Solve for z? 2x-4y+4z-6w=4 6w-4x+4y-4z=-12 6w+4x-2y+6z=64 4z+2w+6y-4x=56

z ≈ 9.22

Step-by-step explanation:

Given the equations

2x-4y+4z-6w=4 1

6w-4x+4y-4z=-12 2

6w+4x-2y+6z=64 3

4z+2w+6y-4x=56 4

We will first need to reduce the equation by cancelling out some variables.

Add equations 1 and 2 will give;

(2x-4x)+(-4y+4y)+(4z-4z)+ (-6w+6w) = 4+12

-2x +0 = 16

-2x = 16

x = -8

Also, equation 2 minus 3

6w-6w+(-4x-4x)+4y+2y+(-4z-6z) = 12-64

-8x+6y-10z = -52

-8(-8)+6y-10z = -52

64+6y-10z = -52

6y-10z  = -52-64

6y-10z = -116

3y-5z = -58 ... 5

Equation 3 * 1 and eqn 4 * 3

6w+4x-2y+6z=64 3

4z+2w+6y-4x=56 4

6w+4x-2y+6z=64

12z+6w+18y-12x= 168

Subtracting both equations;

16x-20y-6z = -104

8x-10y-3z = -52

8(-8)-10y-3z = -52

-64-10y-3z = -52

-10y-3z = -52+64

-10y-3z = 12 6

equating 5 and 6 and solving simultaneously;

3y-5z = -58 ... 5 * 10

-10y-3z = 12 6 * 3

30y-50z = -580

-30y-9z = 36

Add both equations

-50z-9z = -580+36

-59z = -544

z = -544/-59

z = 9.22

Hence z ≈ 9.22

the value of v.w= -12.

step-by-step explanation:

we are given three vector r,v and w as:

r = < 5, -5, -2> ; v = < 2, -8, -8> ; w = < -2, 6, -5>

we know that r,v and w could also be represented as:

r=5i-5j-2k

v=2i-8j-8k

and w= -2i+6j-5k

where i,j and k are the unit vectors of x,y and z-axis respectively.

also we know that it has the following property:

i.i=1 ; j.j=1; k.k=1

and i.=j.i=i.k=k.i=j.k=k.j=0

i.e. when we multiply two vectors or we find it's dot product we need to only multiply the components of x coordinates with each other; y with each other and z with each other.

hence, v.w= (2i-8j- -2i+6j-5k)

                  = 2×(-2)+(-8)×(6)+(-8)×(-5)

                  = -4-48+40

                  = -12

hence, v.w= -12

step-by-step explanation: