Question 126: Motion in 2D & Vectors

Question

$\mathbf{P}$ and $\mathbf{Q}$ are two non-zero vectors inclined to each other at an angle ' $	heta$ '. ' $p$ ' and ' $q$ ' are unit vectors along $\mathbf{P}$ and $\mathbf{Q}$ respectively. The component of $\mathbf{Q}$ in the direction of $\mathbf{Q}$ will be

Accepted Answer

Let two vectors P and Q are represented by graph as below
Here, $\mathbf{Q}_p$ is a vector in the direction of $\mathbf{P}$.
Then, from the right angle triangle, we get
$$\begin{aligned}
\cos 	heta & =\frac{Q_p}{Q} \quad	ext{.... (i)}\
\Rightarrow \quad Q_p & =Q \cos 	heta \
	ext { Also, } \cos 	heta & =\frac{\mathbf{P} \cdot \mathbf{Q}}{P Q} \Rightarrow \frac{Q_p}{Q}=\frac{\mathbf{P} \cdot \mathbf{Q}}{P Q} \
\Rightarrow \quad Q_p & =\frac{\mathbf{P} \cdot \mathbf{Q}}{P}\quad	ext{.... (ii)}
\end{aligned}$$
As given that, $\hat{P}$ is the unit vector along $\mathbf{P}$, then
$$\hat{\mathrm{P}}=\frac{\mathrm{P}}{P}\quad	ext{.... (iii)}$$
Putting the value of $P$ from Eq. (iii) to Eq. (ii), we get
$$Q_p=\hat{\mathbf{P}} \cdot \mathbf{Q}$$

Answer

$P \cdot Q$

Answer

$\frac{P 	imes Q}{P}$

Answer

$\frac{P \cdot Q}{Q}$

Answer

$p \cdot q$

Motion in 2D & Vectors

\mathbf{P} and \mathbf{Q} are two non-zero vectors inclined to each other at an angle ' \theta '. ' p ' and ' q ' are unit vectors along \mathbf{P} and \mathbf{Q} respectively. The component of \mathbf{Q} in the direction of \mathbf{Q} will be

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