Question 29238: Let V be the plane 6x1 + 4x2 - 2x3 = 0 in R^3 and let x= (6 -2 4)
a) find the orthogonal decomposition of x with respect to V
b) find the point in V which is closest to x
Answer by venugopalramana(3286)   (Show Source):
You can put this solution on YOUR website! Let V be the plane 6x1 + 4x2 - 2x3 = 0 in R^3 and let x= (6 -2 4)
HENCE THE NORMAL TO THE PLANE IS REPRESENTED BY (6,4,-2).LET US CALL IT N.
LET P BE THE POINT GIVEN BY X=(6,-2,4)
LET Q (H,K,L)BE THE POINT ON THE PLANE V NEAREST TO P .
HENCE PQ BEING THE SHORTEST DISTANCE FROM P TO THE PLANE IS PERPENDICULAR TO THE PLANE.
a) find the orthogonal decomposition of x with respect to V
I THINK YOU MEAN DECOMPOSITION OF X ALONG NORMAL TO PLANE V THAT IS (6,4,-2) AS INDICATED ABOVE (LET US CALL IT VN ).IF SO
VN = (N/|N|) { (X.N) / | N | } =(6,4,-2) { (6,-2,4).(6,4,-2) / | (6,4,-2) |^2 = (6,4,-2) {(36-8-8)/(36+16+4)} = (6,4,-2)(5/14)
SIMILARLY DECOMPOSITION ALONG A LINE PARALLEL TO THE PLANE ,LET US CALL IT VP CAN BE FOUND AS FOLLOWS
FROM THE ANSWER GIVEN BELOW FOR (b)SECOND PART,PERPENDICULAR FROM P TO THE PLANE V IS PQ WHERE Q IS ON THE PLANE WITH (27/7,-24/7,33/7).HENCE CALLING DECOMPOSITION PARALLEL TO THE PLANE AS VP WE GET
SO..VP=(Q/|Q|){(X.Q) / | Q | } = (27/7,-24/7,33/7) {(6,-2,4).(27/7,-24/7,33/7) } / |(27/7,-24/7,33/7)|^2 = (27/7,-24/7,33/7){ (6*27+2*24+4*33)/7}/{(27^2+24^2+33^2)/49}=(27/7,-24/7,33/7)
HENCE THE DECOMPOSITION OF X W.R.T. PLANE V IS
VN=(6,4,-2)(5/14)....AND....
VP=(27/7,-24/7,33/7)
ALITER...WE CAN ALSO USE GRAM SCHMIDT ORTHOGONALISATION METHOD TO DECOMPOSE THE VECTOR.
LET US CALL Z1= (6,4,-2)..NORMAL TO PLANE...6X1+4X2-2X3=0
Z2=(6,-2,4).....THE POINT P.
LET US ORTHOGONALISE Z2 INTO Y1 AND Y2..PARALLEL AND PERPENDICULAR TO THE PLANE.
LET Y1=Z1=(6,4,-2)
LET Y2=Z2+AY1..WE HAVE TO FIND A.DOT WITH Y1 BOTH SIDES..
Y2.Y1=0=Z2.Y1+AY1.Y1...SINCE Y1 AND Y2 ARE ORTHOGONAL..Y2.Y1=0
A=-(Z2.Y1)/(Y1.Y1)=-(6,-2,4).(6,4,-2)/(6,4,-2).(6,4,-2)= - 20/56= -5/14..HENCE..
Y2=Z2+AY1=(6,-2,4)-(5/14)(6,4,-2)=(1/14)(6*14-5*6,-2*14-5*4,4*14-5(-2))
=(1/14)(54,-48,66)=(1/7)(27,-24,33)
Z2=Y2-AY1...OR.....(6,-2,4)=(1/7)(27,-24,33)-(5/14)(6,4,-2)
b) find the point in V which is closest to x
WE HAVE PQ PERPENDICULAR TO PLANE V OR PARALLEL TO NORMAL N.
PQ IS GIVEN BY {(H-6),(K+2),(L-4)}
HENCE
(H-6) / 6 = (K+2) / 4 = ( L-4) / -2 = R SAY
.
H = 6R+6
K = 4R-2
..L = -2R+4
..SINCE THIS LIES ON PLANE V
WE GET
6(6R+6)+4(4R-2)-2( 4-2R) = 0
..OR
..56R + 20 = 0
.R= -20 / 56 = - 5 / 14
HENCE Q IS GIVEN BY
H = 6* (-5/14) + 6 =27 / 7
.K = 4*( -5/14) - 2 = - 24 / 7
L = 4-(2*-5/14) = 33/7
..
SO THE NEAREST POINT TO P ON THE PLANE (V) IS GIVEN BY Q (27/7,-24/7,33/7)
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