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

r → = 3 q → − 3 p → . q → p → . p → p →

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

r → = 3 q → − 3 p → . q → p → . p → p →

r → = − 3 q → + 3 p → . q → p → . p →

r → = 3 q → − 3 p → . q → p → . p → p →

r → = − q → + p → . q → p → . p → p →

r → = q → − p → . q → p → . p → p →

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Let ABCD be a parallelogram such that $\vec{A B} = \vec{q}, \vec{A D} = \vec{p} a n d ∠ B A D$ be an acute angle. If $\vec{r}$ is the vector that coincides with the altitude directed from the vertex B to the side AD, then is given by

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$\vec{r} = - 3 \vec{q} + \frac{3 (\vec{p} . \vec{q})}{(\vec{p} . \vec{p})}$

$\vec{r} = 3 \vec{q} - \frac{3 (\vec{p} . \vec{q})}{(\vec{p} . \vec{p})} \vec{p}$

$\vec{r} = - \vec{q} + \frac{(\vec{p} . \vec{q})}{(\vec{p} . \vec{p})} \vec{p}$

$\vec{r} = \vec{q} - \frac{(\vec{p} . \vec{q})}{(\vec{p} . \vec{p})} \vec{p}$

answer is B.

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Detailed Solution

$G i v e n A B C D b e a p a r a l l e \log r a m a n d A B = q a n d A D = p l e t B E ⊥ A D a n d B E = r N o w A E = v e c t o r c o m p o n e n t o f r o n p = \frac{(p . q) p}{p . p} f r o m ∆ A B E A B + B E = A E \Rightarrow \bar{q} + \bar{r} = \frac{(p . q) p}{p . p} \Rightarrow r = - q + \frac{(p . q) p}{p . p}$ Question Image