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2012
P1 Q5
Topical
Vectors I
12 P1 Q5
2012 H2 Mathematics Paper 1 Question 5
Vectors I: Basics, Dot and Cross Products
Answers
(i)
μ
=
6
{\mu = 6}
μ
=
6
(ii)
C
(
5
,
7
,
1
)
or
C
(
17
3
,
19
3
,
5
3
)
C \left( 5, 7, 1 \right) \allowbreak \textrm{ or } \allowbreak C \left( \frac{17}{3}, \frac{19}{3}, \frac{5}{3} \right)
C
(
5
,
7
,
1
)
or
C
(
3
17
,
3
19
,
3
5
)
Full solutions
(i)
area of
△
O
A
C
=
126
1
2
∣
O
A
→
×
O
C
→
∣
=
126
∣
a
×
(
λ
a
+
μ
b
)
∣
=
2
126
\begin{align*} \textrm{ area of } \triangle OAC &= \sqrt{126} \\ \frac{1}{2} \left| \overrightarrow{OA} \times \overrightarrow{OC} \right| &= \sqrt{126} \\ \left| \mathbf{a} \times (\lambda \mathbf{a} + \mu \mathbf{b}) \right |&= 2 \sqrt{126} \\ \end{align*}
area of
△
O
A
C
2
1
O
A
×
OC
∣
a
×
(
λ
a
+
μ
b
)
∣
=
126
=
126
=
2
126
∣
λ
(
a
×
a
)
+
μ
(
a
×
b
)
∣
=
2
126
∣
μ
(
1
−
1
1
)
×
(
1
2
0
)
∣
=
2
126
\begin{align*} \left| \lambda (\mathbf{a}\times\mathbf{a}) + \mu (\mathbf{a}\times\mathbf{b}) \right |&= 2 \sqrt{126} \\ \left| \mu \begin{pmatrix} 1 \\ - 1 \\ 1 \end{pmatrix} \times \begin{pmatrix} 1 \\ 2 \\ 0 \end{pmatrix} \right | &= 2 \sqrt{126} \\ \end{align*}
∣
λ
(
a
×
a
)
+
μ
(
a
×
b
)
∣
μ
1
−
1
1
×
1
2
0
=
2
126
=
2
126
∣
μ
(
−
2
1
3
)
∣
=
2
126
∣
μ
∣
4
+
1
+
9
=
2
126
∣
μ
∣
=
6
\begin{align*} \left| \mu \begin{pmatrix} - 2 \\ 1 \\ 3 \end{pmatrix} \right | &= 2 \sqrt{126} \\ \left|\mu\right| \sqrt{4 + 1 + 9} &= 2 \sqrt{126} \\ \left|\mu\right| &= 6 \\ \end{align*}
μ
−
2
1
3
∣
μ
∣
4
+
1
+
9
∣
μ
∣
=
2
126
=
2
126
=
6
Since
μ
{\mu}
μ
is positive,
μ
=
6
■
{\mu = 6 \; \blacksquare}
μ
=
6
■
(ii)
Using
μ
=
4
,
{\mu = 4,}
μ
=
4
,
O
C
→
=
λ
(
1
−
1
1
)
+
4
(
1
2
0
)
=
(
λ
+
4
−
λ
+
8
λ
)
\begin{align*} \overrightarrow{OC} &= \lambda \begin{pmatrix} 1 \\ - 1 \\ 1 \end{pmatrix} + 4 \begin{pmatrix} 1 \\ 2 \\ 0 \end{pmatrix} \\ &= \begin{pmatrix} \lambda + 4 \\ - \lambda + 8 \\ \lambda \end{pmatrix} \end{align*}
OC
=
λ
1
−
1
1
+
4
1
2
0
=
λ
+
4
−
λ
+
8
λ
∣
O
C
→
∣
=
5
3
3
λ
2
−
8
λ
+
80
=
5
3
3
λ
2
−
8
λ
+
80
=
75
3
λ
2
−
8
λ
+
5
=
0
(
λ
−
1
)
(
3
λ
−
5
)
=
0
\begin{align*} \left | \overrightarrow{OC} \right| &= 5 \sqrt{3} \\ \sqrt{3 \lambda^2 - 8 \lambda + 80} &= 5 \sqrt{3} \\ 3 \lambda^2 - 8 \lambda + 80 &= 75 \\ 3 \lambda^2 - 8 \lambda + 5 &= 0 \\ (\lambda - 1)(3 \lambda - 5) &= 0 \\ \end{align*}
OC
3
λ
2
−
8
λ
+
80
3
λ
2
−
8
λ
+
80
3
λ
2
−
8
λ
+
5
(
λ
−
1
)
(
3
λ
−
5
)
=
5
3
=
5
3
=
75
=
0
=
0
λ
=
1
or
λ
=
5
3
\lambda = 1 \textrm{ or } \lambda = \frac{5}{3}
λ
=
1
or
λ
=
3
5
O
C
→
=
(
5
7
1
)
or
O
C
→
=
(
17
3
19
3
5
3
)
\overrightarrow{OC} = \begin{pmatrix} 5 \\ 7 \\ 1 \end{pmatrix} \textrm{ or } \overrightarrow{OC} = \begin{pmatrix} \frac{17}{3} \\ \frac{19}{3} \\ \frac{5}{3} \end{pmatrix}
OC
=
5
7
1
or
OC
=
3
17
3
19
3
5
Possible coordinates of
C
:
{C:}
C
:
(
5
,
7
,
1
)
■
(
17
3
,
19
3
,
5
3
)
■
{\displaystyle \left( 5, 7, 1 \right) \; \blacksquare \; \left( \frac{17}{3}, \frac{19}{3}, \frac{5}{3} \right) \; \blacksquare}
(
5
,
7
,
1
)
■
(
3
17
,
3
19
,
3
5
)
■
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