Transcription of Chapter 12 Section 5 Lines and Planes in Space
{{id}} {{{paragraph}}}
Example: quiz answers
Chapter 12 Section 5 Lines and Planes in Space
Chapter 12 Section 5 Lines and Planesin SpaceExample 1 Show that the line through the points(0,1,1)and(1, 1,6) is perpendicular to theline through the points ( 4,2,1)and( 1,6,2).Vector equation for the first line:r1(t).=<0,1,1>+t(<1, 1,6> <0,1,1>)=<0,1,1>+t <1, 2,5>Vector equation for the second line:r2(s).=< 4,2,1>+s(< 1,6,2> < 4,2,1>)=< 4,2,1>+s <3,4,1>cos =<1, 2,5> <3,4,1>|<1, 2,5>||<3,4,1>|=(1)(3) + ( 2)(4) + (5)(1) 12+ ( 2)2+ 52 32+ 42+ 12=0 30 26=0 Remark: These two Lines 2(a) Find parametric equations for the line through(5,1,0) that is perpendicular to the plane2x y+z= 1A normal vector to the plane is:n=<2, 1,1>r(t) =<5,1,0>+t <2, 1,1>(b) In what points does this line intersect thecoordinate Planes ?xy-plane: 0 +t1t= 0 r(0) =<5,1,0>yz-plane: 5 +t2t= 52 r( 52) =<0,72, 52>zx-plane: 1 +t( 1)t= 1 r(1) =<7,0,1>Example 3 Parallelism, intersection for:L1:x 12=y1=z 14r1(t) =<1,0,1>+t <2,1,4>=<1 + 2t, t,1 + 4t >L2:x1=y+ 22=z+ 23r2(s) =<0, 2, 2>+s <1,2,3>=< s, 2 + 2s, 2 + 3s > <2,1,4> <1,2,3>=< 5, 2,3>6=0 L1 L2r1(t).
Example 2 (a) Find parametric equations for the line through (5,1,0) that is perpendicular to the plane 2x − y + z = 1 A normal vector to the plane is:
Loading..
Information
Domain:
Source:
Link to this page:
Please notify us if you found a problem with this document:
Related search queries
Before you go When you arrive, 3-1-1, Motivational Interviewing Treatment Integrity 3.1.1, 3.1.1, Motivational Interviewing Treatment Integrity, MQTT Version 3, Mqtt, EQUIVALENTS, MEASUREMENTS and YIELDS, Solutions to Exercises on Mathematical Induction, 3 ARTICLE 3. PUBLICATION OF NOTICES, LAFS: Language Arts Florida Standards 3