gsragator9

2022-09-30

Find the value of k for which the plane $kx+4y+2z-6=0$ is parallel to the line $\frac{x-3}{5}=\frac{y}{1}=\frac{z}{-3}$

So based on this I know the normal to the plane is $(k,4,2)$. I also know the direction vector of the line is $(5,1,-3)$

I need to find a vector that is perpendicular to the normal, and then somehow use that with the given direction vector to find k, but I'm not quite sure how.

So based on this I know the normal to the plane is $(k,4,2)$. I also know the direction vector of the line is $(5,1,-3)$

I need to find a vector that is perpendicular to the normal, and then somehow use that with the given direction vector to find k, but I'm not quite sure how.

Matteo Estes

Beginner2022-10-01Added 9 answers

Hint: Compute the dot product

$[k;4;2]\cdot [5;1;-3]$

This product must be zero, then you will get your $k$

$[k;4;2]\cdot [5;1;-3]$

This product must be zero, then you will get your $k$

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