Question

Find the angle between vector u=2i+sqrt(11)j and v=-3i-2j to the nearest degree

Answer 1

Answer: Last option 155°

Step-by-step explanation:

We have the components of the vectors u and v.

Then, to find the angle between them, perform the following steps:

1) Calculate the scalar product u * v

If $u = 2i + \sqrt{11}j$  and $v = -3i-2j$

Then, the product scalar u * v is:

$u * v = (2)(-3) + (\sqrt{11})(-2)$

$u * v = -6-2\sqrt{11}$

$u * v = -12.633$

2) Calculation of the magnitude of both vectors.

$| u | = \sqrt{(2) ^ 2 + (\sqrt{11})^2}\\\\| u | = \sqrt{4+11}\\\\| u | = \sqrt{15}$

$| v | = \sqrt{(- 3) ^ 2 + (- 2) ^ 2}\\\\| v | = \sqrt{9 +4}\\\\| v | = \sqrt{13}$

3) Now that you know the product point between the two vectors and the magnitude of each, then use the following formula to find an angle

$u * v = | u || v |cos(\alpha)$

$-12.633 = \sqrt{15}\sqrt{13}*cos(\alpha)\\\\cos(\alpha) = \frac{-12.633}{\sqrt{15}\sqrt{13}}\\\\arcos(\frac{-12.633}{\sqrt{15}\sqrt{13}}) = \alpha\\\\\alpha =155\°$