Answer to Question #209804 in Analytic Geometry for Faith

Question

1.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 the xy plane has a length ||b||=1/3 and points in the positive y-direction.

1.1.find a.b.

1.2.calculate the distance between the point (-1,√3) and the line 2x-2y-5=0.

Answer 1

$$\vec a=\langle3\cos135\degree, 3\sin 135\degree\rangle,$$

$$\vec b=\langle0,\dfrac{1}{3}\rangle$$

$$\vec a\cdot \vec b=3(-\dfrac{\sqrt{2}}{2})(0)+3(\dfrac{\sqrt{2}}{2})(\dfrac{1}{3})=\dfrac{\sqrt{2}}{2}$$

Or

Angle between two vectors $\angle(\vec a, \vec b)=135\degree-90\degree =45\degree$

$$\vec a\cdot \vec b=|\vec a||\vec b|\cos(\angle(\vec a\cdot \vec b))$$

$$=3(\dfrac{1}{3})\cos 45\degree=\dfrac{\sqrt{2}}{2}$$

$\vec a\cdot \vec b=\dfrac{\sqrt{2}}{2}$

b)

$$d=\dfrac{|Ax_0+By_0+C|}{\sqrt{A^2+B^2}}$$

Point $(-1, 3): x_0=-1, y_0=3$

Line $2x - 2y-5=0$

$$d=\dfrac{|2(-1)-2(3)-5|}{\sqrt{(2)^2+(-2)^2}}=\dfrac{13\sqrt{2}}{4}$$

The distance is $\dfrac{13\sqrt{2}}{4}.$