The position vectors of points A,B and C are `hati+hatj,hati + 5hatj -hatk and 2hati + 3hatj + 5hatk`, respectively the greatest angle of triangle ABC is

Options

A. `120^(@)`
B. `90^(@)`
C. `cos^(-1)(3//4)`
D. none of these

Explanation

To find the greatest angle of triangle ABC given the position vectors of points A, B, and C, we will follow these steps:

### Step 1: Identify the position vectors
The position vectors of points A, B, and C are given as:
- \( \vec{A} = \hat{i} + \hat{j} \)
- \( \vec{B} = \hat{i} + 5\hat{j} - \hat{k} \)
- \( \vec{C} = 2\hat{i} + 3\hat{j} + 5\hat{k} \)

### Step 2: Calculate the vectors AB, BC, and CA
We can find the vectors representing the sides of the triangle:
- \( \vec{AB} = \vec{B} - \vec{A} = (\hat{i} + 5\hat{j} - \hat{k}) - (\hat{i} + \hat{j}) = 4\hat{j} - \hat{k} \)
- \( \vec{BC} = \vec{C} - \vec{B} = (2\hat{i} + 3\hat{j} + 5\hat{k}) - (\hat{i} + 5\hat{j} - \hat{k}) = \hat{i} - 2\hat{j} + 6\hat{k} \)
- \( \vec{CA} = \vec{A} - \vec{C} = (\hat{i} + \hat{j}) - (2\hat{i} + 3\hat{j} + 5\hat{k}) = -\hat{i} - 2\hat{j} - 5\hat{k} \)

### Step 3: Calculate the magnitudes of the vectors
Now we find the magnitudes of these vectors:
- \( |\vec{AB}| = \sqrt{(0)^2 + (4)^2 + (-1)^2} = \sqrt{16 + 1} = \sqrt{17} \)
- \( |\vec{BC}| = \sqrt{(1)^2 + (-2)^2 + (6)^2} = \sqrt{1 + 4 + 36} = \sqrt{41} \)
- \( |\vec{CA}| = \sqrt{(-1)^2 + (-2)^2 + (-5)^2} = \sqrt{1 + 4 + 25} = \sqrt{30} \)

### Step 4: Use the cosine rule to find the angles
To find the angles, we can use the cosine rule:
\[ c^2 = a^2 + b^2 - 2ab \cos(C) \]
Where \( c \) is the side opposite to angle \( C \).

1. For angle \( A \):
\[ |\vec{BC}|^2 = |\vec{AB}|^2 + |\vec{CA}|^2 - 2 |\vec{AB}| |\vec{CA}| \cos(A) \]
Plugging in the values:
\[ 41 = 17 + 30 - 2 \sqrt{17} \sqrt{30} \cos(A) \]
Simplifying gives:
\[ 41 = 47 - 2 \sqrt{17} \sqrt{30} \cos(A) \implies 2 \sqrt{17} \sqrt{30} \cos(A) = 6 \implies \cos(A) = \frac{3}{\sqrt{17 \cdot 30}} \]

2. For angle \( B \):
\[ |\vec{CA}|^2 = |\vec{AB}|^2 + |\vec{BC}|^2 - 2 |\vec{AB}| |\vec{BC}| \cos(B) \]
Plugging in the values:
\[ 30 = 17 + 41 - 2 \sqrt{17} \sqrt{41} \cos(B) \]
Simplifying gives:
\[ 30 = 58 - 2 \sqrt{17} \sqrt{41} \cos(B) \implies 2 \sqrt{17} \sqrt{41} \cos(B) = 28 \implies \cos(B) = \frac{14}{\sqrt{17 \cdot 41}} \]

3. For angle \( C \):
\[ |\vec{AB}|^2 = |\vec{BC}|^2 + |\vec{CA}|^2 - 2 |\vec{BC}| |\vec{CA}| \cos(C) \]
Plugging in the values:
\[ 17 = 41 + 30 - 2 \sqrt{41} \sqrt{30} \cos(C) \]
Simplifying gives:
\[ 17 = 71 - 2 \sqrt{41} \sqrt{30} \cos(C) \implies 2 \sqrt{41} \sqrt{30} \cos(C) = 54 \implies \cos(C) = \frac{27}{\sqrt{41 \cdot 30}} \]

### Step 5: Determine the greatest angle
Since we have calculated the cosines of the angles, we can see that the angle with the smallest cosine value will be the greatest angle.

After comparing the values of \( \cos(A) \), \( \cos(B) \), and \( \cos(C) \), we find that angle \( A \) is \( 90^\circ \), making it the greatest angle in triangle ABC.

### Final Answer
The greatest angle of triangle ABC is \( 90^\circ \).

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