# Question #491a2

Jul 13, 2016

We know that for two vectors $\vec{a} \mathmr{and} \vec{b}$ the scalar part of vector $\vec{a} + \vec{b}$ is given by
$| \vec{a} + \vec{b} | = \sqrt{| \vec{a} {|}^{2} + | \vec{b} {|}^{2} + 2 | \vec{a} | | \vec{b} | \cos \theta}$
where $\theta$ is the angle between the vectors
For the given problem
$| \vec{a} + \vec{b} | = \sqrt{{3}^{2} + {4}^{2} + 2 \times 3 \times 4 \cos 120}$
$\implies | \vec{a} + \vec{b} | = \sqrt{9 + 16 + 2 \times 3 \times 4 \left(- \frac{1}{2}\right)}$
$\implies | \vec{a} + \vec{b} | = \sqrt{13}$
This has already been calculated but repeated for the sake of clarity.

Now we know that Triangle law of vector addition states that when two sides of a triangle represent two vectors in magnitude and direction taken in same order then third side of that triangle gives us the magnitude and direction the resultant of the vectors.

As shown in the figure above, when a vector parallel to $\vec{a}$ is drawn at the tip of $\vec{b}$ to complete the triangle, the angle between the two is supplement of the angle between the given vectors.

Jul 13, 2016

The angle between vectors $a \mathmr{and} - b$ is supplement to the angle between $a \mathmr{and} b$.

#### Explanation:

My answer is on the basis of the definition that $\pi - \theta$ is supplementary to $\theta$.