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!!!