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2011
P1 Q7
Topical
Vectors I
11 P1 Q7
2011 H2 Mathematics Paper 1 Question 7
Vectors I: Basics, Dot and Cross Products
Answers
(i)
O
M
→
=
1
6
a
+
3
10
b
{\overrightarrow{OM}=\frac{1}{6} \mathbf{a} + \frac{3}{10} \mathbf{b}}
OM
=
6
1
a
+
10
3
b
Area of
△
O
M
P
{\triangle OMP}
△
OMP
=
1
20
∣
a
×
b
∣
{=\frac{1}{20} \left| \mathbf{a} \times \mathbf{b} \right|}
=
20
1
∣
a
×
b
∣
k
=
1
20
{k=\frac{1}{20}}
k
=
20
1
(iia)
p
=
1
7
{p=\frac{1}{7}}
p
=
7
1
(iib)
∣
a
⋅
b
∣
{\left|\mathbf{a}\cdot\mathbf{b}\right|}
∣
a
⋅
b
∣
is the length of projection of
b
{\mathbf{b}}
b
on
a
{\mathbf{a}}
a
(iic)
1
7
(
9
7
8
)
{\displaystyle \frac{1}{7} \begin{pmatrix} 9 \\ 7 \\ 8 \end{pmatrix}}
7
1
9
7
8
Full solutions
(i)
Since
O
P
:
P
A
=
1
:
2
,
{OP:PA = 1:2,}
OP
:
P
A
=
1
:
2
,
O
P
→
=
1
3
a
\overrightarrow{OP} = \frac{1}{3}\mathbf{a}
OP
=
3
1
a
Since
O
Q
:
Q
B
=
3
:
2
,
{OQ:QB = 3:2,}
OQ
:
QB
=
3
:
2
,
O
Q
→
=
3
5
b
\overrightarrow{OQ} = \frac{3}{5}\mathbf{b}
OQ
=
5
3
b
By ratio theorem,
O
M
→
=
O
P
→
+
O
Q
→
2
=
1
3
a
+
3
5
b
2
=
1
6
a
+
3
10
b
■
\begin{align*} \overrightarrow{OM} &= \frac{\overrightarrow{OP}+\overrightarrow{OQ}}{2} \\ &= \frac{\frac{1}{3}\mathbf{a}+\frac{3}{5}\mathbf{b}}{2} \\ &= \frac{1}{6} \mathbf{a} + \frac{3}{10} \mathbf{b} \; \blacksquare \end{align*}
OM
=
2
OP
+
OQ
=
2
3
1
a
+
5
3
b
=
6
1
a
+
10
3
b
■
Area of
△
O
M
P
=
1
2
∣
O
P
→
×
O
M
→
∣
=
1
2
∣
(
1
6
a
+
3
10
b
)
×
1
3
a
∣
=
1
2
∣
1
18
a
×
a
+
1
10
b
×
a
∣
=
1
2
∣
0
+
1
10
b
×
a
∣
=
1
2
∣
−
1
10
a
×
b
∣
=
1
20
∣
a
×
b
∣
■
\begin{align*} & \textrm{Area of } \triangle OMP \\ &= \frac{1}{2} \left| \overrightarrow{OP} \times \overrightarrow{OM} \right| \\ &= \frac{1}{2} \left| \left(\frac{1}{6} \mathbf{a} + \frac{3}{10} \mathbf{b}\right) \times \frac{1}{3} \mathbf{a} \right| \\ &= \frac{1}{2} \left| \frac{1}{18} \mathbf{a} \times \mathbf{a} + \frac{1}{10} \mathbf{b} \times \mathbf{a} \right| \\ &= \frac{1}{2} \left| \mathbf{0} + \frac{1}{10} \mathbf{b} \times \mathbf{a} \right| \\ &= \frac{1}{2} \left| - \frac{1}{10} \mathbf{a} \times \mathbf{b} \right| \\ &= \frac{1}{20} \left| \mathbf{a} \times \mathbf{b} \right| \; \blacksquare \end{align*}
Area of
△
OMP
=
2
1
OP
×
OM
=
2
1
(
6
1
a
+
10
3
b
)
×
3
1
a
=
2
1
18
1
a
×
a
+
10
1
b
×
a
=
2
1
0
+
10
1
b
×
a
=
2
1
−
10
1
a
×
b
=
20
1
∣
a
×
b
∣
■
(iia)
Since
a
{\mathbf{a}}
a
is a unit vector,
∣
a
∣
=
1
4
p
2
+
36
p
2
+
9
p
2
=
1
49
p
2
=
0
p
=
1
7
■
since
p
is a positive constant
\begin{align*} \left| \mathbf{a} \right| &= 1 \\ \sqrt{4p^2 + 36p^2 + 9p^2} &= 1 \\ 49 p^2 &= 0 \\ p &= \frac{1}{7} \; \blacksquare \\ & \quad \textrm{since } p \textrm{ is a positive constant} \end{align*}
∣
a
∣
4
p
2
+
36
p
2
+
9
p
2
49
p
2
p
=
1
=
1
=
0
=
7
1
■
since
p
is a positive constant
(iib)
∣
a
⋅
b
∣
{\left|\mathbf{a}\cdot\mathbf{b}\right|}
∣
a
⋅
b
∣
is the length of projection of
b
{\mathbf{b}}
b
on
a
■
{\mathbf{a} \; \blacksquare}
a
■
(iic)
a
×
b
=
1
7
(
2
−
6
3
)
×
(
1
1
−
2
)
=
1
7
(
9
7
8
)
■
\begin{align*} & \mathbf{a} \times \mathbf{b} \\ &= \frac{1}{7} \begin{pmatrix} 2 \\ - 6 \\ 3 \end{pmatrix} \times \begin{pmatrix} 1 \\ 1 \\ - 2 \end{pmatrix} \\ &= \frac{1}{7} \begin{pmatrix} 9 \\ 7 \\ 8 \end{pmatrix} \; \blacksquare \\ \end{align*}
a
×
b
=
7
1
2
−
6
3
×
1
1
−
2
=
7
1
9
7
8
■
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