu \in \mathbb{R} \\ &= \mathbf{a} + \mu (\frac{5}{11} \mathbf{b} - \mathbf{a}) \\ l_{AD}: \mathbf{r} &= (1-\mu)\mathbf{a} + \frac{5}{11}\mu\mathbf{b} \; \blacksquare \end{align*}
r
l
A
D
:
r
=
O
A
+
μ
A
D
,
μ
∈
R
=
a
+
μ
(
11
5
b
−
a
)
=
(
1
−
μ
)
a
+
11
5
μ
b
■
(iii)
When
B
C
{BC}
BC
and
A
D
{AD}
A
D
meet,
3
5
λ
a
+
(
1
−
λ
)
b
=
(
1
−
μ
)
a
+
5
11
μ
b
\frac{3}{5}\lambda\mathbf{a} + (1-\lambda)\mathbf{b} = (1-\mu)\mathbf{a} + \frac{5}{11}\mu\mathbf{b}
5
3
λ
a
+
(
1
−
λ
)
b
=
(
1
−
μ
)
a
+
11
5
μ
b
Comparing,
3
5
λ
=
1
−
μ
1
−
λ
=
5
11
μ
\begin{align} && \quad \frac{3}{5}\lambda &= 1-\mu \\ && \quad 1-\lambda &= \frac{5}{11}\mu \end{align}
5
3
λ
1
−
λ
=
1
−
μ
=
11
5
μ
Solving with a GC,
λ
=
3
4
,
μ
=
11
20
\lambda = \frac{3}{4}, \mu = \frac{11}{20}
λ
=
4
3
,
μ
=
20
11
O
E
→
=
3
5
(
3
4
)
a
+
(
1
−
3
4
)
b
=
9
20
a
+
1
4
b
■
\begin{align*} \overrightarrow{OE} &= \frac{3}{5}\left(\frac{3}{4}\right)\mathbf{a} + \left(1-\frac{3}{4}\right)\mathbf{b} \\ &= \frac{9}{20}\mathbf{a}+\frac{1}{4}\mathbf{b} \; \blacksquare \end{align*}
OE
=
5
3
(
4
3
)
a
+
(
1
−
4
3
)
b
=
20
9
a
+
4
1
b
■
A
E
→
=
O
E
→
−
O
A
→
=
9
20
a
+
1
4
b
−
a
=
−
11
20
a
+
1
4
b
=
1
20
(
−
11
a
+
5
b
)
\begin{align*} \overrightarrow{AE} &= \overrightarrow{OE} - \overrightarrow{OA} \\ &= \frac{9}{20}\mathbf{a}+\frac{1}{4}\mathbf{b} - \mathbf{a} \\ &= - \frac{11}{20}\mathbf{a} + \frac{1}{4}\mathbf{b} \\ &= \frac{1}{20}\left(- 11\mathbf{a}+5\mathbf{b}\right) \end{align*}
A
E
=
OE
−
O
A
=
20
9
a
+
4
1
b
−
a
=
−
20
11
a
+
4
1
b
=
20
1
(
−
11
a
+
5
b
)
E
D
→
=
O
D
→
−
O
E
→
=
5
11
b
−
(
9
20
a
+
1
4
b
)
=
−
9
20
a
+
9
44
b
=
9
220
(
−
11
a
+
5
b
)
\begin{align*} \overrightarrow{ED} &= \overrightarrow{OD} - \overrightarrow{OE} \\ &= \frac{5}{11} \mathbf{b} - \left(\frac{9}{20}\mathbf{a}+\frac{1}{4}\mathbf{b}\right) \\ &= - \frac{9}{20}\mathbf{a} + \frac{9}{44}\mathbf{b} \\ &= \frac{9}{220}\left(- 11\mathbf{a}+5\mathbf{b}\right) \end{align*}
E
D
=
O
D
−
OE
=
11
5
b
−
(
20
9
a
+
4
1
b
)
=
−
20
9
a
+
44
9
b
=
220
9
(
−
11
a
+
5
b
)
A
E
E
D
=
1
20
÷
9
220
=
11
9
\begin{align*} \frac{AE}{ED} &= \frac{1}{20} \div \frac{9}{220}\\ &= \frac{11}{9} \end{align*}
E
D
A
E
=
20
1
÷
220
9
=
9
11
Hence
A
E
:
E
D
=
11
:
9
■
{AE:ED = 11:9 \; \blacksquare}
A
E
:
E
D
=
11
:
9
■
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