1.x Normal Vectors

What is a normal vector

• A vector that is normal/perpendicular to all the vectors

• How to find the normal vector

  • We can find the normal vector directly from the scalar equation

    • $ax+by=c$ we can take (a,b) as a normal vector
    • $ax+by+cz=d$ we can take the (a,b,c) as the normal vector
      • (we can take the coefficents as the normal vector)

EX.

• Find a scalar equation for the plane with a normal vector $\vec{n}=(1,2,3)$ that passes through the point $\vec{p}=(0,-1,2)$

$$\begin{array}{r}\vec{n}\ast \vec{x}=\vec{n}\ast \vec{p}⟹ax+by+cz=a{p}_1+b{p}_2+c{p}_3\\ \vec{n}=(1,2,3),\vec{p}=(p_1,p_2,p_3)=(0,-1,2)\\ x+2y+3z=(1)(0)+2(-1)+3(2)⟹x+2y+3z+4\end{array}$$

• Find a scalar equation for the plane that passes through the three points $P(1,1,1),Q(2,2,3),R(3,1,2)$

$$\begin{array}{r}\vec{P}\vec{Q}=Q-P=(1,1,2)\\ \vec{P}\vec{R}=R-P=(2,0,1)\\ \vec{x}=P+s\vec{P}\vec{Q}+t\vec{P}\vec{R}=\left[\begin{array}{c}1\\ 1\\ 1\end{array}\right]+s\left[\begin{array}{c}1\\ 1\\ 2\end{array}\right]+t\left[\begin{array}{c}2\\ 0\\ 1\end{array}\right]\end{array}$$

• To find a normal vector, take $\vec{n}=\vec{P}\vec{Q}cross\vec{P}\vec{R}$

$$\begin{array}{r}\vec{n}=(1,1,2)cross(2,0,1)=\left[\begin{array}{c}1\\ 3\\ -2\end{array}\right]\\ \vec{n}=(a,b,c)=(1,3,-2),P=(1,1,1)\\ \vec{n}\ast \vec{x}=\vec{n}\ast \vec{p}⟹x+3y-2z=2\end{array}$$