Question

vectors a+b+c = 0 . magnitude of a b and c are 3,5 and 7 respectively.
find angle between a and b.. solve with process

Answer 1

Answer:

$The \:angle\: between\: \vec{a}\;and\:\vec{b}\:is\:\frac{\pi}{3}$

Step-by-step explanation:

Formula used:

The scalar product of $\vec{a}\;and\:\vec{b}$  is defined as

$\vec{a}.\vec{b}=|\vec{a}|\:|\vec{b}|\:cos\theta$

Here, $\theta$ is the angle between $\vec{a}\;and\:\vec{b}$

$|\vec{a}+\vec{b}|^2=|\vec{a}|^2+|\vec{b}|^2+2\vec{a}.\vec{b}$

$Given:\\\\\vec{a}+\vec{b}+\vec{c}=\vec{0}\\\\and\\\\|\vec{a}|=3,\:|\vec{b}|=5,\:|\vec{c}|=7$

$Now,\\\\\vec{a}+\vec{b}=-\vec{c}$

squaring on both sides we get

$(\vec{a}+\vec{b})^2=\vec{c}^2\\\\|\vec{a}|^2+|\vec{b}|^2+2\vec{a}.\vec{b}=|\vec{c}|^2$

$|\vec{a}|^2+|\vec{b}|^2+2|\vec{a}|\:|\vec{b}|\:cos\theta=|\vec{c}|^2$

$(3)^2+(5)^2+2(3)(5)\:cos\theta=(7)^2\\\\9+25+30\:cos\theta=49\\\\34+30\:cos\theta=49\\\\30\:cos\theta=15\\\\cos\theta=\frac{15}{30}\\\\cos\theta=\frac{1}{2}\\\\\theta=\frac{\pi}{3}$

Answer 2

Answer:

angle between a and b = 60°

Step-by-step explanation:

vectors a+b+c = 0 . magnitude of a b and c are 3,5 and 7 respectively.

find angle between a and b.. solve with process

R² = a² + b² +2abcosθ  

R is resultant of a & b

as a + b + c = 0

so R = -c

| R | = | c |

=> c² = a² + b² +2abcosθ

=> 7² = 3² + 5² + 2×3×5Cosθ

=>49 = 9 + 25 + 30Cosθ

=>  15 = 30Cosθ

=>  Cosθ = 1/2

=> θ = 60°

angle between a and b = 60°