MathematicsMedium202×since 2002Q5764Let ABCDABCDABCD be a parallelogram such that AB→=q→,AD→=p→\overrightarrow {AB} = \overrightarrow q ,\overrightarrow {AD} = \overrightarrow pAB=q,AD=p and ∠BAD\angle BAD∠BAD be an acute angle. If r→\overrightarrow rr is the vector that coincide with the altitude directed from the vertex BBB to the side AD,AD,AD, then r→\overrightarrow rr is given by :Ar→=3q→−3(p→.q→)(p→.p→)p→\overrightarrow r = 3\overrightarrow q - {{3\left( {\overrightarrow p .\overrightarrow q } \right)} \over {\left( {\overrightarrow p .\overrightarrow p } \right)}}\overrightarrow pr=3q−(p.p)3(p.q)pBr→=−q→+(p→.q→)(p→.p→)p→\overrightarrow r = - \overrightarrow q + {{\left( {\overrightarrow p .\overrightarrow q } \right)} \over {\left( {\overrightarrow p .\overrightarrow p } \right)}}\overrightarrow pr=−q+(p.p)(p.q)pCr⃗=q⃗−(p⃗.q⃗)(p⃗.p⃗)p⃗\vec r = \vec q - {{\left( {\vec p.\vec q} \right)} \over {\left( {\vec p.\vec p} \right)}}\vec pr=q−(p.p)(p.q)pDr→=−3q→−3(p→.q→)(p→.p→)\overrightarrow r = - 3\overrightarrow q - {{3\left( {\overrightarrow p .\overrightarrow q } \right)} \over {\left( {\overrightarrow p .\overrightarrow p } \right)}}r=−3q−(p.p)3(p.q)Check answerSkip