The vector `2 hat(j)- hat(k)` lies

Options

A. In the plane of XY
B. In the plane of YZ
C. In the plane of XZ
D. Along the X-axis

Answer

To determine the plane in which the vector \( \mathbf{v} = 2\hat{j} - \hat{k} \) lies, we can analyze its components.

### Step-by-Step Solution:

1. **Identify the Components of the Vector**:
The vector can be expressed in terms of its components along the x, y, and z axes. Here, we have:
\[ \mathbf{v} = 0\hat{i} + 2\hat{j} - 1\hat{k} \]
This means:
- The x-component is \( 0 \) (since there is no \( \hat{i} \) term).
- The y-component is \( 2 \) (from the \( 2\hat{j} \) term).
- The z-component is \( -1 \) (from the \( -\hat{k} \) term).

2. **Determine the Relevant Planes**:
- The **xy-plane** is defined by vectors that have no z-component (i.e., \( z = 0 \)).
- The **yz-plane** is defined by vectors that have no x-component (i.e., \( x = 0 \)).
- The **xz-plane** is defined by vectors that have no y-component (i.e., \( y = 0 \)).
- The **x-axis** consists of vectors where both y and z components are zero (i.e., \( y = 0 \) and \( z = 0 \)).

3. **Analyze the Vector**:
Since the x-component is \( 0 \), the vector lies in the yz-plane. The presence of non-zero y and z components confirms that it does not lie along the x-axis or in the xy-plane.

4. **Conclusion**:
Therefore, the vector \( 2\hat{j} - \hat{k} \) lies in the **yz-plane**.

### Final Answer:
The vector \( 2\hat{j} - \hat{k} \) lies in the **yz-plane**.

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