# Linear Algebra: Finding A k-value

## 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

## The Attempt at a Solution

I 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:

$\vec{n_1} \times \vec{n_2} = (16-6k)\hat{i} + (3k-8)\hat{k}$

$0=320 -240k + 39k^2$

$k = \frac{120 \pm 8 \sqrt{30}}{39}$

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?

Mark44
Mentor

## 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

## The Attempt at a Solution

I 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:

$\vec{n_1} \times \vec{n_2} = (16-6k)\hat{i} + (3k-8)\hat{k}$

$0=320 -240k + 39k^2$

$k = \frac{120 \pm 8 \sqrt{30}}{39}$

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.

• 1 person
LCKurtz
Homework Helper
Gold Member
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?

Last edited:
• 1 person
LCKurtz
Homework Helper
Gold Member

## 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

## The Attempt at a Solution

I 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:

$\vec{n_1} \times \vec{n_2} = (16-6k)\hat{i} + (3k-8)\hat{k}$

$0=320 -240k + \color{red}{39}{k^2}$

Also, that would have worked except you have 39 where you should have 45.

• 1 person
Thank you both for your help.

mfb
Mentor
Here are some parts of my work:

$\vec{n_1} \times \vec{n_2} = (16-6k)\hat{i} + (3k-8)\hat{k}$

$0=320 -240k + 39k^2$
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.