Finding Components of a vector Along and Perpendicular to another Vector is considered one the most difficult concept.
8 Questions around this concept.

Solve by difficulty

EASY

Let ABCD be a parallelogram such that $\overrightarrow{A B}=\vec{q}, \overrightarrow{A D}=\vec{p}$ and $\angle B A D$ 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 $\vec{r}$ is given by

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

$\vec{r}=-\vec{q}+\left(\frac{\vec{p} \cdot \vec{q}}{\vec{p} \cdot \vec{p}}\right) \vec{p}$

$\vec{r}=\vec{q}-\left(\frac{\vec{p} \cdot \vec{q}}{\vec{p} \cdot \vec{p}}\right) \vec{p}$

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

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Projection vector of $\vec{a}$ on $\vec{b}$ is

$\left(\frac{\vec{a} \cdot \vec{b}}{|\vec{b}|^{2}}\right) \vec{b}$

$\frac{\overrightarrow{\mathrm{a}} \cdot \overrightarrow{\mathrm{b}}}{|\overrightarrow{\mathrm{b}}|}$

$\frac{\overrightarrow{\mathrm{a}} \cdot \overrightarrow{\mathrm{b}}}{|\overrightarrow{\mathrm{a}}|}$

$\left(\frac{\vec{a} \cdot \vec{b}}{|\vec{a}|^{2}}\right) \hat{b}$

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Concepts Covered - 1

Finding Components of a vector Along and Perpendicular to another Vector

Let $\vec{a}$ and $\vec{b}$ be two vectors represented by $\overrightarrow{O A}$ and $\overrightarrow{O B}$ respectively and let $\Theta$ be the angle between $\vec{a}$ and $\vec{b}$. Then,
$
\vec{b}=\overrightarrow{O M}+\overrightarrow{M B}
$

$
\vec{b}=\overrightarrow{O M}+\overrightarrow{M B}
$

Also,
$
\begin{aligned}
\overrightarrow{O M} & =(O M) \hat{a} \\
& =(O B \cos \theta) \hat{a}==(|\vec{b}| \cos \theta) \hat{a} \\
& =\left(\frac{(\vec{a} \cdot \vec{b})}{|\vec{a}|}\right) \hat{a}=\left(\frac{(\vec{a} \cdot \vec{b})}{|\vec{a}||\vec{a}|}\right) \vec{a} \quad\left[\because \hat{a}=\frac{\vec{a}}{|\vec{a}|}\right] \\
& =\left(\frac{(\vec{a} \cdot \vec{b})}{|\vec{a}|^2}\right) \vec{a}
\end{aligned}
$

As,
$
\begin{aligned}
\vec{b} & =\overrightarrow{O M}+\overrightarrow{M B} \\
\overrightarrow{M B} & =\vec{b}-\overrightarrow{O M}=\vec{b}-\left(\frac{(\vec{a} \cdot \vec{b})}{|\vec{a}|^2}\right) \vec{a}
\end{aligned}
$

Thus, the components of $\vec{b}$ along and perpendicular to $\vec{a}$ are $\left(\frac{(\vec{a} \cdot \vec{b})}{|\vec{a}|^2}\right) \vec{a}$ and $\vec{b}-\left(\frac{(\vec{a} \cdot \vec{b})}{|\vec{a}|^2}\right) \vec{a}$, respectively.