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If A (\vec{a}), B(\vec{b}) , C(\vec{c}) are position vectors of vertices A,B,C of triangle ABC such that length of perpendicular dropped from A to BC  is k\frac{\left |\vec{a}*\vec{b}+\vec{b}*\vec{c}+\vec{c}*\vec{a}\right |}{\left | \vec{b}-\vec{c} \right |}then K equals

Option: 1

\frac{1}{2}

 


Option: 2

1


Option: 3

\frac{3}{2}


Option: 4

2


Answers (1)

best_answer

As we learn @5544

Vector Area -

area of \bigtriangleup ABC

=\frac{1}{2}\left [ \vec{a} \times \vec{b}+\vec{b} \times \vec{c}+\vec{c} \times \vec{a}\right ]

- wherein

\vec{a}, \vec{b} and \vec{c} are three vectors.

 

 Area =\frac{1}{2}\left |\vec{a}*\vec{b}+\vec{b}*\vec{c}+\vec{c}*\vec{a}\right |=\frac{1}{2}*\left | \vec{b}-\vec{c} \right |*k\frac{\left |\vec{a}*\vec{b}+\vec{b}*\vec{c}+\vec{c}*\vec{a}\right |}{\left | \vec{b}-\vec{c} \right |}

(area=\frac{1}{2}*base*length)

K = 1

 

Posted by

Rishi

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