Linear Algebra: Finding A k-value

In summary: The Attempt at a SolutionI figured that if I calculated \vec{n_1} \times \vec{n_2} = \vec{0}, and then take the magnitude of this vector equation, I could procure the correct k-value.Here are some parts of my work:\vec{n_1} \times \vec{n_2} = (16-6k)\hat{i} + (3k-8)\hat{k}0=320 -240k + 39k^2k = \frac{120 \pm 8 \sqrt{30}}{39}When I graphed the two planes, it was quite
  • #1
Bashyboy
1,421
5

Homework Statement


Find the value of k such that no solutions are shared between the planes

x + 2y + kz = 6

and

3x + 6y + 8z = 4

Homework Equations


The Attempt at a Solution



I figured that if I calculated [itex]\vec{n_1} \times \vec{n_2} = \vec{0}[/itex], and then take the magnitude of this vector equation, I could procure the correct k-value.

Here are some parts of my work:

[itex]\vec{n_1} \times \vec{n_2} = (16-6k)\hat{i} + (3k-8)\hat{k}[/itex]

[itex]0=320 -240k + 39k^2[/itex]

[itex]k = \frac{120 \pm 8 \sqrt{30}}{39}[/itex]

When I graphed the two planes, it was quite conspicuous that the two planes were not parallel.

What is wrong with my work and reasoning?
 
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  • #2
Bashyboy said:

Homework Statement


Find the value of k such that no solutions are shared between the planes

x + 2y + kz = 6

and

3x + 6y + 8z = 4

Homework Equations


The Attempt at a Solution



I figured that if I calculated [itex]\vec{n_1} \times \vec{n_2} = \vec{0}[/itex], and then take the magnitude of this vector equation, I could procure the correct k-value.

Here are some parts of my work:

[itex]\vec{n_1} \times \vec{n_2} = (16-6k)\hat{i} + (3k-8)\hat{k}[/itex]

[itex]0=320 -240k + 39k^2[/itex]

[itex]k = \frac{120 \pm 8 \sqrt{30}}{39}[/itex]

When I graphed the two planes, it was quite conspicuous that the two planes were not parallel.

What is wrong with my work and reasoning?

You're making this much harder than it needs to be. The only way the planes won't share any points is if they are parallel and distinct (not coplanar). This problem can be solved by inspection; you don't need cross products or much else for this problem.
 
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  • #3
You want ##\vec n_1 \times \vec n_2## to be the zero vector, so just set its components equal to 0.

Wouldn't it be easier to note that the planes are parallel if the normal vectors are proportional?
 
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  • #4
Bashyboy said:

Homework Statement


Find the value of k such that no solutions are shared between the planes

x + 2y + kz = 6

and

3x + 6y + 8z = 4


Homework Equations





The Attempt at a Solution



I figured that if I calculated [itex]\vec{n_1} \times \vec{n_2} = \vec{0}[/itex], and then take the magnitude of this vector equation, I could procure the correct k-value.

Here are some parts of my work:

[itex]\vec{n_1} \times \vec{n_2} = (16-6k)\hat{i} + (3k-8)\hat{k}[/itex]

[itex]0=320 -240k + \color{red}{39}{k^2}[/itex]

Also, that would have worked except you have 39 where you should have 45.
 
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  • #5
Thank you both for your help.
 
  • #6
Bashyboy said:
Here are some parts of my work:

[itex]\vec{n_1} \times \vec{n_2} = (16-6k)\hat{i} + (3k-8)\hat{k}[/itex]

[itex]0=320 -240k + 39k^2[/itex]
k (scalar parameter) and ##\hat{k}## (direction in space) are completely unrelated, they just share the same letter here. Both brackets have to be zero.

I agree with Mark44, this can be solved just by looking at it.
 

1. What is a k-value in linear algebra?

A k-value in linear algebra refers to the constant multiplier in front of a variable in an equation. It is also known as the slope or rate of change, and it is used to determine how the variable changes in relation to the other variables in the equation.

2. How do you find the k-value in a linear equation?

To find the k-value in a linear equation, you can use the slope formula: k = (y2 - y1)/(x2 - x1). This formula uses two points on the line to calculate the slope, which is the k-value.

3. Why is finding the k-value important in linear algebra?

Finding the k-value is important in linear algebra because it allows us to understand how the variables in an equation are related to each other. It also helps us to graph the equation and make predictions about the behavior of the variables.

4. Can the k-value be negative in a linear equation?

Yes, the k-value can be negative in a linear equation. This indicates that the line is decreasing or sloping downwards from left to right on a graph. A positive k-value indicates an increasing or upward sloping line.

5. How does changing the k-value affect the graph of a linear equation?

Changing the k-value in a linear equation affects the slope or steepness of the line. A larger k-value will result in a steeper line, while a smaller k-value will result in a flatter line. A negative k-value will also result in a line that slopes downwards, while a positive k-value will result in a line that slopes upwards.

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