Question

(1) Equation of a Plane Let P(1,1,-1), Q(1,2,0), R(-2,2,2). (la) Compute PQxPR. (1b) Find the equation of the plane through P, Q and R in the form ax+by+cz=d. (10) What is the angle formed by this plane and the xy-plane?

Answer 1

Given three points, PCX, 4 , 2 ) = P(1, 1,-1) Q(x2, 4,, z)= Q(1,2,0) R (X, Y, 3 Zz) = R(2, 2, 2) So, x = 1 9 9 Z = -1 41 = 1 92 = 2 x = 1, Z₂ = 0 x₂ = 2 9₂ = 2 . O (12) Vector between two points P and a is, Pa = (x₂-x) i + i + (92-9₂) j + (22 - 2,) k. = (1-1) i + (2-1)j + (0-(-1)] km = 0 i + 1 + 1k PR Similarly, Vector between two points p and R is, (x₃ - x, )i + (43-4) j + (Z - Z) k (-2-1) i + (2-1) j + [2-(-1)] k + 1] +3K = -i 4) two vectors, Vector cross product between A = a i + Azj + azk and B = 6, it baj + b₂ k is given by

à B = { jk a, A2 Az bi b2 b₃ i Vecton cross product PQ x PŘ is given by, PQ X PR i j K o 1 1 -3 1 3 - i (3-1)-; (0-(-3)) + (0-(-3)) 2i-j (3) + (3) - 24–3j+3K (16) PQX PŘ Po Q I PT PR P AR a Х

We need to collinear. If first check if the thorul points are PEX, X₂183) thoree points du ) We need to first check if the three points are collinear. If three points P Ca. 9,2), Q(x2, Yz, 22) and R (X3, Y , Z ) collinear then determinant of their co-ordinates will be zero. 17 determinant not equal to Zero) this these points not collinear. 2, y, Y₂ 223 Ya are 21 22 22 O 23 -2 2 2 +1(4-0)-1(2-0)-1 (2+4) = 4-2-6 = - 4 to The three points P, Q and R arre non collinear.

Consider the above figure. Let us consider three non-collinear points P, Q and R. Let the position vector of these points be pa and T. Cross Product of the vellon Pa and PŘ is perpendicular to the plane containing P, a and R. So, PQ x Pk is perpendicular to the plane cont containing P, Q and R. us also assume Let an arbitany point T(x, y, z) in the plane containing P, Q and R. So, the vector, pŤ is also in the plane containing P, Q and R. Equation of vector 87 is, p = (x-21 ) i + (y-g, ) ) + (2-2,)k! = (x-1) i + (y- 1)j + [z-(-1)] k (x-1) i + (y- 1)j +(2+1) k 7 Vectons are containing Pia Since Vello it is in the plane containing P, Q and id R, and Pax P is perpendicular to the plane and R, these two perpen dicular. That is pi and Pax PŘ is each other. So their scalar product will be Zeno- PT. CP& PR ) = perpendicular

we Substitute expression for velton Pt and value of (På & PR ) get, [(ex-1){ +(4-1) j+(z+1)x]. ( 21-3j+ 3x) = 0 2(x-1)-3C7-1+3C2+0 = 0 23-2-3y + 3 + 3z + 3 2x-3y + 32 +4 2x-3y +32 =0 = -4 This is the required equation of the plane containing points P, Q and R. We can also use the vector tripple product to find the equation of plane. For that consider the scalar tripple products of the three vectons,

A as it an je dok Bobilt bejt bok and C - Gi + Cejr esk is given by A. (B x 2) = a, A2 a3 b2 b3 8 C C2 C3 2 . In the same way, we can calculate vecton tripple product of PT. (Pax PR), by using expression of PT PQ and A Ps. from equation ©,0 and ® respectively. P Q = (x2-x1) i + (Y₂-4, )j + (22-2, JK PŘ = (x3 - x) i +(93-y, )j + (23-2, JK PT (x-x1) i + (y-y,dj +(2-2,) k. Using equ- , we of the plane as |(x-a1) Cy-y,) (2-2) (222-21) (92-9, (22-2,) (23-xi) (9₃-9,) (23-2) get and Ta we get, equation Substitute values of x's, y's and z's, we get C you this, substitute Values from equation ,@@&0)

(x-1) (9.1) (2+1)/ o 1 1 -3 1 3 (x-1) (3-1)-CyCo--3)+(2+1) (0--3) -- 2(3-1)-3(y-+3 (2+1) - 22 - 2 - 3y+3+3 2 +3 = 0 23 - 3y +32 -2 +3 +3 = 0 29 - 3y+ 3 2x - 3y + 32 2-4 +32+ 4 =0 This is the required equation of the plane in ax+by + cz ad form. (10) Equation of the xy-plane is, z=0. Consider two planes A,x+ By + C, z =D, and A₂X + B 2 y + Gz = Dz. Angle between these two planes A,A2+ B, B₂ + C, C₂ A2+B+07:1A+8 given by, Coso = So, angle Z=OTS, between 2x-3y +32=-4 and 2x0+ €3x 0 + 3x1 (2²+(-3) 2+ 32 102+02+12 LOS 3 √4 +9+9.55 3 122

So, angle between the concerned plane and xy-plane is given by, O= cost 50.24 122 3 M