The angle between two vectors `-2hat(i)+3hat(j)+hat(k)` and `2hat(i)+2hat(j)-4hat(k)` is

Options

A. obtuse
B. right angle
C. acute
D. can't say

Explanation

To find the angle between the two vectors \(\vec{A} = -2\hat{i} + 3\hat{j} + \hat{k}\) and \(\vec{B} = 2\hat{i} + 2\hat{j} - 4\hat{k}\), we can use the dot product formula.

### Step 1: Calculate the Dot Product
The dot product of two vectors \(\vec{A}\) and \(\vec{B}\) is given by:

\[ \vec{A} \cdot \vec{B} = A_x B_x + A_y B_y + A_z B_z \]

Substituting the components of \(\vec{A}\) and \(\vec{B}\):

\[ \vec{A} \cdot \vec{B} = (-2)(2) + (3)(2) + (1)(-4) \]

Calculating each term:

\[ = -4 + 6 - 4 \]

Now, summing these values:

\[ = -4 + 6 - 4 = -2 \]

### Step 2: Calculate the Magnitudes of the Vectors
Next, we need to find the magnitudes of both vectors.

The magnitude of vector \(\vec{A}\) is:

\[ |\vec{A}| = \sqrt{(-2)^2 + 3^2 + 1^2} = \sqrt{4 + 9 + 1} = \sqrt{14} \]

The magnitude of vector \(\vec{B}\) is:

\[ |\vec{B}| = \sqrt{(2)^2 + (2)^2 + (-4)^2} = \sqrt{4 + 4 + 16} = \sqrt{24} = 2\sqrt{6} \]

### Step 3: Use the Dot Product to Find the Angle
The angle \(\theta\) between the two vectors can be found using the formula:

\[ \vec{A} \cdot \vec{B} = |\vec{A}| |\vec{B}| \cos(\theta) \]

Substituting the known values:

\[ -2 = \sqrt{14} \cdot 2\sqrt{6} \cos(\theta) \]

Calculating the right side:

\[ \sqrt{14} \cdot 2\sqrt{6} = 2\sqrt{84} = 2 \cdot 2\sqrt{21} = 4\sqrt{21} \]

So we have:

\[ -2 = 4\sqrt{21} \cos(\theta) \]

### Step 4: Solve for \(\cos(\theta)\)

\[ \cos(\theta) = \frac{-2}{4\sqrt{21}} = \frac{-1}{2\sqrt{21}} \]

### Step 5: Determine the Nature of the Angle
Since \(\cos(\theta)\) is negative, this indicates that the angle \(\theta\) is obtuse (greater than 90 degrees and less than 180 degrees).

### Final Answer
The angle between the two vectors is obtuse.

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