Question
Let $$ABCD$$ be a parallelogram such that $$\overrightarrow {AB} = \vec q,\,\overrightarrow {AD} = \vec p$$ and $$\angle BAD$$ bean acute angle. If $${\vec r}$$ is the vector that coincide with the altitude directed from the vertex $$B$$ to the side $$AD,$$ then $${\vec r}$$ is given by :
A.
$$\vec r = 3\vec q - \frac{{\left( {\vec p.\vec q} \right)}}{{\left( {\vec p.\vec p} \right)}}\vec p$$
B.
$$\vec r = - \vec q + \frac{{\left( {\vec p.\vec q} \right)}}{{\left( {\vec p.\vec p} \right)}}\vec p$$
C.
$$\vec r = \vec q - \frac{{\left( {\vec p.\vec q} \right)}}{{\left( {\vec p.\vec p} \right)}}\vec p$$
D.
$$\vec r = - 3\vec q - \frac{{3\left( {\vec p.\vec q} \right)}}{{\left( {\vec p.\vec p} \right)}}\vec p$$
Answer :
$$\vec r = - \vec q + \frac{{\left( {\vec p.\vec q} \right)}}{{\left( {\vec p.\vec p} \right)}}\vec p$$
Solution :
Let $$ABCD$$ be a parallelogram such that $$\overrightarrow {AB} = \vec q,\,\overrightarrow {AD} = \vec p$$ and $$\angle BAD$$ be an acute angle.
We have
$$\eqalign{ & \overrightarrow {AX} = \left( {\frac{{\vec p.\vec q}}{{\left| {\vec p} \right|}}} \right)\left( {\frac{{\vec p}}{{\left| {\vec p} \right|}}} \right) = \frac{{\vec p.\vec q}}{{{{\left| {\vec p} \right|}^2}}}\vec p \cr & {\text{Let }}\vec r = \overrightarrow {BX} = \overrightarrow {BA} + \overrightarrow {AX} = - \vec q + \frac{{\vec p.\vec q}}{{{{\left| {\vec p} \right|}^2}}}\vec p \cr} $$
Let $$ABCD$$ be a parallelogram such that $$\overrightarrow {AB} = \vec q,\,\overrightarrow {AD} = \vec p$$ and $$\angle BAD$$ be an acute angle.
We have
$$\eqalign{ & \overrightarrow {AX} = \left( {\frac{{\vec p.\vec q}}{{\left| {\vec p} \right|}}} \right)\left( {\frac{{\vec p}}{{\left| {\vec p} \right|}}} \right) = \frac{{\vec p.\vec q}}{{{{\left| {\vec p} \right|}^2}}}\vec p \cr & {\text{Let }}\vec r = \overrightarrow {BX} = \overrightarrow {BA} + \overrightarrow {AX} = - \vec q + \frac{{\vec p.\vec q}}{{{{\left| {\vec p} \right|}^2}}}\vec p \cr} $$