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2008
P1 Q3
Topical
Vectors I
08 P1 Q3
2008 H2 Mathematics Paper 1 Question 3
Vectors I: Basics, Dot and Cross Products
Answers
(i)
O
P
→
=
(
6
3
−
3
)
{\overrightarrow{OP}=\begin{pmatrix} 6 \\ 3 \\ - 3 \end{pmatrix}}
OP
=
6
3
−
3
(ii)
∠
A
O
B
=
87.8
∘
{\angle AOB = 87.8^\circ}
∠
A
OB
=
87.
8
∘
(iii)
Area of parallelogram
O
A
P
B
=
15
3
units
2
OAPB \allowbreak {= 15 \sqrt{3} \textrm{ units}^2}
O
A
PB
=
15
3
units
2
Full solutions
(i)
Since
O
A
P
B
{OAPB}
O
A
PB
is a parallelogram,
O
B
→
=
A
P
→
{\overrightarrow{OB}=\overrightarrow{AP}}
OB
=
A
P
O
P
→
=
O
A
→
+
A
P
→
=
O
A
→
+
O
B
→
=
(
1
4
−
3
)
+
(
5
−
1
0
)
=
(
6
3
−
3
)
■
\begin{align*} \overrightarrow{OP} &= \overrightarrow{OA} + \overrightarrow{AP} \\ &= \overrightarrow{OA} + \overrightarrow{OB} \\ &= \begin{pmatrix} 1 \\ 4 \\ - 3 \end{pmatrix} + \begin{pmatrix} 5 \\ - 1 \\ 0 \end{pmatrix} \\ &= \begin{pmatrix} 6 \\ 3 \\ - 3 \end{pmatrix} \; \blacksquare \end{align*}
OP
=
O
A
+
A
P
=
O
A
+
OB
=
1
4
−
3
+
5
−
1
0
=
6
3
−
3
■
(ii)
Let
a
,
b
{\mathbf{a}, \mathbf{b}}
a
,
b
and
θ
{\theta}
θ
denote
O
A
→
,
O
B
→
{\overrightarrow{OA}, \overrightarrow{OB}}
O
A
,
OB
and
∠
A
O
B
{\angle AOB}
∠
A
OB
respectively
∣
a
⋅
b
∣
=
∣
a
∣
∣
b
∣
cos
θ
\left|\mathbf{a} \cdot \mathbf{b}\right| = \left| \mathbf{a} \right| \left| \mathbf{b} \right| \cos \theta
∣
a
⋅
b
∣
=
∣
a
∣
∣
b
∣
cos
θ
∣
(
1
4
−
3
)
⋅
(
5
−
1
0
)
∣
=
∣
(
1
4
−
3
)
∣
∣
(
5
−
1
0
)
∣
cos
θ
∣
1
∣
=
(
26
)
(
26
)
cos
θ
\begin{align*} \left|\begin{pmatrix} 1 \\ 4 \\ - 3 \end{pmatrix} \cdot \begin{pmatrix} 5 \\ - 1 \\ 0 \end{pmatrix} \right| &= \left| \begin{pmatrix} 1 \\ 4 \\ - 3 \end{pmatrix} \right| \left| \begin{pmatrix} 5 \\ - 1 \\ 0 \end{pmatrix} \right| \cos \theta \\ \left|1 \right| &= (\sqrt{26}) (\sqrt{26}) \cos \theta \end{align*}
1
4
−
3
⋅
5
−
1
0
∣
1
∣
=
1
4
−
3
5
−
1
0
cos
θ
=
(
26
)
(
26
)
cos
θ
cos
θ
=
1
(
26
)
(
26
)
θ
=
87.8
∘
■
\begin{align*} \cos \theta &= \frac{1}{(\sqrt{26})(\sqrt{26})} \\ \theta &= 87.8^\circ \; \blacksquare \end{align*}
cos
θ
θ
=
(
26
)
(
26
)
1
=
87.
8
∘
■
(iii)
Area of parallelogram
O
A
P
B
=
∣
O
A
→
×
O
B
→
∣
=
∣
(
1
4
−
3
)
×
(
5
−
1
0
)
∣
=
∣
(
−
3
−
15
−
21
)
∣
=
9
+
225
+
441
=
675
=
15
3
units
2
■
\begin{align*} & \textrm{Area of parallelogram } OAPB \\ &= \left| \overrightarrow{OA} \times \overrightarrow{OB} \right| \\ &= \left|\begin{pmatrix} 1 \\ 4 \\ - 3 \end{pmatrix} \times \begin{pmatrix} 5 \\ - 1 \\ 0 \end{pmatrix}\right| \\ &= \left|\begin{pmatrix} - 3 \\ - 15 \\ - 21 \end{pmatrix} \right| \\ &= \sqrt{9 + 225 + 441} \\ &= \sqrt{675} \\ &= 15 \sqrt{3} \textrm{ units}^2 \; \blacksquare \end{align*}
Area of parallelogram
O
A
PB
=
O
A
×
OB
=
1
4
−
3
×
5
−
1
0
=
−
3
−
15
−
21
=
9
+
225
+
441
=
675
=
15
3
units
2
■
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