Question

Three vectors a, b and c are such that a=b+c and their magnitudes are 5,4 and 2 respectively. Find the angle between a and c

Answer 1

Given that [math]|\vec A|=5[/math] , [math]|\vec B|=4[/math], [math]|\vec C|=3[/math] &

[math]\vec A=\vec B+\vec C[/math]

[math]\vec A-\vec B=\vec C[/math]

Taking self scalar product on both the sides as follows

[math](\vec A-\vec B)\cdot(\vec A-\vec B)=\vec C\cdot \vec C[/math]

[math]\vec A\cdot \vec A-\vec B\cdot \vec A-\vec A\cdot \vec B+\vec B\cdot \vec B=|\vec C|^2[/math]

[math]|\vec A|^2-2\vec A\cdot \vec B+|\vec B|^2=|\vec C|^2[/math]

[math]5^2-2|\vec A||\vec B|\cos\theta+4^2=3^2[/math]

[math]41-2\cdot 5\cdot 4\cos\theta=9[/math]

[math]32-40\cos\theta=0[/math]

[math]\cos\theta=\frac{4}{5}[/math]

[math]\theta=\cos^{-1}(\frac45)[/math]

[math]=36.86989764584401^\circ[/math]

Hence, the angle ([math]\theta[/math]) between given vectors [math]\vec A[/math] & [math]\vec B[/math] is [math]36.86989764584401^\circ[/math]