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2019
P2 Q5
Topical
Vectors I
19 P2 Q5
2019 H2 Mathematics Paper 2 Question 5
Vectors I: Basics, Dot and Cross Products
Answers
(i)
O
X
→
=
5
4
a
+
b
{\overrightarrow{OX}=\frac{5}{4} \mathbf{a} + \mathbf{b}}
OX
=
4
5
a
+
b
(ii)
O
Y
→
=
10
3
a
+
8
3
b
{\overrightarrow{OY} = \frac{10}{3} \mathbf{a} + \frac{8}{3} \mathbf{b}}
O
Y
=
3
10
a
+
3
8
b
O
X
:
O
Y
=
3
:
8
{OX:OY = 3:8}
OX
:
O
Y
=
3
:
8
Full solutions
(i)
B
D
→
=
O
D
→
−
O
B
→
=
b
+
5
a
−
b
=
5
a
\begin{align*} \overrightarrow{BD} &= \overrightarrow{OD} - \overrightarrow{OB} \\ &= \mathbf{b} + 5 \mathbf{a} - \mathbf{b} \\ &= 5 \mathbf{a} \end{align*}
B
D
=
O
D
−
OB
=
b
+
5
a
−
b
=
5
a
Equation of line
B
D
:
{BD:}
B
D
:
r
=
O
B
→
+
λ
B
D
→
,
λ
∈
R
=
b
+
λ
(
5
a
)
l
B
D
:
r
=
5
λ
a
+
b
\begin{align*} \mathbf{r} &= \overrightarrow{OB} + \lambda \overrightarrow{BD}, \; \lambda \in \mathbb{R} \\ &= \mathbf{b} + \lambda (5 \mathbf{a}) \\ l_{BD}: \mathbf{r} &= 5 \lambda \mathbf{a} + \mathbf{b} \end{align*}
r
l
B
D
:
r
=
OB
+
λ
B
D
,
λ
∈
R
=
b
+
λ
(
5
a
)
=
5
λ
a
+
b
A
C
→
=
O
C
→
−
O
A
→
=
2
a
+
4
b
−
a
=
a
+
4
b
\begin{align*} \overrightarrow{AC} &= \overrightarrow{OC} - \overrightarrow{OA} \\ &= 2 \mathbf{a} + 4 \mathbf{b} - \mathbf{a} \\ &= \mathbf{a} + 4 \mathbf{b} \end{align*}
A
C
=
OC
−
O
A
=
2
a
+
4
b
−
a
=
a
+
4
b
Equation of line
A
C
:
{AC:}
A
C
:
r
=
O
A
→
+
μ
A
C
→
,
μ
∈
R
=
a
+
μ
(
a
+
4
b
)
l
A
C
:
r
=
(
1
+
μ
)
a
+
4
μ
b
■
\begin{align*} \mathbf{r} &= \overrightarrow{OA} + \mu \overrightarrow{AC}, \; \mu \in \mathbb{R} \\ &= \mathbf{a} + \mu (\mathbf{a} + 4 \mathbf{b}) \\ l_{AC}: \mathbf{r} &= (1+\mu)\mathbf{a} + 4\mu\mathbf{b} \; \blacksquare \end{align*}
r
l
A
C
:
r
=
O
A
+
μ
A
C
,
μ
∈
R
=
a
+
μ
(
a
+
4
b
)
=
(
1
+
μ
)
a
+
4
μ
b
■
When
B
D
{BD}
B
D
and
A
C
{AC}
A
C
meet at
X
{X}
X
,
5
λ
a
+
b
=
(
1
+
μ
)
a
+
4
μ
b
5 \lambda \mathbf{a} + \mathbf{b} = (1+\mu)\mathbf{a} + 4\mu\mathbf{b}
5
λ
a
+
b
=
(
1
+
μ
)
a
+
4
μ
b
Comparing
b
{\mathbf{b}}
b
,
1
=
4
μ
μ
=
1
4
\begin{align*} 1 &= 4\mu \\ \mu &= \frac{1}{4} \end{align*}
1
μ
=
4
μ
=
4
1
O
X
→
=
(
1
+
1
4
)
a
+
4
(
1
4
)
b
=
5
4
a
+
b
■
\begin{align*} \overrightarrow{OX} &= \left(1+\frac{1}{4}\right)\mathbf{a} + 4\left(\frac{1}{4}\right)\mathbf{b} \\ &= \frac{5}{4} \mathbf{a} + \mathbf{b} \; \blacksquare \end{align*}
OX
=
(
1
+
4
1
)
a
+
4
(
4
1
)
b
=
4
5
a
+
b
■
(ii)
C
D
→
=
O
D
→
−
O
C
→
=
b
+
5
a
−
2
a
+
4
b
=
3
a
−
3
b
\begin{align*} \overrightarrow{CD} &= \overrightarrow{OD} - \overrightarrow{OC} \\ &= \mathbf{b} + 5 \mathbf{a} - 2 \mathbf{a} + 4 \mathbf{b} \\ &= 3 \mathbf{a} - 3 \mathbf{b} \end{align*}
C
D
=
O
D
−
OC
=
b
+
5
a
−
2
a
+
4
b
=
3
a
−
3
b
Equation of line
C
D
:
{CD:}
C
D
:
r
=
O
C
→
+
ν
C
D
→
,
ν
∈
R
=
2
a
+
4
b
+
ν
(
3
a
−
3
b
)
l
C
D
:
r
=
(
2
+
3
ν
)
a
+
(
4
−
3
ν
)
b
\begin{align*} \mathbf{r} &= \overrightarrow{OC} + \nu \overrightarrow{CD}, \; \nu \in \mathbb{R} \\ &= 2 \mathbf{a} + 4 \mathbf{b} + \nu (3 \mathbf{a} - 3 \mathbf{b}) \\ l_{CD}: \mathbf{r} &= (2+3\nu)\mathbf{a} + (4-3\nu) \mathbf{b} \end{align*}
r
l
C
D
:
r
=
OC
+
ν
C
D
,
ν
∈
R
=
2
a
+
4
b
+
ν
(
3
a
−
3
b
)
=
(
2
+
3
ν
)
a
+
(
4
−
3
ν
)
b
Since
Y
{Y}
Y
lies on
C
D
,
{CD,}
C
D
,
O
Y
→
=
(
2
+
3
ν
)
a
+
(
4
−
3
ν
)
b
\overrightarrow{OY} = (2+3\nu)\mathbf{a} + (4-3\nu) \mathbf{b}
O
Y
=
(
2
+
3
ν
)
a
+
(
4
−
3
ν
)
b
Since
O
,
X
{O,X}
O
,
X
and
Y
{Y}
Y
are collinear,
O
Y
→
=
k
O
X
→
(
2
+
3
ν
)
a
+
(
4
−
3
ν
)
b
=
5
4
k
a
+
k
b
\begin{gather*} \overrightarrow{OY} = k \overrightarrow{OX} \\ (2+3\nu)\mathbf{a} + (4-3\nu) \mathbf{b} = \frac{5}{4}k\mathbf{a} + k\mathbf{b} \\ \end{gather*}
O
Y
=
k
OX
(
2
+
3
ν
)
a
+
(
4
−
3
ν
)
b
=
4
5
k
a
+
k
b
Comparing,
2
+
3
ν
=
5
4
k
4
−
3
ν
=
k
\begin{align} && \quad 2 + 3 \nu &= \frac{5}{4}k \\ && \quad 4 - 3 \nu &= k \end{align}
2
+
3
ν
4
−
3
ν
=
4
5
k
=
k
Solving with a GC,
ν
=
4
9
,
k
=
8
3
\nu = \frac{4}{9}, \; k = \frac{8}{3}
ν
=
9
4
,
k
=
3
8
O
Y
→
=
10
3
a
+
8
3
b
■
\overrightarrow{OY} = \frac{10}{3} \mathbf{a} + \frac{8}{3} \mathbf{b} \; \blacksquare
O
Y
=
3
10
a
+
3
8
b
■
O
X
→
:
O
Y
→
=
1
:
8
3
=
3
:
8
■
\begin{align*} \overrightarrow{OX} : \overrightarrow{OY} &= 1 : \frac{8}{3} \\ &= 3 : 8 \; \blacksquare \end{align*}
OX
:
O
Y
=
1
:
3
8
=
3
:
8
■
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