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Yearly
2009
P1 Q10
Topical
Vectors II
09 P1 Q10
2009 H2 Mathematics Paper 1 Question 10
Vectors II: Lines and Planes
Answers
(i)
70.9
∘
{70.9^\circ}
70.
9
∘
(ii)
l
:
r
=
(
0
1
0
)
+
λ
(
−
1
−
1
1
)
,
λ
∈
R
{l: \mathbf{r} = \begin{pmatrix} 0 \\ 1 \\ 0 \end{pmatrix} + \lambda \begin{pmatrix} - 1 \\ - 1 \\ 1 \end{pmatrix}, \lambda \in \mathbb{R}}
l
:
r
=
0
1
0
+
λ
−
1
−
1
1
,
λ
∈
R
(iii)
x
−
y
=
−
1
{x - y = - 1}
x
−
y
=
−
1
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
1
3
)
⋅
(
−
1
2
1
)
∣
=
∣
(
2
1
3
)
∣
∣
(
−
1
2
1
)
∣
cos
θ
\left|\begin{pmatrix} 2 \\ 1 \\ 3 \end{pmatrix} \cdot \begin{pmatrix} - 1 \\ 2 \\ 1 \end{pmatrix} \right| = \left| \begin{pmatrix} 2 \\ 1 \\ 3 \end{pmatrix} \right| \left| \begin{pmatrix} - 1 \\ 2 \\ 1 \end{pmatrix} \right| \cos \theta
2
1
3
⋅
−
1
2
1
=
2
1
3
−
1
2
1
cos
θ
∣
3
∣
=
14
6
cos
θ
\left|3 \right| = \sqrt{14} \sqrt{6} \cos \theta
∣
3
∣
=
14
6
cos
θ
cos
θ
=
3
14
6
\cos \theta = \frac{3}{\sqrt{14}\sqrt{6}}
cos
θ
=
14
6
3
θ
=
70.9
∘
■
\theta = 70.9^\circ \; \blacksquare
θ
=
70.
9
∘
■
(ii)
2
x
+
y
+
3
z
=
1
−
x
+
2
y
+
z
=
2
\begin{align} \qquad 2x + y + 3z &= 1 \\ \qquad -x + 2y + z &= 2 \end{align}
2
x
+
y
+
3
z
−
x
+
2
y
+
z
=
1
=
2
Solving using a GC,
l
:
r
=
(
0
1
0
)
+
λ
(
−
1
−
1
1
)
,
λ
∈
R
■
l: \mathbf{r} = \begin{pmatrix} 0 \\ 1 \\ 0 \end{pmatrix} + \lambda \begin{pmatrix} - 1 \\ - 1 \\ 1 \end{pmatrix}, \lambda \in \mathbb{R} \; \blacksquare
l
:
r
=
0
1
0
+
λ
−
1
−
1
1
,
λ
∈
R
■
(iii)
2
x
+
y
+
3
z
−
1
+
k
(
−
x
+
2
y
+
z
−
2
)
=
0
2x+y+3z -1 +k(-x+2y+z-2) = 0
2
x
+
y
+
3
z
−
1
+
k
(
−
x
+
2
y
+
z
−
2
)
=
0
(
2
−
k
)
x
+
(
1
+
2
k
)
y
+
(
3
+
k
)
z
=
1
+
2
k
(2-k)x+(1+2k)y+(3+k)z = 1 +2k
(
2
−
k
)
x
+
(
1
+
2
k
)
y
+
(
3
+
k
)
z
=
1
+
2
k
r
⋅
(
2
−
k
1
+
2
k
3
+
k
)
=
1
+
2
k
\mathbf{r} \cdot \begin{pmatrix}2-k\\1+2k\\3+k\end{pmatrix} = 1 +2k
r
⋅
2
−
k
1
+
2
k
3
+
k
=
1
+
2
k
Substituting equation of
l
{l}
l
into equation of
p
3
{p_3}
p
3
,
LHS
=
(
−
λ
1
−
λ
λ
)
⋅
(
2
−
k
1
+
2
k
3
+
k
)
=
−
λ
(
2
−
k
)
+
(
1
−
λ
)
(
1
+
2
k
)
+
λ
(
3
+
k
)
=
−
2
λ
+
λ
k
+
1
+
2
k
−
λ
−
2
λ
k
+
3
λ
+
λ
k
=
1
+
2
k
=
RHS
\begin{align*} & \textrm{LHS} \\ = & \begin{pmatrix} - \lambda \\ 1 - \lambda \\ \lambda \end{pmatrix} \cdot \begin{pmatrix}2-k\\1+2k\\3+k\end{pmatrix} \\ = & -\lambda(2-k) + (1-\lambda)(1+2k) + \lambda(3+k) \\ = & -2\lambda + \lambda k + 1 + 2k - \lambda - 2 \lambda k + 3 \lambda + \lambda k \\ = & 1 + 2k \\ = & \textrm{RHS} \end{align*}
=
=
=
=
=
LHS
−
λ
1
−
λ
λ
⋅
2
−
k
1
+
2
k
3
+
k
−
λ
(
2
−
k
)
+
(
1
−
λ
)
(
1
+
2
k
)
+
λ
(
3
+
k
)
−
2
λ
+
λk
+
1
+
2
k
−
λ
−
2
λk
+
3
λ
+
λk
1
+
2
k
RHS
Hence
l
{l}
l
lies in
p
3
{p_3}
p
3
for any
k
.
■
{k. \; \blacksquare}
k
.
■
Substituting
(
2
,
3
,
4
)
{(2,3,4)}
(
2
,
3
,
4
)
into equation of
p
3
{p_3}
p
3
,
2
(
2
)
+
3
+
3
(
4
)
−
1
+
k
(
−
2
+
2
(
3
)
+
4
−
2
)
=
0
2(2)+3+3(4) - 1 + k(-2+2(3)+4-2) = 0
2
(
2
)
+
3
+
3
(
4
)
−
1
+
k
(
−
2
+
2
(
3
)
+
4
−
2
)
=
0
18
+
6
k
=
0
k
=
−
3
\begin{gather*}18 + 6k = 0 \\ k=-3\end{gather*}
18
+
6
k
=
0
k
=
−
3
Substituting
k
=
−
3
{k=-3}
k
=
−
3
into equation of
p
3
{p_3}
p
3
,
2
x
+
y
+
3
z
−
1
−
3
(
−
x
+
2
y
+
z
−
2
)
=
0
2x+y+3z - 1 -3(-x+2y+z-2) = 0
2
x
+
y
+
3
z
−
1
−
3
(
−
x
+
2
y
+
z
−
2
)
=
0
5
x
−
5
y
−
5
=
0
5x-5y - 5 = 0
5
x
−
5
y
−
5
=
0
Cartesian equation required:
x
−
y
=
1
■
{x-y = 1 \; \blacksquare}
x
−
y
=
1
■
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