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Response Details:
Question Details:
Find parametric equations for line of intersection of planes P............. -x + 2y + z = 0 ...........1
and Q....... x + z = -1...............................2
LET L BE THE LINE OF INTERSECTION OF PLANES P AND Q
DRS OF NORMAL TO PLANE P ARE [-1,2,1]
DRS OF NORMAL TO PLANE Q ARE [1,0,1]
LET DRS OF LINE L BE A,B,C ..THEN SINCE L IS ON P AND Q , IT WILL BE NORMAL TO THE NORMALS OF PLANES P AND Q.HENCE
-1A+2B+1C=0.............................3
1A+1C=0..............A= - C...........4
PUTTING IN EQN.3 ...........2B+2C=0........B= - C................5
A:B:C = - C : - C : C = -1 : - 1 : 1
NOW LET US FIND A POINT [X1,Y1,Z1]ON L .WE CAN PUT SAY X1=0 AND FIND Z1=-1 FROM EQN.2..AND THEN PUTTING IN EQN.1, GIVES Y1=0.5...SO [0,0.5,-1] IS A POINT ON L.
HENCE EQN. IS
X=-T
Y=0.5-T
Z=T-1 IS THE PARAMETRIC FORM OF EQN. OF L
a) Then find one point on the line. Show it is on both planes.
PUT T=0 TO GET ONE POINT ON L
X=0
Y=0.5
Z=-1
PUTTING IN EQN.1
-x + 2y + z = 0 ...........1...WE GET
0+1-1=0......OK
PUTTING IN EQN.2
.. x + z = -1.........................2...... WE GET
0-1=-1 ........OK
b) Then find the equation of the plane through the point P(2,3,-2) [X',Y',Z']and
perpendicular to the line of intersection of the planes mention above.
THE PLANE WILL HAVE SAME DRS AS THE LINE HENCE EQN. OF PLANE REQD. IS
A(X-X')+B(Y-Y')+C(Z-Z')=0
-1[X-2]-1[Y-3]+1[Z+2]=0
-X-Y+Z+7=0
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