st3he1d0t

2022-10-05

A plane through the origin is perpendicular to the plane 2x−y−z=5 and parallel to the line joining the points (1,2,3) and (4,−1,2). Find the equation of the plane.

vakleraarrc

Beginner2022-10-06Added 6 answers

We know that the required plane's normal is perpendicular to both the given plane's normal (2,−1,−1) and the line's direction vector (3,−3,−1). We can indeed perform a cross product to get the required plane's normal, since its result is perpendicular to both its inputs:

$(2,-1,-1)\times (3,-3,1)=\dots $

$(2,-1,-1)\times (3,-3,1)=\dots $

Riya Andrews

Beginner2022-10-07Added 4 answers

You could as well find the parametric form of the plane:

$x=\lambda \left(\begin{array}{c}2\\ -1\\ -1\end{array}\right)+\mu \left(\begin{array}{c}3\\ -3\\ 1\end{array}\right),\lambda ,\mu \in \mathbb{R}$

where the first direction vector ( the normal vector of the plane) indicates this plane is perpendicular to the given one and the second is the direction vector of the given line to which it is parallel and thus as pointed out by Parcly Taxel a normal vector of the desired plane is given by

$\left(\begin{array}{c}2\\ -1\\ -1\end{array}\right)\times \left(\begin{array}{c}3\\ -3\\ 1\end{array}\right)=\left(\begin{array}{c}-4\\ -5\\ -3\end{array}\right)$

from what You get the coordinate equation of the plane:

$4x+5y+3z=0.$

$x=\lambda \left(\begin{array}{c}2\\ -1\\ -1\end{array}\right)+\mu \left(\begin{array}{c}3\\ -3\\ 1\end{array}\right),\lambda ,\mu \in \mathbb{R}$

where the first direction vector ( the normal vector of the plane) indicates this plane is perpendicular to the given one and the second is the direction vector of the given line to which it is parallel and thus as pointed out by Parcly Taxel a normal vector of the desired plane is given by

$\left(\begin{array}{c}2\\ -1\\ -1\end{array}\right)\times \left(\begin{array}{c}3\\ -3\\ 1\end{array}\right)=\left(\begin{array}{c}-4\\ -5\\ -3\end{array}\right)$

from what You get the coordinate equation of the plane:

$4x+5y+3z=0.$

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