Q.Find a vector in the direction of vector $\vec{a}=\hat{i}-2\hat{j}$ that has magnitude 7 units.

The key idea is Unit Vector Scaling: any vector in a given direction is a scalar multiple of the unit vector along that direction.

Step 1: Find the magnitude of $\vec{a}$.

Step 2: The unit vector in the direction of $\vec{a}$ is

$\hat{a} = \frac{\vec{a}}{|\vec{a}|} = \frac{1}{\sqrt{5}}(\hat{i} - 2\hat{j})$ …

Why This Works: The Unit Vector Scaling Principle

When you need a vector in the direction of a given vector but with a different magnitude, you are essentially asking: "What vector points the same way, but is longer or shorter?" The cleanest way to do this is to first find the unit vector — the vector of length 1 that points exactly in the given direction. Once you have that, you can simply multiply it by the desired magnitude (7 units) to stretch it to the right length.

Think of it like a compass needle. If $\vec{a}$ points northeast, its unit vector is just the direction "northeast" without any length. Multiplying by 7 gives you a 7-unit-long arrow still pointing northeast.

Step-by-Step Solution

1. Find the magnitude of $\vec{a}$.

The vector is $\vec{a} = \hat{i} - 2\hat{j}$. Its magnitude is the square root of the sum of the squares of its components:

2. Construct the unit vector in the direction of $\vec{a}$.

The unit vector $\hat{a}$ is obtained by dividing $\vec{a}$ by its magnitude:

This vector has magnitude 1 and points exactly in the same direction as $\vec{a}$. …

Method: Building a vector of given magnitude in a specified direction

Use this whenever you need a vector that points along a given $\vec{a}$ but has a prescribed length $m$ (here $m\neq|\vec{a}|$).

Steps

Step 1: Find the unit vector along $\vec{a}$.

This strips away the original length, leaving pure direction.

Step 2: Scale the unit vector by the desired magnitude.

$\vec{v}=m\,\hat{a}=m\,\frac{\vec{a}}{|\vec{a}|}.$ …

Common Mistakes

Mistake 1: Multiplying $\vec{a}$ by $7$ directly to "make its magnitude $7$".

Why it's wrong: $7\vec{a}$ has magnitude $7|\vec{a}|=7\sqrt5$, not $7$ — you must scale the unit vector, not the original. Correct approach: first normalise ($\hat{a}=\vec{a}/\sqrt5$), then multiply by $7$ to get $\frac{7}{\sqrt5}(\hat{i}-2\hat{j})$. …