Verified Solution • 2025–26 Session moderate

QJEEM12Maa6ac: JEE Main 2012 Physics - Class 11 JEEM12Maa6ac Solution: Let be a parallelogram such that and be an acute angle. If is…

Let ABCDABCD be a parallelogram such that AB→=q→,AD→=p→\overrightarrow {AB} = \overrightarrow q ,\overrightarrow {AD} = \overrightarrow p and ∠BAD\angle BAD be an acute angle. If r→\overrightarrow r is the vector that coincide with the altitude directed from the vertex BB to the side AD,AD, then r→\overrightarrow r is given by :

Multiple Choice Options

  • A r→=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 p
  • B r→=−q→+(p→.q→)(p→.p→)p→\overrightarrow r = - \overrightarrow q + {{\left( {\overrightarrow p .\overrightarrow q } \right)} \over {\left( {\overrightarrow p .\overrightarrow p } \right)}}\overrightarrow p
  • C r⃗=q⃗−(p⃗.q⃗)(p⃗.p⃗)p⃗\vec r = \vec q - {{\left( {\vec p.\vec q} \right)} \over {\left( {\vec p.\vec p} \right)}}\vec p
  • D r→=−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)}}

Step-by-Step Verified Solution

Let ABCDABCD be a parallelogram such that

AB→=q→,AD→=p→\overrightarrow {AB} = \overrightarrow q ,\overrightarrow {AD} = \overrightarrow p and ∠BAD\angle BAD be an acute angle.

We have

AIEEE 2012 Mathematics - Vector Algebra Question 210 English Explanation

AX→=(p→.q→∣p→∣)(p→∣p→∣)=p→.q→∣p→∣2p→\overrightarrow {AX} = \left( {{{\overrightarrow p .\overrightarrow q } \over {\left| {\overrightarrow p } \right|}}} \right)\left( {{{\overrightarrow p } \over {\left| {\overrightarrow p } \right|}}} \right) = {{\overrightarrow p .\overrightarrow q } \over {{{\left| {\overrightarrow p } \right|}^2}}}\overrightarrow p

Let r→=BX→=BA→+AX→\overrightarrow r = \overrightarrow {BX} = \overrightarrow {BA} + \overrightarrow {AX}

                 =−q→+p→.q→∣p→∣2p→\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = - \overrightarrow q + {{\overrightarrow p .\overrightarrow q } \over {{{\left| {\overrightarrow p } \right|}^2}}}\overrightarrow p

Related Learning Resources & Exam Preparation