Consider a system of two vectors
`overset(rarr)a=3hat i +4 hat j, overset(rarr)b=hat i+hatj` ,

Options

A. `overset(rarr)r=sqrt(2)overset(rarr)a+5overset(rarr)b` will be making equal angles with and `overset(rarr)a` and `overset(rarr)b`
B. The angle made by `overset(rarr)r=overset(rarr)a+overset(rarr)b` with `overset(rarr)a` will be smaller as compared to angle made by with `overset(rarr)r` with `overset(rarr)b`
C. Angle made by `overset(rarr)r=overset(rarr)a+overset(rarr)b` with be greater as compared to angle made by with `overset(rarr)b`
D. Incomplete information to predict any of the above options

Explanation

To solve the problem, we will analyze the two vectors given and check the options provided. We will focus on the first option as described in the video transcript.

### Step-by-Step Solution:

1. **Define the Vectors**:
- Let \(\vec{a} = 3\hat{i} + 4\hat{j}\)
- Let \(\vec{b} = \hat{i} + \hat{j}\)

2. **Calculate the Resultant Vector**:
- According to the first option, the resultant vector \(\vec{r}\) is given by:
\[ \vec{r} = \sqrt{2} \vec{a} + 5 \vec{b} \]
- Substitute the values of \(\vec{a}\) and \(\vec{b}\):
\[ \vec{r} = \sqrt{2}(3\hat{i} + 4\hat{j}) + 5(\hat{i} + \hat{j}) \]
- This simplifies to:
\[ \vec{r} = 3\sqrt{2}\hat{i} + 4\sqrt{2}\hat{j} + 5\hat{i} + 5\hat{j} \]
- Combine the components:
\[ \vec{r} = (3\sqrt{2} + 5)\hat{i} + (4\sqrt{2} + 5)\hat{j} \]

3. **Calculate the Magnitudes**:
- The magnitude of vector \(\vec{a}\) is:
\[ |\vec{a}| = \sqrt{3^2 + 4^2} = \sqrt{9 + 16} = 5 \]
- The magnitude of vector \(\vec{b}\) is:
\[ |\vec{b}| = \sqrt{1^2 + 1^2} = \sqrt{2} \]
- The magnitude of vector \(\vec{r}\) is:
\[ |\vec{r}| = \sqrt{(3\sqrt{2} + 5)^2 + (4\sqrt{2} + 5)^2} \]

4. **Calculate the Dot Products**:
- For \(\cos \theta_1\) (angle between \(\vec{r}\) and \(\vec{a}\)):
\[ \cos \theta_1 = \frac{\vec{r} \cdot \vec{a}}{|\vec{r}| |\vec{a}|} \]
- Calculate \(\vec{r} \cdot \vec{a}\):
\[ \vec{r} \cdot \vec{a} = (3\sqrt{2} + 5) \cdot 3 + (4\sqrt{2} + 5) \cdot 4 \]
- For \(\cos \theta_2\) (angle between \(\vec{r}\) and \(\vec{b}\)):
\[ \cos \theta_2 = \frac{\vec{r} \cdot \vec{b}}{|\vec{r}| |\vec{b}|} \]
- Calculate \(\vec{r} \cdot \vec{b}\):
\[ \vec{r} \cdot \vec{b} = (3\sqrt{2} + 5) + (4\sqrt{2} + 5) \]

5. **Compare the Angles**:
- Since \(\cos \theta_1\) and \(\cos \theta_2\) are calculated, we can compare them to determine which angle is smaller.
- If \(\cos \theta_1 > \cos \theta_2\), then \(\theta_1 < \theta_2\), meaning \(\vec{r}\) makes a smaller angle with \(\vec{a}\) than with \(\vec{b}\).