Any plane passing through (-2, 2, 2) is A(x+2) + B(y – 2)+C(z- 2) = 0 : It passes through (2, –2, –2) — 4А — 4B — 4С 3D0 .(i) It is parallel to 9x- 13y - 3z = 0 . 9A – 13B – 3C = 0 .(i) Solving Eqs.(i) and (ii), A B C 12 - 52 -36 + 12 -52 + 36 A B C -40 -24 -16 .. Required equation of plane is -40(x+2)– 24(y – 2)– 16(z – 2) = 0 →-40x – 80 – 24y + 48 – 16z + 32 = 0 = 40x + 24y + 16z = 0 → 5x+ 3y + 2z = 0

Calculus: Early Transcendentals
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Author:James Stewart
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Chapter1: Functions And Models
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Problem 1RCC: (a) What is a function? What are its domain and range? (b) What is the graph of a function? (c) How...
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can please someon explain a detailed solution on how to get A,B,C   (please write on a paper)

Any plane passing through (-2, 2, 2) is A(x+2) + B(y – 2)+C(z- 2) = 0
: It passes through (2, –2, –2)
— 4А — 4B — 4С 3D0
.(i)
It is parallel to 9x- 13y - 3z = 0
. 9A – 13B – 3C = 0
.(i)
Solving Eqs.(i) and (ii),
A
B
C
12 - 52
-36 + 12
-52 + 36
A
B
C
-40
-24 -16
.. Required equation of plane is -40(x+2)– 24(y – 2)– 16(z – 2) = 0
→-40x – 80 – 24y + 48 – 16z + 32 = 0
= 40x + 24y + 16z = 0
→ 5x+ 3y + 2z = 0
Transcribed Image Text:Any plane passing through (-2, 2, 2) is A(x+2) + B(y – 2)+C(z- 2) = 0 : It passes through (2, –2, –2) — 4А — 4B — 4С 3D0 .(i) It is parallel to 9x- 13y - 3z = 0 . 9A – 13B – 3C = 0 .(i) Solving Eqs.(i) and (ii), A B C 12 - 52 -36 + 12 -52 + 36 A B C -40 -24 -16 .. Required equation of plane is -40(x+2)– 24(y – 2)– 16(z – 2) = 0 →-40x – 80 – 24y + 48 – 16z + 32 = 0 = 40x + 24y + 16z = 0 → 5x+ 3y + 2z = 0
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