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2016
P1 Q5
Topical
Vectors I
16 P1 Q5
2016 H2 Mathematics Paper 1 Question 5
Vectors I: Basics, Dot and Cross Products
Answers
(i)
(
2
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4
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4
a
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2
a
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{\begin{pmatrix}2b\\4b-4a\\-2a\end{pmatrix}}
2
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4
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4
a
−
2
a
(ii)
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×
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\left(\mathbf{u} + \mathbf{v}\right) \times \left(\mathbf{u} - \mathbf{v}\right) \allowbreak {= \begin{pmatrix}-2a\\-8a\\-2a\end{pmatrix}}
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+
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×
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v
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=
−
2
a
−
8
a
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2
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a
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1
12
2
{a=\pm\frac{1}{12} \sqrt{2}}
a
=
±
12
1
2
(iii)
∣
v
∣
=
3
{\left|\mathbf{v}\right| = 3}
∣
v
∣
=
3
Full solutions
(i)
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×
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=
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a
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■
\begin{align*} & \left(\mathbf{u} + \mathbf{v}\right) \times \left(\mathbf{u} - \mathbf{v}\right) \\ &= \mathbf{u} \times \mathbf{u} - \mathbf{u} \times \mathbf{v} + \mathbf{v} \times \mathbf{u} - \mathbf{v} \times \mathbf{v} \\ &= \mathbf{0} - \mathbf{u} \times \mathbf{v} - \mathbf{u} \times \mathbf{v} + \mathbf{0} \\ &= - 2 \mathbf{u} \times \mathbf{v} \\ &= -2 \begin{pmatrix} 2 \\ - 1 \\ 2 \end{pmatrix} \times \begin{pmatrix}a\\0\\b\end{pmatrix} \\ &= -2 \begin{pmatrix}-b\\-(2b-2a)\\a\end{pmatrix} \\ &= \begin{pmatrix}2b\\4b-4a\\-2a\end{pmatrix} \; \blacksquare \end{align*}
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×
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=
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v
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0
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2
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a
0
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−
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a
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4
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4
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■
(ii)
Since the
i
-
{\mathbf{i}\textrm{-}}
i
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and
k
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{\mathbf{k}\textrm{-}}
k
-
components are equal,
2
b
=
−
2
a
b
=
−
a
\begin{align*} 2b &= -2a \\ b &= -a \end{align*}
2
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b
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−
2
a
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−
a
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×
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=
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−
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8
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■
\begin{align*} & \left(\mathbf{u} + \mathbf{v}\right) \times \left(\mathbf{u} - \mathbf{v}\right) \\ &= \begin{pmatrix}-2a\\-8a\\-2a\end{pmatrix} \; \blacksquare \end{align*}
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+
v
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×
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v
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=
−
2
a
−
8
a
−
2
a
■
If
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v
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×
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v
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{ \left(\mathbf{u} + \mathbf{v}\right) \times \left(\mathbf{u} - \mathbf{v}\right)}
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v
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×
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u
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v
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is a unit vector,
∣
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×
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∣
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1
∣
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−
2
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∣
=
1
∣
2
a
(
1
4
1
)
∣
=
1
∣
2
a
∣
18
=
1
\begin{align*} \left| \left(\mathbf{u} + \mathbf{v}\right) \times \left(\mathbf{u} - \mathbf{v}\right) \right| &= 1 \\ \left|\begin{pmatrix}-2a\\-8a\\-2a\end{pmatrix} \right| &= 1 \\ \left|2a \begin{pmatrix}1\\4\\1\end{pmatrix} \right| &= 1 \\ \left|2a\right|\sqrt{18} &= 1 \\ \end{align*}
∣
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×
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18
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1
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1
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18
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■
\begin{align*} \left|a\right| &= \frac{1}{2\sqrt{18}} \\ &= \frac{1}{12} \sqrt{2} \\ a &= \pm \frac{1}{12} \sqrt{2} \; \blacksquare \end{align*}
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■
(iii)
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⋅
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=
0
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\begin{align*} (\mathbf{u}+\mathbf{v})\cdot(\mathbf{u}-\mathbf{v}) &= 0 \\ \mathbf{u}\cdot\mathbf{u} + \mathbf{v}\cdot\mathbf{u} - \mathbf{u}\cdot\mathbf{v} - \mathbf{v}\cdot\mathbf{v} &= 0 \\ \left|\mathbf{u}\right|^2 + \mathbf{u}\cdot\mathbf{v} - \mathbf{u}\cdot\mathbf{v} - \left|\mathbf{v}\right|^2 &= 0 \\ \end{align*}
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⋅
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\left|\mathbf{u}\right|^2 - \left|\mathbf{v}\right|^2 = 0
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2
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2
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=
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4
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3
■
\begin{align*} \left|\mathbf{v}\right| &= \left|\mathbf{u}\right| \\ &= \left|\begin{pmatrix} 2 \\ - 1 \\ 2 \end{pmatrix}\right| \\ &= \sqrt{4 + 1 + 4} \\ &= 3 \; \blacksquare \end{align*}
∣
v
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2
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1
2
=
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3
■
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