Hi, first poster here.(adsbygoogle = window.adsbygoogle || []).push({});

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 vectorbparallel to the line L.

Would the equation for L in vector form just be [(8-7r)/3]i+ [(5r-1)/3]j+ rk?

Vectorbis parallel to that if you just add a constant, correct?

(c) (The tricky one).

Write down normal vectorsn1,n2,n3for the planes P1, P2, P3 respectively. Obtain a vectorcthat lies along the line of intersection of P1 and P2, by using the vectorsn1andn2.

Iscparallel to the vectorbin (b) above? Should it be?

I have gotn1,n2andn3through 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 vectorcand if it should be parallel to vectorbas in Part B.

All help is GREATLY appreciated!!!

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# Homework Help: Planes and Vectors Help

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