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2013
P2 Q4
Topical
Vectors II
13 P2 Q4
2013 H2 Mathematics Paper 2 Question 4
Vectors II: Lines and Planes
Answers
(i)
40.4
∘
{40.4^\circ}
40.
4
∘
(ii)
l
:
r
=
(
−
1
6
−
2
3
0
)
+
λ
(
7
10
6
)
,
λ
∈
R
{l: \mathbf{r} = \begin{pmatrix} - \frac{1}{6} \\ - \frac{2}{3} \\ 0 \end{pmatrix} + \lambda \begin{pmatrix} 7 \\ 10 \\ 6 \end{pmatrix}, \lambda \in \mathbb{R}}
l
:
r
=
−
6
1
−
3
2
0
+
λ
7
10
6
,
λ
∈
R
(iii)
c
=
−
49
{c=-49}
c
=
−
49
or
c
=
35
13
{c=\frac{35}{13}}
c
=
13
35
Full solutions
(i)
∣
n
1
⋅
n
2
∣
=
∣
n
1
∣
∣
n
2
∣
cos
θ
\left|\mathbf{n_1} \cdot \mathbf{n_2}\right| = \left| \mathbf{n_1} \right| \left| \mathbf{n_2} \right| \cos \theta
∣
n
1
⋅
n
2
∣
=
∣
n
1
∣
∣
n
2
∣
cos
θ
∣
(
2
−
2
1
)
⋅
(
−
6
3
2
)
∣
=
∣
(
2
−
2
1
)
∣
∣
(
−
6
3
2
)
∣
cos
θ
∣
−
16
∣
=
(
3
)
(
7
)
cos
θ
\begin{align*} \left|\begin{pmatrix} 2 \\ - 2 \\ 1 \end{pmatrix} \cdot \begin{pmatrix} - 6 \\ 3 \\ 2 \end{pmatrix} \right| &= \left| \begin{pmatrix} 2 \\ - 2 \\ 1 \end{pmatrix} \right| \left| \begin{pmatrix} - 6 \\ 3 \\ 2 \end{pmatrix} \right| \cos \theta \\ \left|- 16 \right| &= (3) (7) \cos \theta \end{align*}
2
−
2
1
⋅
−
6
3
2
∣
−
16
∣
=
2
−
2
1
−
6
3
2
cos
θ
=
(
3
)
(
7
)
cos
θ
cos
θ
=
16
(
3
)
(
7
)
θ
=
40.4
∘
■
\begin{align*} \cos \theta &= \frac{16}{(3)(7)} \\ \theta &= 40.4^\circ \; \blacksquare \end{align*}
cos
θ
θ
=
(
3
)
(
7
)
16
=
40.
4
∘
■
(ii)
p
1
:
2
x
−
2
y
+
z
=
1
p
2
:
−
6
x
+
3
y
+
2
z
=
−
1
\begin{align} \qquad &p_1: 2 x - 2 y + z = 1 \\ \qquad &p_2: - 6 x + 3 y + 2 z = - 1 \end{align}
p
1
:
2
x
−
2
y
+
z
=
1
p
2
:
−
6
x
+
3
y
+
2
z
=
−
1
Using a GC, equation for
l
:
{l:}
l
:
r
=
(
−
1
6
−
2
3
0
)
+
λ
(
7
10
6
)
,
λ
∈
R
■
\mathbf{r} = \begin{pmatrix} - \frac{1}{6} \\ - \frac{2}{3} \\ 0 \end{pmatrix} + \lambda \begin{pmatrix} 7 \\ 10 \\ 6 \end{pmatrix}, \lambda \in \mathbb{R} \; \blacksquare
r
=
−
6
1
−
3
2
0
+
λ
7
10
6
,
λ
∈
R
■
(iii)
Let
B
(
0
,
0
,
1
)
{B \left( 0, 0, 1 \right)}
B
(
0
,
0
,
1
)
and
C
(
0
,
−
1
,
1
)
{C \left( 0, - 1, 1 \right)}
C
(
0
,
−
1
,
1
)
denote points on
p
1
,
p
2
{p_1,p_2}
p
1
,
p
2
respectively
Distance from
A
{A}
A
to
p
1
{p_1}
p
1
=
∣
B
A
→
⋅
n
^
1
∣
{=\left|\overrightarrow{BA} \cdot \mathbf{\hat{n}_1} \right|}
=
B
A
⋅
n
^
1
=
∣
(
4
3
c
−
1
)
⋅
(
2
−
2
1
)
∣
(
2
−
2
1
)
∣
∣
{\displaystyle =\left|\begin{pmatrix}4\\3\\c-1\end{pmatrix} \cdot \frac{\begin{pmatrix} 2 \\ - 2 \\ 1 \end{pmatrix}}{\left| \begin{pmatrix} 2 \\ - 2 \\ 1 \end{pmatrix} \right|} \right|}
=
4
3
c
−
1
⋅
2
−
2
1
2
−
2
1
=
∣
8
−
6
+
c
−
1
3
∣
{\displaystyle =\left|\frac{8 - 6 + c - 1}{3} \right|}
=
3
8
−
6
+
c
−
1
=
∣
c
+
1
∣
3
{\displaystyle = \frac{\left|c+1\right|}{3}}
=
3
∣
c
+
1
∣
Distance from
A
{A}
A
to
p
2
{p_2}
p
2
=
∣
C
A
→
⋅
n
^
2
∣
{=\left|\overrightarrow{CA} \cdot \mathbf{\hat{n}_2} \right|}
=
C
A
⋅
n
^
2
=
∣
(
4
4
c
+
1
)
⋅
(
−
6
3
2
)
∣
(
−
6
3
2
)
∣
∣
{\displaystyle =\left|\begin{pmatrix}4\\4\\c+1\end{pmatrix} \cdot \frac{\begin{pmatrix} - 6 \\ 3 \\ 2 \end{pmatrix}}{\left| \begin{pmatrix} - 6 \\ 3 \\ 2 \end{pmatrix} \right|} \right|}
=
4
4
c
+
1
⋅
−
6
3
2
−
6
3
2
=
∣
−
24
+
12
+
2
c
−
2
7
∣
{\displaystyle =\left|\frac{- 24 + 12 + 2c - 2}{7} \right|}
=
7
−
24
+
12
+
2
c
−
2
=
∣
2
c
−
14
∣
7
{\displaystyle = \frac{\left|2c-14\right|}{7}}
=
7
∣
2
c
−
14
∣
Since
A
{A}
A
is equidistant from
p
1
{p_1}
p
1
and
p
2
,
{p_2,}
p
2
,
∣
c
+
1
∣
3
=
∣
2
c
−
14
∣
7
7
∣
c
+
1
∣
=
3
∣
2
c
−
14
∣
\begin{align*} \frac{\left|c+1\right|}{3} &= \frac{\left|2c-14\right|}{7} \\ 7\left|c+1\right| &= 3\left|2c-14\right| \end{align*}
3
∣
c
+
1
∣
7
∣
c
+
1
∣
=
7
∣
2
c
−
14
∣
=
3
∣
2
c
−
14
∣
7
c
+
7
=
6
c
−
42
or
7
c
+
7
=
−
(
6
x
−
42
)
c
=
−
49
or
c
=
35
13
■
\begin{align*} 7c + 7 = 6c - 42 &\quad \textrm{or} \quad 7c + 7 = -(6x-42) \\ c=-49 &\quad \textrm{or} \quad c = \frac{35}{13} \; \blacksquare \end{align*}
7
c
+
7
=
6
c
−
42
c
=
−
49
or
7
c
+
7
=
−
(
6
x
−
42
)
or
c
=
13
35
■
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