MHT CET 2019 2nd May Evening Shift | Vector Algebra Question 8 | Physics

$\mathbf{P}$ 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

$P \cdot Q$

$\frac{P \times Q}{P}$

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

$p \cdot q$

MHT CET 2019 2nd May Morning Shift

MCQ (Single Correct Answer)

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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 $\left[\tan 45^{\circ}=1\right]$

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

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

$\frac{\pi}{4}$

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

MHT CET 2019 2nd May Morning Shift

MCQ (Single Correct Answer)

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If $\sqrt{A^2+B^2}$ represents the magnitude of resultant of two vectors $(\mathbf{A}+\mathbf{B})$ and $(\mathbf{A}-\mathbf{B})$, then the angle between two vectors is

$\cos ^{-1}\left[-\frac{2\left(A^2-B^2\right)}{\left(A^2+B^2\right)}\right]$

$\cos ^{-1}\left[-\frac{A^2-B^2}{A^2 B^2}\right]$

$\cos ^{-1}\left[-\frac{\left(A^2+B^2\right)}{2\left(A^2-B^2\right)}\right]$

$\cos ^{-1}\left[-\frac{\left(A^2-B^2\right)}{A^2+B^2}\right]$

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