FPM(2023) January Paper 01R Q4 and solution

Solution

Let $O$, $A$, and $B$ be fixed points such that:

$$ \overrightarrow{OA} = p\mathbf{i} + 2p\mathbf{j}, \quad \overrightarrow{OB} = 5\mathbf{i} + 9p\mathbf{j}. $$

We are given that $\overrightarrow{AB}$ is parallel to $(\mathbf{i} - 2\mathbf{j})$.

(a) Find the value of $p$

The vector $\overrightarrow{AB}$ is given by:

$$ \overrightarrow{AB} = \overrightarrow{OB} - \overrightarrow{OA}. $$

Substitute the given values:

$$ \overrightarrow{AB} = (5\mathbf{i} + 9p\mathbf{j}) - (p\mathbf{i} + 2p\mathbf{j}). $$

Simplify:

$$ \overrightarrow{AB} = (5 - p)\mathbf{i} + (9p - 2p)\mathbf{j}. $$

$$ \overrightarrow{AB} = (5 - p)\mathbf{i} + 7p\mathbf{j}. $$

Since $\overrightarrow{AB}$ is parallel to $(\mathbf{i} - 2\mathbf{j})$, there exists a scalar $k$ such that:

$$ \overrightarrow{AB} = k(\mathbf{i} - 2\mathbf{j}). $$

Equating components:

$$ 5 - p = k, \quad 7p = -2k. $$

From the first equation:

$$ k = 5 - p. $$

Substitute $k = 5 - p$ into the second equation:

$$ 7p = -2(5 - p). $$

Expand:

$$ 7p = -10 + 2p. $$

Rearrange:

$$ 7p - 2p = -10 \implies 5p = -10 \implies p = -2. $$

(b) Find $\overrightarrow{AB}$ as a simplified expression in terms of $\mathbf{i}$ and $\mathbf{j}$

Substitute $p = -2$ into

$$ \overrightarrow{AB} = (5 - p)\mathbf{i} + 7p\mathbf{j}: $$

$$ \overrightarrow{AB} = (5 - (-2))\mathbf{i} + 7(-2)\mathbf{j}. $$

$$ \overrightarrow{AB} = (5 + 2)\mathbf{i} - 14\mathbf{j}. $$

$$ \overrightarrow{AB} = 7\mathbf{i} - 14\mathbf{j}. $$

(c) Find a unit vector parallel to $\overrightarrow{OA}$

The vector $\overrightarrow{OA}$ is:

$$ \overrightarrow{OA} = p\mathbf{i} + 2p\mathbf{j}. $$

Substitute $p = -2$:

$$ \overrightarrow{OA} = -2\mathbf{i} + 2(-2)\mathbf{j}. $$

$$ \overrightarrow{OA} = -2\mathbf{i} - 4\mathbf{j}. $$

The magnitude of $\overrightarrow{OA}$ is:

$$ |\overrightarrow{OA}| = \sqrt{(-2)^2 + (-4)^2} = \sqrt{4 + 16} = \sqrt{20} = 2\sqrt{5}. $$

The unit vector parallel to $\overrightarrow{OA}$ is:

$$ \frac{\overrightarrow{OA}}{|\overrightarrow{OA}|} = \frac{-2\mathbf{i} - 4\mathbf{j}}{2\sqrt{5}}. $$

Simplify:

$$ \frac{\overrightarrow{OA}}{|\overrightarrow{OA}|} = \frac{-1}{\sqrt{5}}\mathbf{i} + \frac{-2}{\sqrt{5}}\mathbf{j}. $$

Rewrite in the form

$$ \left(\frac{\sqrt{a}}{5}\right) (b\mathbf{i} + c\mathbf{j}) $$

$$ \frac{\overrightarrow{OA}}{|\overrightarrow{OA}|} = \left( \frac{\sqrt{5}}{5} \right) (-1\mathbf{i} - 2\mathbf{j}). $$

Here:

$$ a = 5, \quad b = -1, \quad c = -2. $$

Final Answers

• (a) $p = -2$
• (b) $\overrightarrow{AB} = 7\mathbf{i} - 14\mathbf{j}$
• (c) The unit vector parallel to $\overrightarrow{OA}$ is:

$$ \left( \frac{\sqrt{5}}{5} \right) (-1\mathbf{i} - 2\mathbf{j}), $$

where

$$ a = 5, \quad b = -1, \quad c = -2. $$