Question #211628

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


1
Expert's answer
2021-06-29T16:46:52-0400

Solution.

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

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

<(a,b)=13590=45o;<(\overrightarrow{a},\overrightarrow{b})=135-90=45^o;

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

ab=31/322=22;\overrightarrow{a}\sdot\overrightarrow{b}=3\sdot1/3\sdot\dfrac{\sqrt{2}}{2}=\dfrac{\sqrt{2}}{2};

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

2x2y5=0;2x-2y-5=0;

d=AMx+BMy+CA2+B2;d=\dfrac{|AM_x+BM_y+C|}{\sqrt{A^2+B^2}};

d=2(1)23522+(2)2=7+2322=72+264;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.ab=22;5.1. \overrightarrow{a}\sdot\overrightarrow{b}=\dfrac{\sqrt{2}}{2};

5.2.d=72+264.5.2.d=\dfrac{7\sqrt{2}+2\sqrt{6}}{4}.



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