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?


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 $$ -3x + 6y + z + 11 = 0. $$


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.