A unit vector making an obtuse angle with x-axis and perpendicular to the plane containing the points `A (1, 2, 3), B(2, 3, 4), C(1, 5, 7)` also make an obtuse angle with

Options

A. y-axis
B. z-axis
C. y and z axes
D. none

Answer

To solve the problem of finding a unit vector that makes an obtuse angle with the x-axis and is perpendicular to the plane containing the points A(1, 2, 3), B(2, 3, 4), and C(1, 5, 7), we can follow these steps:

### Step 1: Find the vectors AB and AC
First, we need to find the vectors AB and AC from the points A, B, and C.

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

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

### Step 2: Compute the cross product AB × AC
Next, we compute the cross product of vectors AB and AC to find a vector that is perpendicular to the plane formed by points A, B, and C.

\[ \vec{AB} \times \vec{AC} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 1 & 1 \\ 0 & 3 & 4 \end{vmatrix} \]

Calculating the determinant:
- For \(\hat{i}\):
\[ \hat{i} \cdot (1 \cdot 4 - 1 \cdot 3) = \hat{i} \cdot (4 - 3) = \hat{i} \]
- For \(\hat{j}\):
\[ -\hat{j} \cdot (1 \cdot 4 - 1 \cdot 0) = -\hat{j} \cdot 4 = -4\hat{j} \]
- For \(\hat{k}\):
\[ \hat{k} \cdot (1 \cdot 3 - 1 \cdot 0) = \hat{k} \cdot 3 = 3\hat{k} \]

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

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

### Step 4: Find the unit vector
To find the unit vector perpendicular to the plane, we divide the cross product by its magnitude:
\[ \hat{n} = \frac{\vec{AB} \times \vec{AC}}{\|\vec{AB} \times \vec{AC}\|} = \frac{1}{\sqrt{26}} \hat{i} - \frac{4}{\sqrt{26}} \hat{j} + \frac{3}{\sqrt{26}} \hat{k} \]

### Step 5: Ensure the vector makes an obtuse angle with the x-axis
The unit vector \(\hat{n}\) makes an obtuse angle with the x-axis if the coefficient of \(\hat{i}\) is negative. The current unit vector is:
\[ \hat{n} = \frac{1}{\sqrt{26}} \hat{i} - \frac{4}{\sqrt{26}} \hat{j} + \frac{3}{\sqrt{26}} \hat{k} \]
Since the coefficient of \(\hat{i}\) is positive, we take the negative of the unit vector:
\[ \hat{n} = -\frac{1}{\sqrt{26}} \hat{i} + \frac{4}{\sqrt{26}} \hat{j} - \frac{3}{\sqrt{26}} \hat{k} \]

### Final Result
The required unit vector is:
\[ \hat{n} = -\frac{1}{\sqrt{26}} \hat{i} + \frac{4}{\sqrt{26}} \hat{j} - \frac{3}{\sqrt{26}} \hat{k} \]