A - - b - and c - are the position vectors of points A- B and C respectively- prove that- a - b - b - x C - c - a - is a vector perpendicular to theplane of triangle A B C-

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Byju's Answer

Standard XII

Mathematics

In Radius

a , → b → and...

Question

$\vec{a,} \vec{b} and \vec{c}$ are the position vectors of points A, B and C respectively, prove that: $\vec{a} \times \vec{b} + \vec{b} \times \vec{c} + \vec{c} \times \vec{a}$ is a vector perpendicular to the plane of triangle ABC.

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Solution

$We know that if any vector is perpendicular to all three sides of ∆ A B C, it must be perpendicular to the plane of ∆ A B C . Now, \vec{AB} = \vec{b} - \vec{a}, \vec{BC} = \vec{c} - \vec{b}, \vec{CA} = \vec{a} - \vec{c} (Position vectors of A, B and C are \vec{a}, \vec{b}, \vec{c}) We have \vec{AB} . (\vec{a} \times \vec{b} + \vec{b} \times \vec{c} + \vec{c} \times \vec{a}) = (\vec{b} - \vec{a}) . (\vec{a} \times \vec{b} + \vec{b} \times \vec{c} + \vec{c} \times \vec{a}) = \vec{b} . (\vec{a} \times \vec{b}) + \vec{b} . (\vec{b} \times \vec{c}) + \vec{b} . (\vec{c} \times \vec{a}) - \vec{a} . (\vec{a} \times \vec{b}) - \vec{a} . (\vec{b} \times \vec{c}) - \vec{a} . (\vec{c} \times \vec{a}) (By distributive law) = [\vec{b} \vec{a} \vec{b}] + [\vec{b} \vec{b} \vec{c}] + [\vec{b} \vec{c} \vec{a}] - [\vec{a} \vec{a} \vec{b}] - [\vec{a} \vec{b} \vec{c}] - [\vec{a} \vec{c} \vec{a}] = 0 + 0 + [\vec{b} \vec{c} \vec{a}] - 0 - [\vec{a} \vec{b} \vec{c}] - 0 = 0 (∵ [\vec{b} \vec{c} \vec{a}] = [\vec{a} \vec{b} \vec{c}]) \vec{BC} . (\vec{a} \times \vec{b} + \vec{b} \times \vec{c} + \vec{c} \times \vec{a}) = (\vec{c} - \vec{b}) . (\vec{a} \times \vec{b} + \vec{b} \times \vec{c} + \vec{c} \times \vec{a}) = \vec{c} . (\vec{a} \times \vec{b}) + \vec{c} . (\vec{b} \times \vec{c}) + \vec{c} . (\vec{c} \times \vec{a}) - \vec{b} . (\vec{a} \times \vec{b}) - \vec{b} . (\vec{b} \times \vec{c}) - \vec{b} . (\vec{c} \times \vec{a}) (By distributive law) = [\vec{c} \vec{a} \vec{b}] + [\vec{c} \vec{b} \vec{c}] + [\vec{c} \vec{c} \vec{a}] - [\vec{b} \vec{a} \vec{b}] - [\vec{b} \vec{b} \vec{c}] - [\vec{b} \vec{c} \vec{a}] = [\vec{c} \vec{a} \vec{b}] + 0 + 0 - 0 - 0 - [\vec{b} \vec{c} \vec{a}] = 0 (∵ [\vec{c} \vec{a} \vec{b}] = [\vec{b} \vec{c} \vec{a}]) Similarly, \vec{CA} . (\vec{a} \times \vec{b} + \vec{b} \times \vec{c} + \vec{c} \times \vec{a}) = (\vec{a} - \vec{c}) . (\vec{a} \times \vec{b} + \vec{b} \times \vec{c} + \vec{c} \times \vec{a}) = \vec{a} . (\vec{a} \times \vec{b}) + \vec{a} . (\vec{b} \times \vec{c}) + \vec{a} (\vec{c} \times \vec{a}) - \vec{c} . (\vec{a} \times \vec{b}) - \vec{c} . (\vec{b} \times \vec{c}) - \vec{c} . (\vec{c} \times \vec{a}) (By distributive law) = [\vec{a} \vec{a} \vec{b}] + [\vec{a} \vec{b} \vec{c}] + [\vec{a} \vec{c} \vec{a}] - [\vec{c} \vec{a} \vec{b}] - [\vec{c} \vec{b} \vec{c}] - [\vec{c} \vec{c} \vec{a}] = 0 + [\vec{a} \vec{b} \vec{c}] + 0 - [\vec{c} \vec{a} \vec{b}] - 0 - 0 = 0 (∵ [\vec{c} \vec{a} \vec{b}] = [\vec{a} \vec{b} \vec{c}]) Hence, vector \vec{a} \times \vec{b} + \vec{b} \times \vec{c} + \vec{c} \times \vec{a} is perpendicular to all sides of ∆ ABC and also perpendicular to the plane of ∆ ABC .$

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a,→ b →and c →are the position vectors of points A, B and C respectively, prove that: a →×b →+b →×c →+c →×a →is a vector perpendicular to the plane of triangle ABC.

$\vec{a,} \vec{b} and \vec{c}$ are the position vectors of points A, B and C respectively, prove that: $\vec{a} \times \vec{b} + \vec{b} \times \vec{c} + \vec{c} \times \vec{a}$ is a vector perpendicular to the plane of triangle ABC.