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In a right angled triangle ABC, the hypo...

In a right angled triangle ABC, the hypotenuse AB =p, then `vec(AB).vec(AC) + vec(BC).vec(BA)+vec(CA).vec(CB)` is equal to:

A

`2p^(2)`

B

`p^(2)/2`

C

`p^(2)`

D

None of these

Text Solution

AI Generated Solution

The correct Answer is:
To solve the problem, we need to analyze the expression \(\vec{AB} \cdot \vec{AC} + \vec{BC} \cdot \vec{BA} + \vec{CA} \cdot \vec{CB}\) in the context of triangle ABC, where AB is the hypotenuse and equals \(p\). ### Step 1: Identify the vectors In triangle ABC: - Let \(\vec{AB} = \vec{B} - \vec{A}\) - Let \(\vec{AC} = \vec{C} - \vec{A}\) - Let \(\vec{BC} = \vec{C} - \vec{B}\) ### Step 2: Calculate each dot product 1. **Calculate \(\vec{AB} \cdot \vec{AC}\)**: \[ \vec{AB} \cdot \vec{AC} = |\vec{AB}| |\vec{AC}| \cos(\theta) \] Here, \(\theta\) is the angle between \(\vec{AB}\) and \(\vec{AC}\). Since triangle ABC is a right triangle, \(\theta\) is the angle at A, which is \(90^\circ\). Thus, \(\cos(90^\circ) = 0\). \[ \vec{AB} \cdot \vec{AC} = 0 \] 2. **Calculate \(\vec{BC} \cdot \vec{BA}\)**: \[ \vec{BC} \cdot \vec{BA} = |\vec{BC}| |\vec{BA}| \cos(\phi) \] Here, \(\phi\) is the angle between \(\vec{BC}\) and \(\vec{BA}\). Since \(\vec{BA} = -\vec{AB}\), we have: \[ \vec{BC} \cdot \vec{BA} = -\vec{BC} \cdot \vec{AB} \] Again, since \(\vec{AB}\) and \(\vec{BC}\) are perpendicular, this dot product is also \(0\). 3. **Calculate \(\vec{CA} \cdot \vec{CB}\)**: \[ \vec{CA} \cdot \vec{CB} = |\vec{CA}| |\vec{CB}| \cos(\psi) \] Here, \(\psi\) is the angle between \(\vec{CA}\) and \(\vec{CB}\). Since \(\vec{CA} = -\vec{AC}\) and \(\vec{CB} = -\vec{BC}\), we have: \[ \vec{CA} \cdot \vec{CB} = -\vec{AC} \cdot (-\vec{BC}) = \vec{AC} \cdot \vec{BC} \] Again, since \(\vec{AC}\) and \(\vec{BC}\) are perpendicular, this dot product is also \(0\). ### Step 3: Combine the results Now, we can combine all the results: \[ \vec{AB} \cdot \vec{AC} + \vec{BC} \cdot \vec{BA} + \vec{CA} \cdot \vec{CB} = 0 + 0 + 0 = 0 \] ### Final Result Thus, the final result is: \[ \vec{AB} \cdot \vec{AC} + \vec{BC} \cdot \vec{BA} + \vec{CA} \cdot \vec{CB} = 0 \] ---
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