A unit vector making an obtuse angle with x-axis and perpendicular to the plane containing the points `hati +2hatj +3hatk, 2hati +3hatj +4hatk and hati +5hatj +7hatk` also makes an obtuse angle with

Options

A. y-axis
B. z-axis
C. y and z axes
D. x and y axes

Explanation

To solve the problem step by step, we need to find a unit vector that is perpendicular to the plane formed by the given points and makes an obtuse angle with the x-axis.

### Step 1: Identify the Points
Let the points be:
- \( A = \hat{i} + 2\hat{j} + 3\hat{k} \)
- \( B = 2\hat{i} + 3\hat{j} + 4\hat{k} \)
- \( C = \hat{i} + 5\hat{j} + 7\hat{k} \)

### Step 2: Find Vectors AB and AC
We need to find the vectors \( \vec{AB} \) and \( \vec{AC} \).

1. **Calculate \( \vec{AB} \)**:
\[ \vec{AB} = \vec{B} - \vec{A} = (2\hat{i} + 3\hat{j} + 4\hat{k}) - (\hat{i} + 2\hat{j} + 3\hat{k}) = (2-1)\hat{i} + (3-2)\hat{j} + (4-3)\hat{k} = \hat{i} + \hat{j} + \hat{k} \]

2. **Calculate \( \vec{AC} \)**:
\[ \vec{AC} = \vec{C} - \vec{A} = (\hat{i} + 5\hat{j} + 7\hat{k}) - (\hat{i} + 2\hat{j} + 3\hat{k}) = (1-1)\hat{i} + (5-2)\hat{j} + (7-3)\hat{k} = 0\hat{i} + 3\hat{j} + 4\hat{k} \]

### Step 3: Compute the Cross Product \( \vec{AB} \times \vec{AC} \)
To find a vector perpendicular to the plane formed by \( A, B, C \), we compute the cross product \( \vec{AB} \times \vec{AC} \).

\[ \vec{AB} = \hat{i} + \hat{j} + \hat{k}, \quad \vec{AC} = 0\hat{i} + 3\hat{j} + 4\hat{k} \]

Using the determinant method to find the cross product:
\[ \vec{AB} \times \vec{AC} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 1 & 1 \\ 0 & 3 & 4 \end{vmatrix} \]

Calculating the determinant:
\[ = \hat{i} \begin{vmatrix} 1 & 1 \\ 3 & 4 \end{vmatrix} - \hat{j} \begin{vmatrix} 1 & 1 \\ 0 & 4 \end{vmatrix} + \hat{k} \begin{vmatrix} 1 & 1 \\ 0 & 3 \end{vmatrix} \]

Calculating each of these:
1. \( \hat{i}(1 \cdot 4 - 1 \cdot 3) = \hat{i}(4 - 3) = \hat{i} \)
2. \( -\hat{j}(1 \cdot 4 - 1 \cdot 0) = -\hat{j}(4) = -4\hat{j} \)
3. \( \hat{k}(1 \cdot 3 - 1 \cdot 0) = \hat{k}(3) = 3\hat{k} \)

Thus,
\[ \vec{AB} \times \vec{AC} = \hat{i} - 4\hat{j} + 3\hat{k} \]

### Step 4: Find the Magnitude of the Cross Product
Now, we find the magnitude of \( \vec{AB} \times \vec{AC} \):
\[ |\vec{AB} \times \vec{AC}| = \sqrt{(1)^2 + (-4)^2 + (3)^2} = \sqrt{1 + 16 + 9} = \sqrt{26} \]

### Step 5: Normalize the Vector
The unit vector perpendicular to the plane is given by:
\[ \hat{n} = \frac{\vec{AB} \times \vec{AC}}{|\vec{AB} \times \vec{AC}|} = \frac{\hat{i} - 4\hat{j} + 3\hat{k}}{\sqrt{26}} \]

### Step 6: Determine the Angle with the X-axis
The unit vector makes an obtuse angle with the x-axis if the x-component is negative. The x-component of \( \hat{n} \) is:
\[ \frac{1}{\sqrt{26}} \]
Since this is positive, we need to take the negative of the unit vector to ensure it makes an obtuse angle with the x-axis:
\[ \hat{n} = -\frac{\hat{i} - 4\hat{j} + 3\hat{k}}{\sqrt{26}} = -\frac{1}{\sqrt{26}}\hat{i} + \frac{4}{\sqrt{26}}\hat{j} - \frac{3}{\sqrt{26}}\hat{k} \]

### Step 7: Check the Components
Now we check the signs of the components:
- The x-component is negative (obtuse angle with x-axis).
- The y-component is positive (acute angle with y-axis).
- The z-component is negative (obtuse angle with z-axis).

### Conclusion
The unit vector makes an obtuse angle with the x-axis and z-axis.

### Final Answer
The unit vector makes an obtuse angle with the **z-axis**.

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