Answer to Question #209779 in Analytic Geometry for Jaguar

Question

Assume that a vector ~a of length ||~a|| = 3 units. In addition, ~a points in a direction that is 135◦

counter-

clockwise from the positive x-axis, and a vector ~b in th xy-plane has a length ||~b|| =

1

3

and points in the

positive y-direction.

(5.1) Find ~a · ~b.

(5.2) Calculate the distance between the point (−1, √

3) and the line 2x − 2y − 5 = 0.

Answer 1

Getting vector a and b

$\vec{a}=3(\cos{135\degree}i+\sin{135\degree}j)\\ \vec{a}=3(\frac{-1}{\sqrt{2}}i+\frac{1}{\sqrt{2}}j)\\ \vec{a}=\frac{-3}{\sqrt{2}}i+\frac{3}{\sqrt{2}}j\\ \vec{b}=13j\\ \vec{a}\cdot \vec{b}=(\frac{-3}{\sqrt{2}}i+\frac{3}{\sqrt{2}}j)(13j)=\frac{39}{\sqrt{2}}\\$

(5.2)

The distance between the point (−1, $\sqrt{3}$ ) and the line 2x − 2y − 5 = 0 is first calculated by the formula

$d=\frac{|ax_o+by_o+c|}{\sqrt{a^2+b^2}}\\ a=2\\ b=-2\\ c=-5\\ x_o=-1\\ y_o=\sqrt{3}\\ d=\frac{|2\times-1-2\times\sqrt{3}-5|}{\sqrt{2^2+(-2)^2}}\\ d=\frac{10.46}{2.83}\\ d=3.7units$