Answer to Question #211628 in Analytic Geometry for prince

Question

Question 5:

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

Solution.

$5.1.||a||=3;$

$||b||=1/3;$

$<(\overrightarrow{a},\overrightarrow{b})=135-90=45^o;$

$\overrightarrow{a}\sdot\overrightarrow{b}=||\overrightarrow{a}||\sdot||\overrightarrow{b}||cos<(\overrightarrow{a},\overrightarrow{b});$

$\overrightarrow{a}\sdot\overrightarrow{b}=3\sdot1/3\sdot\dfrac{\sqrt{2}}{2}=\dfrac{\sqrt{2}}{2};$

$5.2. M(-1;\sqrt{3});$

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

$d=\dfrac{|AM_x+BM_y+C|}{\sqrt{A^2+B^2}};$

$d=\dfrac{|2\sdot(-1)-2\sdot\sqrt{3}-5|}{\sqrt{2^2+(-2)^2}}=\dfrac{7+2\sqrt{3}}{2\sqrt{2}}=\dfrac{7\sqrt{2}+2\sqrt{6}}{4};$

Answer:$5.1. \overrightarrow{a}\sdot\overrightarrow{b}=\dfrac{\sqrt{2}}{2};$

$5.2.d=\dfrac{7\sqrt{2}+2\sqrt{6}}{4}.$