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Question Number 95342 by i jagooll last updated on 24/May/20

determine the value of k such that the point A(4,−2,6) B(0,1,0) C(1,0,−5) and D(1,k,−2) lie on the same plane

$$\mathrm{determine}\:\mathrm{the}\:\mathrm{value}\:\mathrm{of}\:\mathrm{k}\:\mathrm{such}\:\mathrm{that}\: \\ $$$$\mathrm{the}\:\mathrm{point}\:\mathrm{A}\left(\mathrm{4},−\mathrm{2},\mathrm{6}\right)\:\mathrm{B}\left(\mathrm{0},\mathrm{1},\mathrm{0}\right)\:\mathrm{C}\left(\mathrm{1},\mathrm{0},−\mathrm{5}\right) \\ $$$$\mathrm{and}\:\mathrm{D}\left(\mathrm{1},\mathrm{k},−\mathrm{2}\right)\:\mathrm{lie}\:\mathrm{on}\:\mathrm{the}\:\mathrm{same} \\ $$$$\mathrm{plane}\: \\ $$

Answered by john santu last updated on 24/May/20

let u = AB=(−4,3,−6) v=(−3,2,−11) . let n vector normal n = u×v = determinant (((−4 3 −6)),((−3 2 −11))) n = (−21,−26,1) . set P(x,y,z) arbitary point in plane. put BP = (x,y−1,z) then the equation of plane is n.BP = 0 −21x−26(y−1)+z = 0 so take x=1,y=k and z = −2 −21.1−26(k−1)−2=0 −21−26k+26−2=0 26k = 3 ⇒ k = (3/(26)) .

$$\mathrm{let}\:\boldsymbol{\mathrm{u}}\:=\:\boldsymbol{\mathrm{A}}\mathrm{B}=\left(−\mathrm{4},\mathrm{3},−\mathrm{6}\right) \\ $$$$\boldsymbol{\mathrm{v}}=\left(−\mathrm{3},\mathrm{2},−\mathrm{11}\right)\:.\:\mathrm{let}\:\boldsymbol{\mathrm{n}}\:\mathrm{vector}\: \\ $$$$\mathrm{normal}\:\boldsymbol{\mathrm{n}}\:=\:\boldsymbol{\mathrm{u}}×\boldsymbol{\mathrm{v}}\:=\:\begin{vmatrix}{−\mathrm{4}\:\:\:\:\mathrm{3}\:\:\:\:−\mathrm{6}}\\{−\mathrm{3}\:\:\:\:\mathrm{2}\:\:\:\:−\mathrm{11}}\end{vmatrix} \\ $$$$\boldsymbol{\mathrm{n}}\:=\:\left(−\mathrm{21},−\mathrm{26},\mathrm{1}\right)\:.\:\mathrm{set}\:\mathrm{P}\left(\mathrm{x},\mathrm{y},\mathrm{z}\right) \\ $$$$\mathrm{arbitary}\:\mathrm{point}\:\mathrm{in}\:\mathrm{plane}. \\ $$$$\mathrm{put}\:\mathrm{BP}\:=\:\left(\mathrm{x},\mathrm{y}−\mathrm{1},\mathrm{z}\right)\: \\ $$$$\mathrm{then}\:\mathrm{the}\:\mathrm{equation}\:\mathrm{of}\:\mathrm{plane}\:\mathrm{is}\: \\ $$$$\boldsymbol{\mathrm{n}}.\boldsymbol{\mathrm{B}}\mathrm{P}\:=\:\mathrm{0} \\ $$$$−\mathrm{21x}−\mathrm{26}\left(\mathrm{y}−\mathrm{1}\right)+\mathrm{z}\:=\:\mathrm{0} \\ $$$$\mathrm{so}\:\mathrm{take}\:\mathrm{x}=\mathrm{1},\mathrm{y}=\mathrm{k}\:\mathrm{and}\:\mathrm{z}\:=\:−\mathrm{2}\: \\ $$$$−\mathrm{21}.\mathrm{1}−\mathrm{26}\left(\mathrm{k}−\mathrm{1}\right)−\mathrm{2}=\mathrm{0} \\ $$$$−\mathrm{21}−\mathrm{26k}+\mathrm{26}−\mathrm{2}=\mathrm{0} \\ $$$$\mathrm{26k}\:=\:\mathrm{3}\:\Rightarrow\:\mathrm{k}\:=\:\frac{\mathrm{3}}{\mathrm{26}}\:.\: \\ $$

Commented by i jagooll last updated on 24/May/20

thank you

$$\mathrm{thank}\:\mathrm{you} \\ $$

Answered by mr W last updated on 24/May/20

determinant ((4,(−3),6),(1,(−1),(−5)),(1,(k−1),(−2)))=0 determinant ((4,(−3),6),(0,1,(26)),(0,k,3))=0 4(3−26k)=0 ⇒k=(3/(26))

$$\begin{vmatrix}{\mathrm{4}}&{−\mathrm{3}}&{\mathrm{6}}\\{\mathrm{1}}&{−\mathrm{1}}&{−\mathrm{5}}\\{\mathrm{1}}&{{k}−\mathrm{1}}&{−\mathrm{2}}\end{vmatrix}=\mathrm{0} \\ $$$$\begin{vmatrix}{\mathrm{4}}&{−\mathrm{3}}&{\mathrm{6}}\\{\mathrm{0}}&{\mathrm{1}}&{\mathrm{26}}\\{\mathrm{0}}&{{k}}&{\mathrm{3}}\end{vmatrix}=\mathrm{0} \\ $$$$\mathrm{4}\left(\mathrm{3}−\mathrm{26}{k}\right)=\mathrm{0} \\ $$$$\Rightarrow{k}=\frac{\mathrm{3}}{\mathrm{26}} \\ $$

Commented by i jagooll last updated on 24/May/20

thank you

$$\mathrm{thank}\:\mathrm{you} \\ $$

Commented by john santu last updated on 24/May/20

how you get (1,−1,−5) and (1,k−1,2) ?

$$\mathrm{how}\:\mathrm{you}\:\mathrm{get}\:\left(\mathrm{1},−\mathrm{1},−\mathrm{5}\right)\:\mathrm{and}\: \\ $$$$\left(\mathrm{1},\mathrm{k}−\mathrm{1},\mathrm{2}\right)\:? \\ $$

Commented by PRITHWISH SEN 2 last updated on 24/May/20

by substracting points

$$\mathrm{by}\:\mathrm{substracting}\:\mathrm{points} \\ $$

Commented by john santu last updated on 24/May/20

ooo thank you

$$\mathrm{ooo}\:\mathrm{thank}\:\mathrm{you} \\ $$

Commented by mr W last updated on 24/May/20

vectors AB, CB and DB are in the same plane.

$${vectors}\:{AB},\:{CB}\:{and}\:{DB}\:{are}\:{in}\:{the} \\ $$$${same}\:{plane}. \\ $$

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