- #1
Muzly
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Hi, first poster here.
I'm stuck on a question in first-year University Maths. I'll put down what I know where I've done it.
Question: Consider these linear equations:
x + 2y - z = 2
2x + y + 3z = 5
x + 5y - 6z = 1
for the planes P1, P2 and P3 respectively.
(a) Using Gaussian elimination, show that there is one free variable, and that the solutions lie along a line L in three dimensions.
Obviously, I'm not going to put down the whole algorithm, but my solution set is as follows:
(S1,S2,S3) = ( (8-7r)/3 , (5r-1)/3 , r )
Where r is our free variable.
Any ideas on how to show (prove) that the equation lies along a line in three dimensions? Or will that be enough?
(b) Write the equation for L in vector form, and find a vector b parallel to the line L.
Would the equation for L in vector form just be [(8-7r)/3]i + [(5r-1)/3]j + rk?
Vector b is parallel to that if you just add a constant, correct?
(c) (The tricky one).
Write down normal vectors n1, n2, n3 for the planes P1, P2, P3 respectively. Obtain a vector c that lies along the line of intersection of P1 and P2, by using the vectors n1 and n2.
Is c parallel to the vector b in (b) above? Should it be?
I have got n1, n2 and n3 through finding the vector equation for P1, P2 and P3, and using cross-multiplication.
The vector forms are:
P1 - 2i + j - 2k
P2 - (2/5)i + (1/5)j + (3/5)k
P3 - i + (1/5)j - (1/6)k
Using X, Y and Z intercepts, and labelling each point A, B, and C respectively, I obtained the following vectors (I'll only show for normal vector for P1, too much to type here!)
A->B = (-2, 1, 0)
A->C = (-2, 0, -2)
Cross multiply these and you'll get:
n1 = 2i + 4j - 2k
Also,
n2 = (3/25)i + (6/25)j + (2/25)k
n3 = (-1/30)i - (1/6)j + (1/5)k
Is anyone able to just check my work (to make sure the response to my problem is correct), and able to explain to me clearly about how to obtain the vector c and if it should be parallel to vector b as in Part B.
All help is GREATLY appreciated!
I'm stuck on a question in first-year University Maths. I'll put down what I know where I've done it.
Question: Consider these linear equations:
x + 2y - z = 2
2x + y + 3z = 5
x + 5y - 6z = 1
for the planes P1, P2 and P3 respectively.
(a) Using Gaussian elimination, show that there is one free variable, and that the solutions lie along a line L in three dimensions.
Obviously, I'm not going to put down the whole algorithm, but my solution set is as follows:
(S1,S2,S3) = ( (8-7r)/3 , (5r-1)/3 , r )
Where r is our free variable.
Any ideas on how to show (prove) that the equation lies along a line in three dimensions? Or will that be enough?
(b) Write the equation for L in vector form, and find a vector b parallel to the line L.
Would the equation for L in vector form just be [(8-7r)/3]i + [(5r-1)/3]j + rk?
Vector b is parallel to that if you just add a constant, correct?
(c) (The tricky one).
Write down normal vectors n1, n2, n3 for the planes P1, P2, P3 respectively. Obtain a vector c that lies along the line of intersection of P1 and P2, by using the vectors n1 and n2.
Is c parallel to the vector b in (b) above? Should it be?
I have got n1, n2 and n3 through finding the vector equation for P1, P2 and P3, and using cross-multiplication.
The vector forms are:
P1 - 2i + j - 2k
P2 - (2/5)i + (1/5)j + (3/5)k
P3 - i + (1/5)j - (1/6)k
Using X, Y and Z intercepts, and labelling each point A, B, and C respectively, I obtained the following vectors (I'll only show for normal vector for P1, too much to type here!)
A->B = (-2, 1, 0)
A->C = (-2, 0, -2)
Cross multiply these and you'll get:
n1 = 2i + 4j - 2k
Also,
n2 = (3/25)i + (6/25)j + (2/25)k
n3 = (-1/30)i - (1/6)j + (1/5)k
Is anyone able to just check my work (to make sure the response to my problem is correct), and able to explain to me clearly about how to obtain the vector c and if it should be parallel to vector b as in Part B.
All help is GREATLY appreciated!