Why is the vector $v = i+j+4k$ perpendicular to the plane $x+y+4z=0$?

Question

Why is the vector $v = i+j+4k$ perpendicular to the plane given by $x+y+4z=0$ ?

I understand that given an equation defining a plane, you can just read off the coefficients for each variable to get a normal vector. It's been about 60 years since I learned this and would appreciate a refresher... thank you and God Bless!

Answer 1

If you have two vectors $a_1i+a_2j+a_3k$ and $b_1i+b_2j+b_3k$ in $\Bbb R^3$, then the angle $\theta$ between them is such that$$\cos\theta=\frac{a_1b_1+a_2b_2+a_3b_3}{\sqrt{a_1^{\,2}+a_2^{\,2}+a_3^{\,2}}\sqrt{b_1^{\,2}+b_2^{\,2}+b_3^{\,2}}}.$$

Now, take any element $xi+yj+zk$ from that plane. Then the angle $\theta$ between $xi+yj+zk$ and $i+j+4k$ is such that$$\cos\theta=\frac{x+y+4z}{\sqrt{x^2+y^2+z^2}\sqrt{6}}=0.$$So, $\theta=90^\circ$.

Answer 2

Let us consider the plane $P = \{(x,y,z)\in\mathbb{R}^{3} \mid x + y + 4z = 0\}$.

Clearly, $O = (0,0,0)\in P$, $A = (1,-1,0)\in P$ and $B = (0,-4,1)\in P$.

Let us denote $w = A - O = (1,-1,0)$ and $z = B - O = (0,-4,1)$.

Thus the lines \begin{align*} \begin{cases} L_{1} = \{(x,y,z)\in\mathbb{R}^{3} \mid (x,y,z) = O + tw\}\\\\ L_{2} = \{(x,y,z)\in\mathbb{R}^{3} \mid (x,y,z) = O + tz\} \end{cases} \end{align*} are contained in $P$, are distinct and are not parallel.

Moreover, we do also have that \begin{align*} \begin{cases} \langle v,w\rangle = \langle(1,1,4),(1,-1,0)\rangle = 1 - 1 + 0 = 0\\\\ \langle v,z\rangle = \langle(1,1,4),(0,-4,1)\rangle = 0 - 4 + 4 + 0 = 0 \end{cases} \end{align*}

Consequently, since $v\perp L_{1}$ and $v\perp L_{2}$, we conclude that $v\perp P$, and we are done.