Q 1 the points p1 11q 2 41 and r1 4 2 lie on a plane axbyczd...

Q. 1 The points p=(1,-1,1),q=(-2,-4,1) and r=(1,-4,-2) lie on a plane ax+by+cz=d in 3 dimensions. (a) Compute the displacement vectors vec(pq)=(,)vec(pr)=(,) (b) Compute the cross product vector vec(pq)\times vec(pr)=(,) (c) Write down the equation of the plane ◻x+◻y+◻z= Q. 2 Write down the coordinates (p_(1),p_(2),p_(3)) of one point on the plane 5x-5y-3z=2 in R^(3) (p_(1),p_(2),p_(3))=[,]

Transcript

00:01 In this question, we're given a plane.

00:03 This is the equation of the plane.

00:05 We are to find three points on the plane.

00:08 So this is the plane.

00:11 So three points, p, q and r.

00:14 And we were to explore pq, cross with pr.

00:20 And how is that related to the coefficient of plane? that is 2 .34 here, the coefficient of plane.

00:28 Now first, to find a point on the plane, now, the plane equation is 2x plus 3y plus 4 z is equal to 4.

00:38 We can make an educated guess for a value of x, y and z, sub it into here.

00:45 Okay, evaluate this and see whether it's equals to the right -hand side.

00:50 If it is, then the point comprising of x, y and z value that you're subbed in, will be on the plane.

00:59 Now for pq cross pr, you know that you will get a vector that is perpendicular to both.

01:05 So this is pq cross pr.

01:11 A vector perpendicular to both pq and pr would be also perpendicular to the plane.

01:17 So this pq cross pr is actually parallel to the normal vector of the plane.

01:28 Okay, so let's find a point p.

01:31 So i'm going to let p be 0 -01.

01:37 I'm going to sub it into the left -hand side of the plane equation here.

01:48 This will give me 4, and that is equal to the right -hand side of the plane.

01:53 So the point p is on the plane.

01:57 I'm going to let q be 2 -0 -0.

02:03 Again, sub -it -in...

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