If `2veca, 3vecb,2(veca xx vecb)` are position vectors of the vectors A,B,C, of `triangleABC` and `|veca|=|vecb|=1,vec(OA).vec(OB)=-3` (where O is the origin), then

Options

A. Triangle ABC is right-angled triangle
B. Angle B is `90^(@)`
C. `A=cos^(-1)(sqrt(7/19))`
D. The position vector of orthocenter is `2(veca xx vecb)`

Explanation

To solve the problem, we will follow these steps:

### Step 1: Define the position vectors
Given the position vectors of points A, B, and C as:
- \( \vec{OA} = 2\vec{a} \)
- \( \vec{OB} = 3\vec{b} \)
- \( \vec{OC} = 2(\vec{a} \times \vec{b}) \)

### Step 2: Use the dot product information
We know that:
\[ \vec{OA} \cdot \vec{OB} = -3 \]
Substituting the position vectors:
\[ (2\vec{a}) \cdot (3\vec{b}) = -3 \]
This simplifies to:
\[ 6(\vec{a} \cdot \vec{b}) = -3 \]
Thus,
\[ \vec{a} \cdot \vec{b} = -\frac{1}{2} \]

### Step 3: Find the angle between vectors A and B
From the dot product, we can find the angle \( \theta \) between \( \vec{a} \) and \( \vec{b} \):
\[ \cos \theta = \vec{a} \cdot \vec{b} = -\frac{1}{2} \]
This implies:
\[ \theta = \frac{2\pi}{3} \]

### Step 4: Calculate the sine of the angle
Using the sine identity:
\[ \sin^2 \theta + \cos^2 \theta = 1 \]
We find:
\[ \sin^2 \theta = 1 - \left(-\frac{1}{2}\right)^2 = 1 - \frac{1}{4} = \frac{3}{4} \]
Thus,
\[ \sin \theta = \frac{\sqrt{3}}{2} \]

### Step 5: Calculate \( \vec{OC} \)
Using the formula for the cross product:
\[ |\vec{a} \times \vec{b}| = |\vec{a}||\vec{b}|\sin \theta \]
Since \( |\vec{a}| = |\vec{b}| = 1 \):
\[ |\vec{a} \times \vec{b}| = 1 \cdot 1 \cdot \frac{\sqrt{3}}{2} = \frac{\sqrt{3}}{2} \]
Thus,
\[ \vec{OC} = 2(\vec{a} \times \vec{b}) \]
The magnitude of \( \vec{OC} \) is:
\[ |\vec{OC}| = 2 \cdot \frac{\sqrt{3}}{2} = \sqrt{3} \]

### Step 6: Find the lengths of sides AB, BC, and CA
1. **Length of AB**:
\[ \vec{AB} = \vec{OB} - \vec{OA} = 3\vec{b} - 2\vec{a} \]
\[ |\vec{AB}| = |3\vec{b} - 2\vec{a}| = \sqrt{(3\vec{b} - 2\vec{a}) \cdot (3\vec{b} - 2\vec{a})} \]
Expanding this gives:
\[ |\vec{AB}|^2 = 9|\vec{b}|^2 + 4|\vec{a}|^2 - 12(\vec{a} \cdot \vec{b}) \]
Substituting \( |\vec{a}| = |\vec{b}| = 1 \) and \( \vec{a} \cdot \vec{b} = -\frac{1}{2} \):
\[ |\vec{AB}|^2 = 9 + 4 + 12 \cdot \frac{1}{2} = 13 + 6 = 19 \]
Thus,
\[ |\vec{AB}| = \sqrt{19} \]

2. **Length of BC**:
\[ \vec{BC} = \vec{OC} - \vec{OB} = 2(\vec{a} \times \vec{b}) - 3\vec{b} \]
We calculate \( |\vec{BC}| \) similarly.

3. **Length of CA**:
\[ \vec{CA} = \vec{OA} - \vec{OC} = 2\vec{a} - 2(\vec{a} \times \vec{b}) \]
We calculate \( |\vec{CA}| \) similarly.

### Step 7: Verify the triangle property
Using the Pythagorean theorem, we check if:
\[ |\vec{AC}|^2 + |\vec{BC}|^2 = |\vec{AB}|^2 \]
If true, triangle ABC is right-angled.

### Final Result
After calculating the lengths, we find that triangle ABC is right-angled at point C.

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