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Question: If vectors \(\mathbf{a},b,\mathbf{c}\) satisfy the condition \(|\mathbf{a} - \mathbf{c}| = |\mathbf{...

If vectors a,b,c\mathbf{a},b,\mathbf{c} satisfy the condition ∣a−c∣=∣b−c∣|\mathbf{a} - \mathbf{c}| = |\mathbf{b} - \mathbf{c}|, then (b−a).(c−a+b2)(\mathbf{b} - \mathbf{a}).\left( \mathbf{c} - \frac{\mathbf{a} + \mathbf{b}}{\mathbf{2}} \right)is equal to

A

0

B

–1

C

1

D

2

Answer

0

Explanation

Solution

(b−a).(c−a+b2)=b.c−b.(a+b2)−a.c+a2(a+b)(\mathbf{b} - \mathbf{a}).\left( \mathbf{c} - \frac{\mathbf{a} + \mathbf{b}}{2} \right) = \mathbf{b}.\mathbf{c} - \mathbf{b}.\left( \frac{\mathbf{a} + \mathbf{b}}{2} \right) - \mathbf{a}.\mathbf{c} + \frac{\mathbf{a}}{2}(\mathbf{a} + \mathbf{b})

and ∣a−c∣=∣b−c∣|\mathbf{a} - \mathbf{c}| = |\mathbf{b} - \mathbf{c}| ⇒\Rightarrow ∣a−c∣2=∣b−c∣2|\mathbf{a} - \mathbf{c}|^{2} = |\mathbf{b} - \mathbf{c}|^{2}

∴ a+b=2c\mathbf{a} + \mathbf{b} = 2\mathbf{c}

Therefore, (b−a).(c−a+b2)=0.(\mathbf{b} - \mathbf{a}).\left( \mathbf{c} - \frac{\mathbf{a} + \mathbf{b}}{2} \right) = 0.