If the magnitudes of vectors vec P,vec Q,and | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) Given that the magnitudes are $P = 5$,$Q = 12$,and $R = 13$.Since $5^2 + 12^2 = 25 + 144 = 169 = 13^2$,the vectors form a right-angled triangle wh

Vedclass · Accepted Answer

$\cos^{-1}(\frac{12}{13})$

Vedclass · Answer

(C) Given that the magnitudes are $P = 5$,$Q = 12$,and $R = 13$.Since $5^2 + 12^2 = 25 + 144 = 169 = 13^2$,the vectors form a right-angled triangle where $\vec R$ is the hypotenuse.In the triangle formed by vectors $\vec P$,$\vec Q$,and $\vec R$,the angle $	heta$ is between $\vec Q$ and $\vec R$.Using trigonometry in the right-angled triangle,$\cos 	heta = \frac{	ext{Adjacent side}}{	ext{Hypotenuse}} = \frac{Q}{R}$.Substituting the values,$\cos 	heta = \frac{12}{13}$.Therefore,$	heta = \cos^{-1}(\frac{12}{13})$.

If the magnitudes of vectors $\vec P$,$\vec Q$,and $\vec R$ are $5$,$12$,and $13$ units respectively,and if $\vec P + \vec Q = \vec R$,then the angle between $\vec Q$ and $\vec R$ is:

Similar Questions

The vectors $\vec{A}$ and $\vec{B}$ are such that $|\vec{A}+\vec{B}|=|\vec{A}-\vec{B}|$. The angle between the two vectors is: (in $^{\circ}$)

The magnitude of the sum of the two vectors $\vec{A}$ and $\vec{B}$ is equal to the magnitude of the difference of the two vectors $\vec{A}$ and $\vec{B}$. The angle between $\vec{A}$ and $\vec{B}$ is: (in $^{\circ}$)

The vector sum of two forces is perpendicular to their vector difference. In this case,the forces:

$A$ car moves with a speed of $20 \, m/s$ towards the North. It turns and moves with the same speed towards the West. What is the change in velocity?

If $\overrightarrow A = 2\hat i + \hat j$,$\overrightarrow B = 3\hat j - \hat k$,and $\overrightarrow C = 6\hat i - 2\hat k$,find the value of $\overrightarrow A - 2\overrightarrow B + 3\overrightarrow C$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module