An abstract geometric representation of a three-dimensional plane described by the equation r=(k,3k,12) +s(2,2,3)+t(-1,1,3), where parameters s and t are part of real numbers. The plane passes through the origin. The plane's color is ideally defined by the value of k. Include small 3-dimensional shapes in different colors representing the values a) 2, b) 1, c) 0, d) -1. Ensure no textual information is contained in the image.

Which value of k will make the plane r=(k,3k,12) +s(2,2,3)+t(-1,1,3), s,teR passing through the origin?

a) 2
b) 1
c)0
d) -1

to pass through the origin, r = (0,0,0)

so

k + 2s - t = 0
3k + 2s + t = 0
12 + 3s + 3t = 0 --> 4 + s + t = 0

add 1st and 2nd ---> 4k + 4s = 0 or k = -s
add 1st and 3rd ---> k+4 + 3s = 0

sub in k=-s
-s + 4 + 3s = 0
s = - 2
then k = 2

check:
if k= 2 and s = - 2, then in 3rd
4 - 2 + t = 0
t = -2

r = (2, 6, 12) -2(2,2,3) -2(-1,1,3)
= (2-4+2 , 6-4-2 , 12-6-6 )
= (0,0,0)

yeah!!!

Why did the plane go through therapy? Because it had trouble finding its origin!

To make the plane pass through the origin, we need to find the value of k that makes the coordinates (k, 3k, 12) satisfy the equation (0, 0, 0) = (k, 3k, 12) + s(2, 2, 3) + t(-1, 1, 3).

Setting up the equations, we have:
0 = k + 2s - t
0 = 3k + 2s + t
0 = 12 + 3s + 3t

To solve these equations, we'll use a bit of humor math!

Let's eliminate t from the first two equations by adding them together:
0 = k + 2s - t
0 = 3k + 2s + t
_________________________________
0 = 4k + 4s

Now let's eliminate t from the first and third equations:
0 = k + 2s - t
0 = 12 + 3s + 3t
_________________________________
0 = k + 2s - t - 12 - 3s - 3t
0 = k - 3s - 4t - 12

Now, we have two equations:
0 = 4k + 4s
0 = k - 3s - 4t - 12

To find the value of k, let's eliminate s by multiplying the second equation by -1 and adding them together:

0 = 4k + 4s
0 = -(k - 3s - 4t - 12)
_______________________
0 = 4k + 4s - k + 3s + 4t + 12
0 = 3k + 7s + 4t + 12

Now, we have two equations:
0 = 4k + 4s
0 = 3k + 7s + 4t + 12

To further simplify, we can divide the first equation by 4 to get:
0 = k + s

Substituting this into the second equation:
0 = 3(k+s) + 7s + 4t + 12
0 = 3k + 3s + 7s + 4t + 12
0 = 3k + 10s + 4t + 12

But since we know that k + s = 0, we can replace k+s with 0:
0 = 3(0) + 10s + 4t + 12
0 = 10s + 4t + 12

Now, let's solve for t:
t = -3s - 3

Substitute the value of t back into our original equation:
0 = k + s
0 = 4k + 4s

0 = 4k + 4s
0 = 4k + 4s
____________________
0 = 4k + 4s

So, k could be any value! That means option (a) 2, option (b) 1, option (c) 0, or option (d) -1 could all potentially make the plane pass through the origin.

Hope that humor helped you through this math problem! Keep smiling!

To find the value of k that will make the plane pass through the origin, we need to substitute the origin, (0, 0, 0), into the equation of the plane and solve for k.

Let's substitute (0, 0, 0) into the equation of the plane:

r = (k, 3k, 12) + s(2, 2, 3) + t(-1, 1, 3)

Substituting (0, 0, 0) into the equation of the plane, we get:

(0, 0, 0) = (k, 3k, 12) + s(2, 2, 3) + t(-1, 1, 3)

Simplifying the equation:

(0, 0, 0) = (k, 3k, 12) + (2s, 2s, 3s) + (-t, t, 3t)

Now we can equate the corresponding components:

0 = k + 2s - t (1)
0 = 3k + 2s + t (2)
0 = 12 + 3s + 3t (3)

We can solve this system of linear equations to find the value of k.

From equation (1):
-k = 2s - t

Substituting this into equation (3):
0 = 12 + 3s + 3t
0 = 12 + 3s + 3(2s - t)
0 = 12 + 3s + 6s - 3t
0 = 9s + 6s - 3t + 12
0 = 15s - 3t + 12 (4)

From equation (2):
3k = -t - 2s

Substituting this into equation (4):
0 = 15s - 3(-t - 2s) + 12
0 = 15s + 3t + 6s + 12
0 = 21s + 3t + 12

Now we can use this equation to eliminate t:

From equation (1):
-k = 2s - t

Rearranging it:
t = 2s + k

Substituting this into the equation above:
0 = 21s + 3(2s + k) + 12
0 = 21s + 6s + 3k + 12
0 = 27s + 3k + 12 (5)

Now we have two equations, (4) and (5), with two unknowns (s and k). We can solve this system of equations:

Equation (5) can be rewritten as:
27s = -3k - 12

Substituting this into equation (4):
0 = 15s - 3t + 12
0 = 15(-3k - 12) - 3t + 12
0 = -45k - 180 - 3t + 12
0 = -45k - 3t - 168 (6)

We need to find the value of k that satisfies both equation (5) and (6). Since the question asks for the value of k, we can solve equation (5) for k:

27s = -3k - 12
3k = -27s - 12
k = (-27s - 12) / 3
k = -9s - 4

Substituting this value of k into equation (6):
0 = -45(-9s - 4) - 3t - 168
0 = 405s + 180 - 3t - 168
0 = 405s - 3t + 12 (7)

Now we have equation (7) with s and t as variables. However, we only need to find the value of k, not s and t. Since the equation does not involve k, we can disregard it.

Therefore, the value of k that will make the plane pass through the origin is not determined by the given system of equations. So, none of the given options (a) 2, (b) 1, (c) 0, or (d) -1 are correct.

To determine the value of k that will make the plane pass through the origin, we need to substitute the coordinates of the origin (0, 0, 0) into the equation of the plane.

First, we have the equation of the plane in vector form:
r = (k, 3k, 12) + s(2, 2, 3) + t(-1, 1, 3)

Now, plug in the coordinates of the origin (0, 0, 0) into the equation:
(0, 0, 0) = (k, 3k, 12) + s(2, 2, 3) + t(-1, 1, 3)

Now we can separate the x, y, and z components of the equation:
0 = k + 2s - t
0 = 3k + 2s + t
0 = 12 + 3s + 3t

Now, we have a system of equations:
k + 2s - t = 0
3k + 2s + t = 0
12 + 3s + 3t = 0

To solve this system, we can use matrix representation.

Rewrite the system of equations in matrix form:
| 1 2 -1 | | k | | 0 |
| 3 2 1 | * | s | = | 0 |
| 0 3 3 | | t | | 12 |

To find the values of k, s, and t that satisfy this system, we need to row-reduce the augmented matrix. After performing row reduction, we get:

| 1 0 -1 | | k | | 0 |
| 0 1 1 | * | s | = | 0 |
| 0 0 5 | | t | | 12 |

From the row-reduced echelon form, we can see that the system is consistent if and only if the last row of the coefficient matrix reduces to a non-zero row. In this case, it reduces to [0 0 5], which implies that the system is consistent.

The third equation, 0 = 5t, tells us that t = 0.

Now, substitute t = 0 into the first two equations:
k - s = 0
3k + s = 0

From the first equation, we get k = s.

Substitute k = s into the second equation:
3s + s = 0
4s = 0
s = 0

Since s = 0, we have k = s = 0.

Therefore, the value of k that will make the plane pass through the origin is k = 0.

So, the correct choice is c) 0.