Position Vectors (A) Worksheet | Cazoom Maths Worksheets - Free Printable

Problem Analysis and Solution

The problem involves position vectors and their relationships in a grid of parallelograms. Let's solve it step by step.

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#### Section A

1. Mark the points C to J using the following clues:

The grid is made of identical parallelograms, and we are given position vectors relative to point \( O \). We need to mark points \( C \) to \( J \) based on these vectors.

- Point \( C \): \( \overrightarrow{OC} = -\mathbf{b} \)
- This means point \( C \) is located at a position vector that is the negative of \( \mathbf{b} \). In the grid, this would be one unit "down" from \( O \).

- Point \( D \): \( \overrightarrow{OD} = -\mathbf{a} \)
- This means point \( D \) is located at a position vector that is the negative of \( \mathbf{a} \). In the grid, this would be one unit "left" from \( O \).

- Point \( E \): \( \overrightarrow{OE} = \mathbf{b} - 2\mathbf{a} \)
- This means point \( E \) is located at a position vector that is \( \mathbf{b} \) (one unit "up") minus \( 2\mathbf{a} \) (two units "left"). So, \( E \) is two units left and one unit up from \( O \).

- Point \( F \): \( \overrightarrow{OF} = -\mathbf{b} - 2\mathbf{a} \)
- This means point \( F \) is located at a position vector that is \( -\mathbf{b} \) (one unit "down") minus \( 2\mathbf{a} \) (two units "left"). So, \( F \) is two units left and one unit down from \( O \).

- Point \( G \): \( \overrightarrow{OG} = 2\mathbf{b} + \mathbf{a} \)
- This means point \( G \) is located at a position vector that is \( 2\mathbf{b} \) (two units "up") plus \( \mathbf{a} \) (one unit "right"). So, \( G \) is one unit right and two units up from \( O \).

- Point \( H \): \( \overrightarrow{OH} = \mathbf{b} - 3\mathbf{a} \)
- This means point \( H \) is located at a position vector that is \( \mathbf{b} \) (one unit "up") minus \( 3\mathbf{a} \) (three units "left"). So, \( H \) is three units left and one unit up from \( O \).

- Point \( I \): \( \overrightarrow{OI} = \mathbf{b} + \frac{1}{2}\mathbf{a} \)
- This means point \( I \) is located at a position vector that is \( \mathbf{b} \) (one unit "up") plus \( \frac{1}{2}\mathbf{a} \) (half a unit "right"). So, \( I \) is half a unit right and one unit up from \( O \).

- Point \( J \): \( \overrightarrow{OJ} = -3\mathbf{a} - \frac{1}{2}\mathbf{b} \)
- This means point \( J \) is located at a position vector that is \( -3\mathbf{a} \) (three units "left") minus \( \frac{1}{2}\mathbf{b} \) (half a unit "down"). So, \( J \) is three units left and half a unit down from \( O \).

2. What is the relationship between the vectors \( \overrightarrow{FB} \) and \( \overrightarrow{GB} \)?

- First, find \( \overrightarrow{FB} \):
\[
\overrightarrow{FB} = \overrightarrow{OB} - \overrightarrow{OF}
\]
Given \( \overrightarrow{OB} = \mathbf{b} \) and \( \overrightarrow{OF} = -\mathbf{b} - 2\mathbf{a} \):
\[
\overrightarrow{FB} = \mathbf{b} - (-\mathbf{b} - 2\mathbf{a}) = \mathbf{b} + \mathbf{b} + 2\mathbf{a} = 2\mathbf{b} + 2\mathbf{a}
\]

- Next, find \( \overrightarrow{GB} \):
\[
\overrightarrow{GB} = \overrightarrow{OB} - \overrightarrow{OG}
\]
Given \( \overrightarrow{OB} = \mathbf{b} \) and \( \overrightarrow{OG} = 2\mathbf{b} + \mathbf{a} \):
\[
\overrightarrow{GB} = \mathbf{b} - (2\mathbf{b} + \mathbf{a}) = \mathbf{b} - 2\mathbf{b} - \mathbf{a} = -\mathbf{b} - \mathbf{a}
\]

- Now, compare \( \overrightarrow{FB} \) and \( \overrightarrow{GB} \):
\[
\overrightarrow{FB} = 2\mathbf{b} + 2\mathbf{a}
\]
\[
\overrightarrow{GB} = -\mathbf{b} - \mathbf{a}
\]
Notice that:
\[
\overrightarrow{FB} = -2(\overrightarrow{GB})
\]
Therefore, \( \overrightarrow{FB} \) is parallel to \( \overrightarrow{GB} \) but in the opposite direction, and its magnitude is twice that of \( \overrightarrow{GB} \).

3. Explain how you know that \( D \), \( B \), and \( G \) are collinear?

- First, find \( \overrightarrow{DB} \):
\[
\overrightarrow{DB} = \overrightarrow{OB} - \overrightarrow{OD}
\]
Given \( \overrightarrow{OB} = \mathbf{b} \) and \( \overrightarrow{OD} = -\mathbf{a} \):
\[
\overrightarrow{DB} = \mathbf{b} - (-\mathbf{a}) = \mathbf{b} + \mathbf{a}
\]

- Next, find \( \overrightarrow{BG} \):
\[
\overrightarrow{BG} = \overrightarrow{OG} - \overrightarrow{OB}
\]
Given \( \overrightarrow{OG} = 2\mathbf{b} + \mathbf{a} \) and \( \overrightarrow{OB} = \mathbf{b} \):
\[
\overrightarrow{BG} = (2\mathbf{b} + \mathbf{a}) - \mathbf{b} = \mathbf{b} + \mathbf{a}
\]

- Compare \( \overrightarrow{DB} \) and \( \overrightarrow{BG} \):
\[
\overrightarrow{DB} = \mathbf{b} + \mathbf{a}
\]
\[
\overrightarrow{BG} = \mathbf{b} + \mathbf{a}
\]
Since \( \overrightarrow{DB} = \overrightarrow{BG} \), the vectors are equal, which means points \( D \), \( B \), and \( G \) lie on the same straight line. Therefore, they are collinear.

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#### Section B

1. Write down, in terms of \( \mathbf{a} \), \( \mathbf{b} \), \( \mathbf{c} \), \( \mathbf{d} \), and \( \mathbf{e} \), the vector for:

- \( \overrightarrow{RM} \):
\[
\overrightarrow{RM} = \mathbf{a}
\]

- \( \overrightarrow{QR} \):
\[
\overrightarrow{QR} = \mathbf{e}
\]

- \( \overrightarrow{QP} \):
\[
\overrightarrow{QP} = \overrightarrow{QR} + \overrightarrow{RP} = \mathbf{e} + (-\mathbf{d}) = \mathbf{e} - \mathbf{d}
\]

- \( \overrightarrow{MO} \):
\[
\overrightarrow{MO} = \overrightarrow{MR} + \overrightarrow{RO} = -\mathbf{a} + (\mathbf{b} + \mathbf{c})
\]
\[
\overrightarrow{MO} = -\mathbf{a} + \mathbf{b} + \mathbf{c}
\]

- \( \overrightarrow{MN} \):
\[
\overrightarrow{MN} = \overrightarrow{MR} + \overrightarrow{RN} = \mathbf{a} + \mathbf{b}
\]

- \( \overrightarrow{QN} \):
\[
\overrightarrow{QN} = \overrightarrow{QP} + \overrightarrow{PN} = (\mathbf{e} - \mathbf{d}) + (\mathbf{b} + \mathbf{c})
\]
\[
\overrightarrow{QN} = \mathbf{e} - \mathbf{d} + \mathbf{b} + \mathbf{c}
\]

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Final Answers

#### Section A
1. Points marked based on the given vectors.
2. Relationship between \( \overrightarrow{FB} \) and \( \overrightarrow{GB} \):
\[
\boxed{\overrightarrow{FB} = -2\overrightarrow{GB}}
\]
3. Explanation of collinearity of \( D \), \( B \), and \( G \):
\[
\boxed{\overrightarrow{DB} = \overrightarrow{BG}}
\]

#### Section B
1. Vectors in terms of \( \mathbf{a} \), \( \mathbf{b} \), \( \mathbf{c} \), \( \mathbf{d} \), and \( \mathbf{e} \):
\[
\overrightarrow{RM} = \mathbf{a}, \quad \overrightarrow{QR} = \mathbf{e}, \quad \overrightarrow{QP} = \mathbf{e} - \mathbf{d}, \quad \overrightarrow{MO} = -\mathbf{a} + \mathbf{b} + \mathbf{c}, \quad \overrightarrow{MN} = \mathbf{a} + \mathbf{b}, \quad \overrightarrow{QN} = \mathbf{e} - \mathbf{d} + \mathbf{b} + \mathbf{c}
\]