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Question

Let ABCD be a parallelogram such that AB=q , AD=p and $$\angle BAD$$ be an acute angle. If r is the vector that considies with the altitude directed from the vertex B to the side AD, then r is given by

A
$$r=3q\frac { 3(p.q) }{ (p.p) } p$$
B
$$r=-q+(\frac { p.q }{ p.p } )p$$
C
$$r=q-(\frac { p.q }{ p.p } )p$$
D
$$r=-3q+\frac { 3(p.q) }{ (p.p) } p$$
Solution
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Correct option is B. $$r=-q+(\frac { p.q }{ p.p } )p$$

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