The vector `veca` coplanar with the vectors `hati` and `hatj` perendicular to the vector `vecb=4hati-3hatj+5hatk` such that `|veca|=|vecb|` is

Options

A. `sqrt(2)(3hati+4hatj)` or `-sqrt(2)(3hati+4hatj)`
B. `sqrt(2)(4hati+3hatj)` or `-sqrt(2)(4hati+3hatj)`
C. `sqrt(3)(4hati+5hatj)` ro `-sqrt(3)(4hati+5hatj)`
D. `sqrt(3)(5hati+4hatj)` or `-sqrt(3)(5hati+4hatj)`

Explanation

To solve the problem step by step, we need to find the vector \(\vec{a}\) that is coplanar with the vectors \(\hat{i}\) and \(\hat{j}\), perpendicular to the vector \(\vec{b} = 4\hat{i} - 3\hat{j} + 5\hat{k}\), and has the same magnitude as \(\vec{b}\).

### Step 1: Define the vector \(\vec{a}\)
Since \(\vec{a}\) is coplanar with \(\hat{i}\) and \(\hat{j}\), we can express it as:
\[ \vec{a} = x\hat{i} + y\hat{j} + 0\hat{k} \]
This means that \(z = 0\).

### Step 2: Find the condition for coplanarity
The condition for coplanarity of vectors \(\vec{a}\), \(\hat{i}\), and \(\hat{j}\) can be expressed using the scalar triple product:
\[ \vec{a} \cdot (\hat{i} \times \hat{j}) = 0 \]
Calculating \(\hat{i} \times \hat{j}\) gives us \(\hat{k}\). Thus:
\[ \vec{a} \cdot \hat{k} = 0 \]
Since \(\vec{a}\) has no \(\hat{k}\) component, this condition is satisfied.

### Step 3: Find the perpendicularity condition
Since \(\vec{a}\) is perpendicular to \(\vec{b}\), we have:
\[ \vec{a} \cdot \vec{b} = 0 \]
Calculating the dot product:
\[ (x\hat{i} + y\hat{j}) \cdot (4\hat{i} - 3\hat{j} + 5\hat{k}) = 0 \]
This simplifies to:
\[ 4x - 3y + 0 = 0 \]
From this, we can express \(x\) in terms of \(y\):
\[ 4x = 3y \implies x = \frac{3}{4}y \]

### Step 4: Find the magnitudes
Next, we need to ensure that the magnitudes of \(\vec{a}\) and \(\vec{b}\) are equal. The magnitude of \(\vec{b}\) is:
\[ |\vec{b}| = \sqrt{4^2 + (-3)^2 + 5^2} = \sqrt{16 + 9 + 25} = \sqrt{50} = 5\sqrt{2} \]
The magnitude of \(\vec{a}\) is:
\[ |\vec{a}| = \sqrt{x^2 + y^2 + 0^2} = \sqrt{x^2 + y^2} \]

### Step 5: Substitute \(x\) in terms of \(y\)
Substituting \(x = \frac{3}{4}y\) into the magnitude equation:
\[ |\vec{a}| = \sqrt{\left(\frac{3}{4}y\right)^2 + y^2} = \sqrt{\frac{9}{16}y^2 + y^2} = \sqrt{\frac{9}{16}y^2 + \frac{16}{16}y^2} = \sqrt{\frac{25}{16}y^2} = \frac{5}{4}|y| \]
Setting this equal to the magnitude of \(\vec{b}\):
\[ \frac{5}{4}|y| = 5\sqrt{2} \]
Solving for \(|y|\):
\[ |y| = 4\sqrt{2} \]

### Step 6: Find \(y\) and \(x\)
Thus, \(y = \pm 4\sqrt{2}\). Now substituting back to find \(x\):
\[ x = \frac{3}{4}y = \frac{3}{4}(4\sqrt{2}) = 3\sqrt{2} \]
So we have:
\[ x = \pm 3\sqrt{2}, \quad y = \pm 4\sqrt{2}, \quad z = 0 \]

### Step 7: Write the final vector \(\vec{a}\)
The vector \(\vec{a}\) can be expressed as:
\[ \vec{a} = 3\sqrt{2}\hat{i} + 4\sqrt{2}\hat{j} \quad \text{or} \quad \vec{a} = -3\sqrt{2}\hat{i} - 4\sqrt{2}\hat{j} \]
Thus, the final answer is:
\[ \vec{a} = \pm \sqrt{2}(3\hat{i} + 4\hat{j}) \]