Question

If a+b+c=0 and a =3 b =5 c =7 find the angle between a and b

Answer 1

|a|=3,|b|=4,|c|=5

 

 a+b+c=0

 

 a+b =-c

 

|a+b| =|-c|

 

|a+b|2 =|-c|2

 

|a|2 +|b|2 +2a.b =|c|2

 

 9  + 25  +2a.b=49

 

a.b=3*5

9+25+2ab cos theta=49

15=30cos theta

or cos theta=1/2

theata =60

so angle is 60 degrees

mark brainliest plzzzzz

 

Answer 2

Answer:

60°

Step-by-step explanation:

Given: If a+b+c=0 and |a| =3,  |b| = 5, |c| =7

To find: the angle between a and b

Let angle between a and b be Ф

Formula used: $\theta=\cos^{-1}\dfrac{a\cdot b}{|a||b|}$

a+b+c=0

a + b = - c

Taking square both sides

$(a+b)^2=(-c)^2$

$|a|^2+|b|^2+2a\cdot b=|c|^2$

$3^2+5^2+2a\cdot b=7^2$

$2a\cdot b=49-25-9$

$a\cdot b=\dfrac{15}{2}$

$\theta=\cos^{-1}\dfrac{15}{2\cdot 3\cdot 5}$

$\theta=\cos^{-1}(\dfrac{1}{2})$

$\theta=60^\circ$

hence, the angle between a and b is 60°