Angle between the vectors `vec(a)=-hat(i)+2hat(j)+hat(k)` and `vec(b)=xhat(i)+hat(j)+(x+1)hat(k)`

Options

A. is obtuse angle
B. is acute angle
C. is `90^(@)`
D. depends on x

Explanation

To find the angle between the vectors \(\vec{A} = -\hat{i} + 2\hat{j} + \hat{k}\) and \(\vec{B} = x\hat{i} + \hat{j} + (x + 1)\hat{k}\), we will follow these steps:

### Step 1: Calculate the dot product \(\vec{A} \cdot \vec{B}\)
The dot product of two vectors \(\vec{A}\) and \(\vec{B}\) is given by:
\[ \vec{A} \cdot \vec{B} = A_x B_x + A_y B_y + A_z B_z \]
For our vectors:
- \(A_x = -1\), \(A_y = 2\), \(A_z = 1\)
- \(B_x = x\), \(B_y = 1\), \(B_z = x + 1\)

Substituting these values:
\[ \vec{A} \cdot \vec{B} = (-1)(x) + (2)(1) + (1)(x + 1) \]
\[ = -x + 2 + x + 1 \]
\[ = 3 \]

### Step 2: Calculate the magnitudes of \(\vec{A}\) and \(\vec{B}\)
The magnitude of a vector \(\vec{V} = V_x \hat{i} + V_y \hat{j} + V_z \hat{k}\) is given by:
\[ |\vec{V}| = \sqrt{V_x^2 + V_y^2 + V_z^2} \]

**Magnitude of \(\vec{A}\)**:
\[ |\vec{A}| = \sqrt{(-1)^2 + (2)^2 + (1)^2} = \sqrt{1 + 4 + 1} = \sqrt{6} \]

**Magnitude of \(\vec{B}\)**:
\[ |\vec{B}| = \sqrt{x^2 + (1)^2 + (x + 1)^2} \]
Calculating \((x + 1)^2\):
\[ (x + 1)^2 = x^2 + 2x + 1 \]
Thus,
\[ |\vec{B}| = \sqrt{x^2 + 1 + x^2 + 2x + 1} = \sqrt{2x^2 + 2x + 2} = \sqrt{2(x^2 + x + 1)} \]

### Step 3: Use the dot product to find \(\cos \theta\)
The formula relating the dot product and the angle between two vectors is:
\[ \vec{A} \cdot \vec{B} = |\vec{A}| |\vec{B}| \cos \theta \]
Substituting the values we found:
\[ 3 = \sqrt{6} \cdot \sqrt{2(x^2 + x + 1)} \cdot \cos \theta \]
Thus,
\[ \cos \theta = \frac{3}{\sqrt{6} \cdot \sqrt{2(x^2 + x + 1)}} \]

### Step 4: Analyze the value of \(\cos \theta\)
Since the numerator \(3\) is a positive constant, and the denominator \(\sqrt{6} \cdot \sqrt{2(x^2 + x + 1)}\) is always positive for any real value of \(x\) (as it involves squares), we conclude that:
\[ \cos \theta > 0 \]
This implies that the angle \(\theta\) is always less than \(90^\circ\).

### Conclusion
The angle between the vectors \(\vec{A}\) and \(\vec{B}\) is always less than \(90^\circ\) for any real value of \(x\).

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