Problem 1: a) find the point where the line through the points P(1,0,1) and Q(0,
ID: 2873410 • Letter: P
Question
Problem 1: a) find the point where the line through the points P(1,0,1) and Q(0,3,2) meets the plane 2x-y+3z=6 b) find the angle between the line and the normal to the plane Problem 1: a) find the point where the line through the points P(1,0,1) and Q(0,3,2) meets the plane 2x-y+3z=6 b) find the angle between the line and the normal to the plane a) find the point where the line through the points P(1,0,1) and Q(0,3,2) meets the plane 2x-y+3z=6 b) find the angle between the line and the normal to the planeExplanation / Answer
a)line PQ is P+t*(Q-P)
==>(1,0,1)+t((0,3,2)-(1,0,1))
==>(1,0,1)+t(-1,3,1)
==>(1-t,3t,1+t)
(1-t,3t,1+t) intersects 2x-y+3z=6
==>2(1-t)-3t+3(1+t)=6
==>2-2t-3t+3+3t=6
==>5-2t=6
==>2t=-1
==>t=(-1/2)
point =(1-(-1/2),3(-1/2),1+(-1/2)) =(3/2 ,-3/2,1/2)
point where the line through the points P(1,0,1) and Q(0,3,2) meets the plane 2x-y+3z=6 =(3/2 ,-3/2,1/2)
b) direction vector of line r=(-1,3,1), normal of plane n=(2,-1,3)
cos theta =(r.n)/|r||n| =|(-1,3,1).(2,-1,3)|/sqrt((-1)2+32+12)sqrt(22+(-1)2+32) =2/sqrt154
theta =cos-1(2/sqrt154)
theta=80.70
angle between the line and the normal to the plane =80.70