Matrices homework

[Maarie_Sophie]

Hello everyone ! I really need help with these matrices homework.Thank you all of you for the great work you’re doing and I really need your help please.
The chapters these exercises are from is Orthogonality,Cross product, Norm, Dot Product, and distance in Rn and vector in 2 space, 3 space and n space

 

[Deleted member 4993]

Hello everyone ! I really need help with these matrices homework.Thank you all of you for the great work you’re doing and I really need your help please. The chapters these exercises are from is Orthogonality,Cross product, Norm, Dot Product, and distance in Rn and vector in 2 space, 3 space and n space

You have posted 12 problems and you have not shown any work!!

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Please share your work/thoughts about this problem.

 

[HallsofIvy]

I would also point out that none of these have anything to do with "probability and statistics" and, while you could use "matrices" to solve them, it is not necessary and I certainly would not.

For example, the first problem is
"Find the equation of the plane passing through (-4, -1, -1), (-2, 0, 1), and (1, 1, 2)".

Any plane can be written as Ax+ By+ Cz= 1 for some numbers A, B, and C. To solve for A, B, and C we need three equations and we get those by putting in the x, y, z value of those three points:
A(-4)+ B(-1)+ C(-1)= -4A- B- C= 1.
A(-2)+ B(0)+ C(1)= -2A+ C= 1.
A(1)+ B(1)+ C(2)= A+ B+ 2C= 1.

Seeing that the second equation has no "B" term, while the first equation has "-B" and the third has "+B", I would first eliminate B by adding the first and third equations: -3A+ C=2. Now eliminate C by Subtracting that from -2A+C= 1: A= -1. Then -2A+ C= 2+ C=1 so C=1- 2= -1. And A+ B+ 2C= -1+ B- 2= B- 3= 1 so B= 4.

The plane that passes through (-4, -1, -1), (-2, 0, 1), and (1, 1, 2)
is -x+ 4y- z= 1.

Check:
-(-4)+ 4(-1)- (-1)= 4- 4+ 1= 1.
-(-2)+ 4(0)- (1)= 2- 1= 1.
-(1)+ 4(1)- (2)= -1+ 4- 2= 1.

 

[HallsofIvy]

The second equation asks you to determine whether the given planes are parallel or perpendicular. Do you know what those mean?

The first pair of equations is 4x- y+ 2z= 5 and 7x- 3y+ 4z= 0.
Do you understand that <4, -1, 2> and <7, -3, 4> are the "normal" vectors to the two planes? Are those two vectors perpendicular or parallel?

 

[HallsofIvy]

For problem 3, I presume you know that two lines, or two planes, or a line and a plane, are "parallel" if and only if they do not intersect. So I would start by replacing, x, y, and z in the equation of the plane by their exprssions in "t" from the line. That gives an equation in the single variable, t. If that equation can be solved for t, that that "t" gives the point of itersection. If there is the equation can not be solved then there is no intersection and the line and plane are parallel.

 

[HallsofIvy]

The fourth problem asks you find the line of intersection of 7x- 2y+ 3z= -4 and -3x+ y+ 2z+ 5= 0.

I assume that you know that you can, generally, solve three equations for three values of x, y, and z. But here there are only two equations! So treat one, z, say, as a constant and solve the two equations for the other two (x and y in terms of z). The write z=t so we can express x, y, and z in terms of the parameter, t.