Question 128: Motion in 2D & Vectors

Question

The resultant $\mathbf{R}$ of $\mathbf{P}$ and $\mathbf{Q}$ is perpendicular to $\mathbf{P}$. Also $|\mathbf{P}|=|\mathbf{R}|$. The angle between $\mathbf{P}$ and $\mathbf{Q}$ is $\left[	an 45^{\circ}=1ight]$

Accepted Answer

Given that $\mathbf{R}$ is resultant of $\mathbf{P}$ and $\mathbf{Q}$ as shown in the figure below,
$\mathrm{BC}=\mathrm{P}, \mathrm{CA}=Q$ and $\mathrm{BA}=R$
Given, $B A$ and $B C$ are perpendicular and equal in magnitude.
So, from property of triangle,
$$\angle A C B=45^{\circ}$$
Now, BC has to be extended up to $D$ so, that $C D=P$
Now, CD and CA have the initial point $C$, so the angle between CD and CA;
$$=180^{\circ}-45^{\circ}=135^{\circ}=\frac{3 \pi}{4}$$
So, angle between $P$ and $Q$ is $3 \frac{\pi}{4}$.

Answer

$\frac{5 \pi}{4}$

Answer

$\frac{7 \pi}{4}$

Answer

$\frac{\pi}{4}$

Answer

$\frac{3 \pi}{4}$

Motion in 2D & Vectors

The resultant \mathbf{R} of \mathbf{P} and \mathbf{Q} is perpendicular to \mathbf{P}. Also |\mathbf{P}|=|\mathbf{R}|. The angle between \mathbf{P} and \mathbf{Q} is \left[\tan 45^{\circ}=1\right]

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