If overrightarrow{a}+overrightarrow{b}+overri | vedclass

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

Question

(D) Given that $\overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c}=\overrightarrow{0}$.We can write this as $\overrightarrow{a}+\overrightarrow{

Vedclass · Accepted Answer

$\frac{\pi}{3}$

Vedclass · Answer

(D) Given that $\overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c}=\overrightarrow{0}$.We can write this as $\overrightarrow{a}+\overrightarrow{b}=-\overrightarrow{c}$.Squaring both sides,we get $(\overrightarrow{a}+\overrightarrow{b})^2=(-\overrightarrow{c})^2$.Expanding the dot product,we have $|\overrightarrow{a}|^2+|\overrightarrow{b}|^2+2|\overrightarrow{a}||\overrightarrow{b}| \cos 	heta=|\overrightarrow{c}|^2$,where $	heta$ is the angle between $\overrightarrow{a}$ and $\overrightarrow{b}$.Substituting the given values: $3^2+4^2+2(3)(4) \cos 	heta = (\sqrt{37})^2$.$9+16+24 \cos 	heta = 37$.$25+24 \cos 	heta = 37$.$24 \cos 	heta = 37-25 = 12$.$\cos 	heta = \frac{12}{24} = \frac{1}{2}$.Since $\cos 	heta = \frac{1}{2}$,we have $	heta = \frac{\pi}{3}$.

Element	$M^{3+}/M$	$M^{+}/M$
$Al$	$-1.66$	$+0.55$
$Tl$	$+1.26$	$-0.34$

If $\overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c}=\overrightarrow{0}$ and $|\overrightarrow{a}|=3, |\overrightarrow{b}|=4$ and $|\overrightarrow{c}|=\sqrt{37}$,then the angle between $\overrightarrow{a}$ and $\overrightarrow{b}$ is:

Similar Questions

The area (in square units) of the triangle formed by the lines $x=0$,$y=0$ and $3x+4y=12$ is:

Consider the following standard electrode potentials ($E^o$ in volts) in aqueous solution:

Element $M^{3+}/M$ $M^{+}/M$

$Al$ $-1.66$ $+0.55$

$Tl$ $+1.26$ $-0.34$

Based on these data,which of the following statements is correct?

Which of the following is the simplest amino acid?

If $l$ and $m$ are the degree and the order respectively of the differential equation of the family of all circles in the $XY$ plane with radius $5$ units,then $2l + 3m =$

The angle between the lines $\vec{r}=(2 \hat{i}-3 \hat{j}+\hat{k})+\lambda(\hat{i}+4 \hat{j}+3 \hat{k})$ and $\vec{r}=(\hat{i}-\hat{j}+2 \hat{k})+\mu(\hat{i}+2 \hat{j}-3 \hat{k})$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module