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Find-the-equation-for-the-plane-through-the-point-A-6-2-4-B-2-4-8-C-4-2-2-Vector-Analysis-M-m-




Question Number 182984 by Mastermind last updated on 18/Dec/22
Find the equation for the plane  through the point A(6, 2, −4),  B(−2, 4, 8), C(4, −2, 2). −Vector Analysis    M.m
FindtheequationfortheplanethroughthepointA(6,2,4),B(2,4,8),C(4,2,2).VectorAnalysisM.m
Answered by cortano1 last updated on 18/Dec/22
 n^(→)  = AB^(→)  × AC^(→)  =  determinant (((−8       2          12)),((−2    −4         6)))   = (60, 24, 36 )   AX^(→)  = (x−6, y−2, z+4)   eq of plane ⇒ AX^(→)  . n^(→)  = 0  ⇒60(x−6)+24(y−2)+36(z+4)=0  ⇒5x−30+2y−4+3z+12=0  ⇒5x+2y+3z−22=0
n=AB×AC=|8212246|=(60,24,36)AX=(x6,y2,z+4)eqofplaneAX.n=060(x6)+24(y2)+36(z+4)=05x30+2y4+3z+12=05x+2y+3z22=0
Answered by mr W last updated on 18/Dec/22
Commented by mr W last updated on 18/Dec/22
AB^(→) =(−8, 2, 12)  AC^(→) =(−2, −4, 6)  n^→ =AB^(→) ×AC^(→) =(−8,2,12)×(−2,−4,6)                            =(60,24,36)  say any point on the plane is  P(x,y,z)  AP^(→) =(x−6, y−2, z+4)  we have AP^(→) ⊥n^(→) ,  ⇒AP^(→) ∙n^(→) =0  ⇒60(x−6)+24(y−2)+36(z+4)=0  ⇒10x+4y+3z−22=0 ✓
AB=(8,2,12)AC=(2,4,6)n=AB×AC=(8,2,12)×(2,4,6)=(60,24,36)sayanypointontheplaneisP(x,y,z)AP=(x6,y2,z+4)wehaveAPn,APn=060(x6)+24(y2)+36(z+4)=010x+4y+3z22=0
Commented by cortano1 last updated on 18/Dec/22
why n^(→)  = (60,24,36)?
whyn=(60,24,36)?
Commented by mr W last updated on 18/Dec/22
n^→ =(−8,2,12)×(−2,−4,6)=(60,24,36)  you got a different n^→ , because you  wrongly took AB^(→) =(−8,2,−12).  please recheck.
n=(8,2,12)×(2,4,6)=(60,24,36)yougotadifferentn,becauseyouwronglytookAB=(8,2,12).pleaserecheck.
Commented by cortano1 last updated on 18/Dec/22
 oohhh yes . but your answer have  a typo
oohhhyes.butyouranswerhaveatypo
Commented by mr W last updated on 18/Dec/22
thanks! i have fixed it.
thanks!ihavefixedit.
Commented by Mastermind last updated on 18/Dec/22
Thank you boss
Thankyouboss

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