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

Options

A. `-ahat(i)-bvec(j)`
B. `ahat(i)-b hat(j)`
C. `-ahat(i)+bhat(j)`
D. `bhat(i)-ahat(j)`

Explanation

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

### Step 1: Find the magnitude of the vector \(\vec{r}\)
The magnitude of the vector \(\vec{r}\) is given by:
\[ |\vec{r}| = \sqrt{a^2 + b^2} \]

### Step 2: Determine the condition for perpendicularity
Two vectors \(\vec{u}\) and \(\vec{v}\) are perpendicular if their dot product is zero:
\[ \vec{u} \cdot \vec{v} = 0 \]

### Step 3: Find a vector that is perpendicular to \(\vec{r}\)
To find a vector that is perpendicular to \(\vec{r}\), we can use the property that if \(\vec{r} = a \hat{i} + b \hat{j}\), then a vector perpendicular to it can be expressed as:
\[ \vec{u} = -b \hat{i} + a \hat{j} \]

### Step 4: Check the magnitude of the perpendicular vector
Now, we need to ensure that the magnitude of \(\vec{u}\) is the same as the magnitude of \(\vec{r}\):
\[ |\vec{u}| = \sqrt{(-b)^2 + a^2} = \sqrt{b^2 + a^2} \]
Since \(b^2 + a^2\) is equal to \(a^2 + b^2\), we have:
\[ |\vec{u}| = |\vec{r}| \]

### Conclusion
Thus, the vector that is in the xy-plane, through the origin, perpendicular to \(\vec{r}\), and of the same length is:
\[ \vec{u} = -b \hat{i} + a \hat{j} \]