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Question about linear algebra equation conversion

62 viewed last edited 1 year ago
Anonymous
0

Please press on the link


https://ibb.co/z7LGjFH

I’m doing some homework and I have come across a question that I can’t find any online help for so I’ve come here in hope that someone is very good at this.

The link leads to an image showing how far I’ve gotten, any help would be greatly appreciated.


Thanks and have a great day!


Krishna
2

Step 1:  Recall the Cartesian equation of the plane

            Equation:  Ax + By + Cz + D = 0

            Where A,B,C and D are constants and not all A,B,C are zero, can be taken

            to be an equation of a plane in space. The coefficients A, B and C are the

          components of a normal vector  

              

            

Step 2: Note down the coordinates of the three points from the given equations

              GIVEN \frac{x-4}{5} = \frac{y - 5}{8} = \frac{z - 10}{3}

                            And (6, 9, 4)

              

            POINTS:     \frac{x-4}{5} = \frac{y - 5}{8} = \frac{z - 10}{3} = t       \because assume\ t


                            Now \frac{x-4}{5} = t  \Rightarrow  x = 5t + 4      

            

                                     \frac{y - 5}{8} = t   \Rightarrow y = 8t + 5


                                       \frac{z - 10}{3}= t \Rightarrow z = 3t + 10


                      take t = 0

                  Then,  x = 5t + 4 \Rightarrow x = 5(0) + 4 \Rightarrow x = 4

                              y = 8t + 5 \Rightarrow y = 8(0) + 5 \Rightarrow y = 5

                              z = 3t + 10 \Rightarrow z = 3(0) + 10 \Rightarrow z = 10

                  So the point (x, y, z) = (4, 5, 10)


                    Similarly, take  x = 1      

                      then, the other point is (x, y, z) = (9, 13, 13)


            Therefore, the points are

                    A(4, 5, 10), B(9, 13,13) and C(6, 9, 4)


Step 3: Find the normal of the vectors AB and AC

                        A(4,5,10), B(9, 13, 13)

                       AB = A - B = (-5, -8, -3)      

                                          

                      A(4, 5, 10), C(6, 9, 4)

                        AC =  A - C = (-2, -4, 6)

                                                                                                          

                    normal vector (n) = AB X AC = \left( \begin{array}{ccc} i & j & k \\ -5 & -8 & -3 \\ -2 & -4 & 6 \end{array} \right)

                                              

                                                =[math]i\left[\left(-8\right)\left(6\right)\ -\left(-4)\left(-3\right)\right]\ -\ j\left[\left(-5\right)\left(6\right)-\left(-2\right)\left(-3\right)\right]\ +\ k\left[\left(-5\right)\left(-4\right)\ -\ \left(-8\right)\left(-2\right)\right]\right][/math]


                                                = i\left(-48\ -12\right)-\ j\left(-30\ -\ 6\right)\ +k\left(20\ -16\right)


                                                = (-60i+36j+4k)


                                                = 4\left(-15i+9j+k\right)


Step 4: Determine the Cartesian equation by using the normal

                                Ax + By + Cz + D =0


                                  n =  -15i+9j+k  


                                n=(A,B,C)=(-15,\ 9,\ 1)


                      So, The Cartesian equation

                                  -15x + 9y + z + D = 0


                        At point (6, 9, 4) find D


                                  -15(6)+9(9)+1(4)+D=0


                                  -90+81+4+D=0


                                  -5+D=0


                                  D=5

        

                      The Cartesian equation

                              -15x+9y+z+5=0


Anonymous
1
Thank you .I appreciate this!