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2010
P1 Q1
Topical
Vectors I
10 P1 Q1
2010 H2 Mathematics Paper 1 Question 1
Vectors I: Basics, Dot and Cross Products
Answers
(i)
p
=
3
7
{p=\frac{3}{7}}
p
=
7
3
Full solutions
(i)
49
p
2
=
9
p
2
=
9
49
p
=
3
7
■
since
p
>
0
\begin{align*} 49 p^2 &= 9 \\ p^2 &= \frac{9}{49} \\ p &= \frac{3}{7} \; \blacksquare \quad \textrm{ since } p>0 \\ \end{align*}
49
p
2
p
2
p
=
9
=
49
9
=
7
3
■
since
p
>
0
(ii)
(
a
+
b
)
⋅
(
a
−
b
)
=
a
⋅
a
+
b
⋅
a
−
a
⋅
b
−
b
⋅
b
=
∣
a
∣
2
+
a
⋅
b
−
a
⋅
b
−
∣
b
∣
2
=
0
■
since
∣
a
∣
=
∣
b
∣
\begin{align*} & (\mathbf{a}+\mathbf{b}) \cdot (\mathbf{a}-\mathbf{b}) \\ &= \mathbf{a}\cdot\mathbf{a} +\mathbf{b}\cdot\mathbf{a} - \mathbf{a}\cdot\mathbf{b} - \mathbf{b}\cdot\mathbf{b} \\ &= \left|\mathbf{a}\right|^2 + \mathbf{a}\cdot\mathbf{b} - \mathbf{a}\cdot\mathbf{b} - \left|\mathbf{b}\right|^2 \\ &= 0 \; \blacksquare \quad \textrm{ since } \left|\mathbf{a}\right|=\left|\mathbf{b}\right| \\ \end{align*}
(
a
+
b
)
⋅
(
a
−
b
)
=
a
⋅
a
+
b
⋅
a
−
a
⋅
b
−
b
⋅
b
=
∣
a
∣
2
+
a
⋅
b
−
a
⋅
b
−
∣
b
∣
2
=
0
■
since
∣
a
∣
=
∣
b
∣
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