Sanai Ball

2022-09-07

I need to find the equation of a plane that passes through the point with position vector $A=5i+j+3k$ and the line ${l}_{1}=2i-8j-17k+\lambda \left(i-3j-4k\right)$
I found the normal by finding the cross product of $A\ast \left[\left(5i+j+3k\right)-\left(2i-8j-17k\right)\right]$ which turned out to be ${n}_{1}=-24i-32j+18k$. Then I used $r\cdot n=a\cdot n\to \left(5i+j+3k\right)\cdot \left(-24i-32j+18k\right)=-98$
However, it turns out that the above value should be 49 rather than −98. I realize that it can be divided by -2 to get the answer; however, I don't understand why this happens.

farbhas3t

−98 is correct if you use ${n}_{1}=-24i-32j+18k$ as normal vector. Then the equation of the plane is
$\begin{array}{}\text{(1)}& -24x-32y+18z=-98.\end{array}$
But you can divide the normal vector by −2, obtaining ${n}_{2}=12i+16j-9k$. For such normal vector the correct value is 49 and the equation of the plane is
$\begin{array}{}\text{(2)}& 12x+16y-9z=49.\end{array}$
Of course the equations (1) and (2) are equivalent and they represents the same plane.

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