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Mathematical Tools and Vectors

3 Board Physics previous year questions on Mathematical Tools and Vectors — options free on every question; 1 include the answer & explanation free, the rest unlock with PYQ Pass.

Q1 FREE PREVIEW

PYQ

The angle between vector $\vec{Q}$ and the resultant of $(2\vec{Q}+2\vec{P})$ and $(2\vec{Q}-2\vec{P})$ is

$0^\circ$

${\tan }^{-1}\frac{P}{Q}$

${\tan }^{-1}\left(\frac{2\vec{Q}-2\vec{P}}{2\vec{Q}+2\vec{P}}\right)$

${\tan }^{-1}(2Q/P)$

✓ Correct answer: a)

$0^\circ$

Explanation

Let the resultant of both the vectors be $\vec{R}$, then

$$\begin{aligned}&\vec{R} & =(2 \vec{Q}+2 \vec{P})+(2 \vec{Q}-2 \vec{P}) \\\vec{R} & =4 \vec{Q}$$
&\text { Angle between } \vec{Q} \text { and } \vec{R} \text { is zero }
\end{aligned}

PYQ

What will be the projection of vector $\vec{A}=\hat{i}+\hat{j}+\hat{k}$ on vector $\vec{B}=\hat{i}+\hat{j}$ ?

$2(\hat{i}+\hat{j}+\hat{k})$

$\sqrt{2}(\hat{i}+\hat{j})$

$(\hat{i}+\hat{j})$

$\sqrt{2}(\hat{i}+\hat{j}+\hat{k})$

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PYQ

Two vectors $\vec{P}$ and $\vec{Q}$ have equal magnitudes. If the magnitude of $\vec{P}+\vec{Q}$ is $n$ times the magnitude of $\vec{P}-\vec{Q}$, then angle between $\vec{P}$ and $\vec{Q}$ is:

${\sin }^{-1}\left(\frac{{n}^{2}-1}{{n}^{2}+1}\right)$

${\cos }^{-1}\left(\frac{{n}^{2}-1}{{n}^{2}+1}\right)$

${\sin }^{-1}\left(\frac{n-1}{n+1}\right)$

${\cos }^{-1}\left(\frac{n-1}{n+1}\right)$

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