Vectors in Planes: Solving for m & n, Finding Orthogonal Plane

[Lancelot1]

Hiya,

I have a question in high school math, vectors to be precise:

The point K(3.5,-0.5,-2) is the intersection point of the plane S1: mx-7my+5z+3=0 with the line l1: (n,1,-2)+t(-1,1,0)

a) find the parameters n and m
b) find the equation of the plane that contains the line l1 and is orthogonal to the plane S1.
c) Another line is given, l2: (2,0,-1)+t(3,-3,0). Find the area of the parallelogram which has two vertices on l2 and to other vertices are K, and another point E(1,2,-2) which is on l1.

I managed to solve (a) and found that m=1, n=2 and t=-1.5.

I need your assistance with the other part of the question.

Thank you.

 

[MarkFL]

Hiya,

I have a question in high school math, vectors to be precise:

The point K(3.5,-0.5,-2) is the intersection point of the plane S1: mx-7my+5z+3=0 with the line l1: (n,1,-2)+t(-1,1,0)

a) find the parameters n and m
b) find the equation of the plane that contains the line l1 and is orthogonal to the plane S1.
c) Another line is given, l2: (2,0,-1)+t(3,-3,0). Find the area of the parallelogram which has two vertices on l2 and to other vertices are K, and another point E(1,2,-2) which is on l1.

I managed to solve (a) and found that m=1, n=2 and t=-1.5.

I need your assistance with the other part of the question.

Thank you.

a) To find the parameters $n$ and $m$, we can plug in the coordinates of the given point to the plane and the line:

\(\displaystyle m(3.5)-7m(-0.5)+5(-2)+3=0\)

Solving for $m$, we find:

\(\displaystyle m=1\quad\checkmark\)

\(\displaystyle n-t=3.5\)

\(\displaystyle t+1=-0.5\)

Adding, we obtain:

\(\displaystyle n+1=3\implies n=2\quad\checkmark\)

b) To find a plane, all we need is a point in the plane, and a vector perpendicular to the plane. Can you find those?

 

[HallsofIvy]

For (b) let the unknown plane be Ax+ By+ Cz= D. Since it is perpendicular to mx- 7my+ 5z+ 3= 0, its normal vector must be perpendicular to the normal vector of that plane: mA- 7mB+ 5C= 0. The fact that line (n,1,-2)+t(-1,1,0) lies in the plane means that A(n- t)+ B(1+ t)- 2C= D for all t. We already know that m= 1 and n= 2 so those equations become A- 14B+ 5C= 0 and A(2- t)+ B(1+ t)- 2C= 2A- At+ B+ Bt- 2C= (B- A)t+ 2A+ B- 2C= D for all t. Since that last equation is true for all t, corresponding coefficients must be equal: B- A= 0, 2A+ B- 2C= D.

We have the three equations A- 14B+ 5C= 0, B- A= 0, and 2A+ B- 2C= D. Normally, three equations would not be enough to solve for four unknowns, but we could always add or multiply Ax+ By+ Cz= D by a non-zero number to get another equation for the same plane. In particular, we can assume, say, that D= 1 so we have A- 14B+ 5C= 0, B- A= 0, and 2A+ B- 2C= 1, three equations to solve for A, B, and C.