What is the vector in the XY -plane through origin and perpendicular to the vector `vec(r )= a hat(i) +b hat(j)` and of the same length?

Options

A. `-a hat(i) - b hat(j)`
B. `a hat(i)- b hat(j)`
C. `-a hat(i) + b hat(j)`
D. `b hat(i) - a hat(j)`

Answer

To find the vector in the XY-plane that passes through the origin, is perpendicular to the vector \(\vec{r} = a \hat{i} + b \hat{j}\), and has the same length, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Given Vector**:
The given vector is \(\vec{r} = a \hat{i} + b \hat{j}\).

2. **Calculate the Length of the Given Vector**:
The length (magnitude) of the vector \(\vec{r}\) is calculated using the formula:
\[ |\vec{r}| = \sqrt{a^2 + b^2} \]

3. **Determine the Condition for Perpendicularity**:
For a vector \(\vec{v} = x \hat{i} + y \hat{j}\) to be perpendicular to \(\vec{r}\), their dot product must be zero:
\[ \vec{r} \cdot \vec{v} = (a \hat{i} + b \hat{j}) \cdot (x \hat{i} + y \hat{j}) = ax + by = 0 \]

4. **Find a Perpendicular Vector**:
We can express the perpendicular vector in terms of \(a\) and \(b\). A common choice is:
\[ \vec{v} = -b \hat{i} + a \hat{j} \]
This choice ensures that \(\vec{v}\) is perpendicular to \(\vec{r}\) because:
\[ a(-b) + b(a) = -ab + ab = 0 \]

5. **Check the Length of the Perpendicular Vector**:
Now we need to ensure that \(\vec{v}\) has the same length as \(\vec{r}\):
\[ |\vec{v}| = \sqrt{(-b)^2 + a^2} = \sqrt{b^2 + a^2} \]
Since \(b^2 + a^2 = a^2 + b^2\), we find that:
\[ |\vec{v}| = |\vec{r}| \]

6. **Final Result**:
Thus, the vector that is perpendicular to \(\vec{r}\) and has the same length is:
\[ \vec{v} = -b \hat{i} + a \hat{j} \]