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Let $L$ be a line that passes through points $a = (1,-1,-2)$ and $b =(2,-1,1)$. Let $V_1$ be the plane $x+y-3z+6=0$.

  1. Find the equation for $L$.
  2. Find the equation for the plane $V_2$ that is perpendicular to $V_1$ and contains the line $L$.

I have found $L = (x,y,z)=(b-a)t+a = (1,0,3)t+(1,-1,-2)$.

Thus the parallel vector to $L$ is $ab = (1,0,3)$

The normal vector to the plane $V_1$ can be taken from the coefficients of $x$,$y$ and $z$: $v_1 n = (1,1,-3)$

I now have $ab$ and $v_1 n$ if I find $ab \times v_1n$ I find a vector that is normal to $ab$ and $v_1n$ call this vector $v_2n = (-3,6,1)$.

$v_2n = (-3,6,1)$ is normal to $v_1n$ I tested this with the dot product and must lay in $V_1$ and also be the defining vector for $V_2$ as it is normal to $ab$.

Is my reasoning correct and how do I continue from here?

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Because $v_2 n = v_1 n \times ab$ you already know that $v_2 n$ is orthogonal to both $v_1 n$ and $ab$, so you don’t have to check this again with the dot product. Other than that your calculations so far seem good to me.

As you have already found the normal vector $v_2 n$ of $V_2$ all that’s left is to find $d \in \mathbb{R}$ such that $L$ is conainted in the plane $U_d$ given by $$ -3x + 6y + z + d = 0. $$ Because $ab$ is orthogonoal to $v_2 n$ the line $L$ is at least parallel to $U_d$ for every $d \in \mathbb{R}$. So to make sure that $L$ is acually contained in $U_d$ this plane we just need to make sure that some point of $L$ lies in $U_d$.

We take $(1,-1,-2) \in L$. Because $$ (-3) \cdot 1 + 6 \cdot (-1) + 1 \cdot (-2)+ d = -11 + d $$ we see that the point $(1,-1,-2)$ lies in $U_d$ if and only if $d = 11$. So the plane $V_2$ is $U_{11}$ and therefore given by the equation $$ -3x + 6y + z + 11 = 0. $$

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