I have 4 points – $A(0, 0, 1)$, $B(2, 0, -1)$, $C(-1, 0, 2)$ and $D(1, -1, -2)$. I have to find the plane defined by the lines $AB$ and $CD$.
I started by comptuing the vectors $\vec{AB}$ and $\vec{CD}$ and I got that $\vec{AB} = (2, 0, -2)$ and that $\vec{CD}= (2, -1, -4)$. Then I computed the vector product for $\vec{AB}$ and $\vec{BC}$, which came out as $2 \vec i + 4 \vec j – 2 \vec k$. So the normal of the plane should be $(2, 4, -2)$ and those are $A$, $B$ and $C$ coefficients in the plane equation, which goes as such: $ax + by + cz + d = 0$. I don't understand how will I get the $d$ in the plane equation.
Best Answer
To get $D$, plug in a point on the plane i.e any one of $A,B,C$ or $D$. Also, you should use a different $D$ in your plane equation to not confuse things.