Show that the vectors `vec(a)=2hat(i)+3hat(j)` and `vec(b)=4hat(i)+6hat(j)` are parallel.

Answer

To show that the vectors \(\vec{a} = 2\hat{i} + 3\hat{j}\) and \(\vec{b} = 4\hat{i} + 6\hat{j}\) are parallel, we can use two methods. Here is the step-by-step solution:

### Method 1: Using the Dot Product

1. **Calculate the Dot Product**:
\[ \vec{a} \cdot \vec{b} = (2\hat{i} + 3\hat{j}) \cdot (4\hat{i} + 6\hat{j}) \]
Using the property of dot product:
\[ \vec{a} \cdot \vec{b} = 2 \cdot 4 + 3 \cdot 6 = 8 + 18 = 26 \]

2. **Calculate the Magnitudes of the Vectors**:
- For \(\vec{a}\):
\[ |\vec{a}| = \sqrt{2^2 + 3^2} = \sqrt{4 + 9} = \sqrt{13} \]
- For \(\vec{b}\):
\[ |\vec{b}| = \sqrt{4^2 + 6^2} = \sqrt{16 + 36} = \sqrt{52} = 2\sqrt{13} \]

3. **Use the Dot Product Formula**:
The formula relating the dot product to the angle \(\theta\) between the vectors is:
\[ \vec{a} \cdot \vec{b} = |\vec{a}| |\vec{b}| \cos \theta \]
Substituting the values we found:
\[ 26 = \sqrt{13} \cdot 2\sqrt{13} \cos \theta \]
Simplifying:
\[ 26 = 2 \cdot 13 \cos \theta \implies 26 = 26 \cos \theta \]
Therefore:
\[ \cos \theta = 1 \implies \theta = 0^\circ \]

4. **Conclusion**:
Since the angle \(\theta\) is \(0^\circ\), the vectors \(\vec{a}\) and \(\vec{b}\) are parallel.

### Method 2: Using Direction Ratios

1. **Identify Direction Ratios**:
The direction ratios of \(\vec{a}\) are \(2\) and \(3\), and the direction ratios of \(\vec{b}\) are \(4\) and \(6\).

2. **Check if One Vector is a Scalar Multiple of the Other**:
We can express \(\vec{b}\) in terms of \(\vec{a}\):
\[ \vec{b} = 2 \cdot \vec{a} = 2(2\hat{i} + 3\hat{j}) = 4\hat{i} + 6\hat{j} \]

3. **Conclusion**:
Since \(\vec{b}\) is a scalar multiple of \(\vec{a}\), the vectors are parallel.

### Final Conclusion:
Both methods confirm that the vectors \(\vec{a}\) and \(\vec{b}\) are parallel.

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