If A = 2i - 3 j + 4k its components in yz plane and zx plane are respectively

Options

A. `sqrt(13)` and 5
B. 5 and `2sqrt(5)`
C. `2sqrt(5)` and `sqrt(13)`
D. `sqrt(13)` and `sqrt(29)`

Answer

To find the components of the vector \( \mathbf{A} = 2\mathbf{i} - 3\mathbf{j} + 4\mathbf{k} \) in the yz-plane and zx-plane, we will follow these steps:

### Step 1: Identify the components of the vector
The vector \( \mathbf{A} \) has the following components:
- \( A_x = 2 \) (the coefficient of \( \mathbf{i} \))
- \( A_y = -3 \) (the coefficient of \( \mathbf{j} \))
- \( A_z = 4 \) (the coefficient of \( \mathbf{k} \))

### Step 2: Calculate the magnitude of the vector in the yz-plane
In the yz-plane, we only consider the y and z components of the vector. The formula to find the magnitude of the vector in the yz-plane is:
\[ |\mathbf{A}_{yz}| = \sqrt{A_y^2 + A_z^2} \]
Substituting the values:
\[ |\mathbf{A}_{yz}| = \sqrt{(-3)^2 + (4)^2} = \sqrt{9 + 16} = \sqrt{25} = 5 \]

### Step 3: Calculate the magnitude of the vector in the zx-plane
In the zx-plane, we only consider the z and x components of the vector. The formula to find the magnitude of the vector in the zx-plane is:
\[ |\mathbf{A}_{zx}| = \sqrt{A_z^2 + A_x^2} \]
Substituting the values:
\[ |\mathbf{A}_{zx}| = \sqrt{(4)^2 + (2)^2} = \sqrt{16 + 4} = \sqrt{20} = 2\sqrt{5} \]

### Step 4: Present the final result
The components of the vector \( \mathbf{A} \) in the yz-plane and zx-plane are:
- In the yz-plane: \( 5 \)
- In the zx-plane: \( 2\sqrt{5} \)

Thus, the answer is \( 5 \) and \( 2\sqrt{5} \).