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Question 31574: How do you to treat two equations in part (d)???
Two planes are defined by the equations given below.
x + 3y + z = 1 (1)
2x + 7y − z = 1 (2)
(a) Write down a normal to each of the planes.
(b) How can you tell immediately that the planes are not parallel?
(c) Write a quick test that shows they are not perpendicular.
(d) Write the augmented matrix for the system of two equations, and find the solutions
by row-reducing by hand. State clearly the row operations you use.
(e) Hence find the line of intersection of the planes defined by the two equations. Give a
direction vector for the line, and a point on the line.
Why is it that text books only talk about three equations and to bad if you only have two??? What are the rules? Hours of time spent with no answers, It's madeness.
Answer by venugopalramana(3286)   (Show Source):
You can put this solution on YOUR website! How do you to treat two equations in part (d)???
Two planes are defined by the equations given below.
x + 3y + z = 1 (1)
2x + 7y − z = 1 (2)
(a) Write down a normal to each of the planes.
(b) How can you tell immediately that the planes are not parallel?
(c) Write a quick test that shows they are not perpendicular.
(d) Write the augmented matrix for the system of two equations, and find the solutions
by row-reducing by hand. State clearly the row operations you use.
(e) Hence find the line of intersection of the planes defined by the two equations. Give a
direction vector for the line, and a point on the line.
Why is it that text books only talk about three equations and to bad if you only have two??? What are the rules? Hours of time spent with no answers, It's madeness.
WELL TRY TO UNDERSTAND THE DIFFERENT MEANINGS OF 'SOLVE THE EQNS'.
WE SHALL SEE IT AS WE DO THIS PROBLEM
A.NORMAL TO THE PLANE IS GIVEN BY DIRETION RATIOS (D.R.)WHICH ARE EQUAL TO COEFFICIENTS OF X,Y,Z IN THE EQN.SO NORMAL TO PLANE 1..IS....(1,3,1)
NORMAL TO PLANE 2 IS (2,7,-1)
B.IF THE D.R.S ARE IN PROPORTION THEN THE PLANES ARE PARALLEL.HERE WE FIND
1/2 IS NOT EQUAL TO 3/7 ETC...SO THEY ARE NOT PARALLEL.
C.IF SUM OF PRODUCT OF D.R S IS ZERO THEN THE PLANES ARE PERPENDICULAR.HERE
1*2+3*7-1*1 IS NOT ZERO.HENCE THE PLANES ARE NOT PERPENDICULAR.
D.NOW FINDING THE SOLUTION..THERE ARE 2 EQNS. ONLY AND THERE ARE 3 UNKNOWNS.
AS YOU WONDERED,IN SUCH A CASE WE CAN ONLY EXPRESS TWO OF THE UNKNOWNS INTERMS OF THE THIRD.SO LET US DO IT USING MATRICES AS YOU WANTED.
R1 1 3 1 1
R2 2 7 -1 1
NR2=R2-2*R1
1 3 1 1
0 1 -3 -1
NR1=R1-3*R2
1 0 10 4
0 1 -3 -1
THAT IS WE GOT
X+10Z=4
.OR
X=4-10Z.....OR......Z=(X-4)/-10................III
Y-3Z=-1
OR
.Y=3Z-1......OR......Z=(Y+1)/3............IV
E.EQN.OF LINE IS ALREADY GIVEN BY THE COMBINATION OF THE 2 PLANES EQNS.GIVEN ORIGINALY,SINCE THEIR INTERSECTION IS THE STRAIGHT LINE REQUIRED.
HOWEVER IF YOU MEN EQN. OF LINE IN PARAMETRIC FORM THEN ANSWER IS OBTAINED FROM ABOVE EQNS.III AND IV.....BY EQUATING COMMON VARIABLE Z .
(X-4)/-10=(Y+1)/3 = Z/1....................................V...............
WHICH TELLS US THAT THE LINE HAS D.R.S OF (-10,3,1)....THAT IS EQUAL TO THE DENOMINATORS IN THE EQN.V.
AND IT PASSES THROUGH THE POINT X-4=0..OR..X=4;Y+1=0..OR..Y=-1 AND Z-0=0... ....OR.....Z=0.....THAT IS BY EQUATING THE NUMERATORS TO ZERO IN THE EQN.V.
HOPE YOU GOT IT...IF IN DOUBT PLEASE COME BACK.
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