The points O, P, Q and R have coordinates (0,0,0), (2,6,-2), (2,4,-1) and (4,6,0) respectively. Find a vector perpendicular to both PQ and QR and hence or otherwise, determine an equation for the plane PQR. The line (x-2) = y = (z-1)/3 meets the plane PQR at the point T and the point S on this line such T is the midpoint of OS. Find
a) the cosine of the acute angle between the line and the plane PQR, and
b) the perpendicular distance from S to the plane PQR.
vector PQ = (2,4,-1) - (2,6,-2) = (0,-2,1)
vector QR = (4,6,0) - (2,4,-1) - (2,2,1)
vector perpendicular to PQ and QR = (0,-2,1)X(2,2,1)=(-4,2,4) ;cross product
vector equation of plane PQR: r.(-2,1,2) = (4,6,0).(-2,1,2) = -2
converting the equation of line from cartesian to vector form:
r = (2,0,1) + λ(1,1,3)
substituting the equation of the line into the equation of PQR to find point of intersection, T : λ = 0
therefore vector OT= (2,0,1)
this is where i'm stuck, im unable to find the cosine of the acute angle between the line and the plane PQR. Here's my attempt.
let the angle between the line and plane be theta.
cos(90-theta) = |(1,1,3).(-2,1,2)| / root(11) root (9)
using sin^2 (theta) + cos^2 (theta) = 1
cos(theta) = root(74)/3root(11)
b) perpendicular distance = TS X (-2,1,2)/root(9)
(2,0,1) X (-2,1,2)/root(9)
the answer for this part is 2/3.. can someone find my error?