Recall the definition of the Unit Normal Vector.

$\displaystyle V_n=\frac{V}{||V||}$

Let $\displaystyle V=(3,3,4)$

$\displaystyle ||V||=\sqrt{3^2+3^2+4^2}=\sqrt{34}$

$\displaystyle V_n=(\frac{3}{\sqrt{34}}, \frac{3}{\sqrt{34}}, \frac{4}{\sqrt{34}} )$

To find the unit normal vector, you must first find the unit tangent vector.  The equation for the unit tangent vector, $\displaystyle T(t)$,  is

$\displaystyle T(t)=\frac{r(t)}{\left \| r(t)\right \|}$

where $\displaystyle r(t)$ is the vector and $\displaystyle \left \| r(t)\right \|$ is the magnitude of the vector.

The equation for the unit normal vector,$\displaystyle N(t)$,  is 

$\displaystyle N(t)=\frac{T'(t)}{\left \| T'(t)\right \|}$

where $\displaystyle T'(t)$ is the derivative of the unit tangent vector and $\displaystyle \left \| T'(t)\right \|$ is the magnitude of the derivative of the unit vector.

For this problem

$\displaystyle v(t)=t^{2}\hat{i}+3t^{2}\hat{j}$

$\displaystyle \left \| v(t)\right \|=\sqrt{\left (t^{2} \right )^{2}+\left (3t^{2} \right )^{2}}=\sqrt{t^{4}+9t^{4}}=\sqrt{10t^{4}}=t^{2}\sqrt{10}$

$\displaystyle T(t)=\frac{v(t)}{\left \| v(t)\right \|}=\frac{t^{2}\hat{i}+3t^{2}\hat{j}}{t^{2}\sqrt{10}}=\frac{1}{\sqrt{10}}\hat{i}+\frac{3}{\sqrt{10}}\hat{j}$

$\displaystyle T'(t)=\frac{d}{dt}\left ( \frac{1}{\sqrt{10}}\hat{i}+\frac{3}{\sqrt{10}}\hat{j} \right )=0$

$\displaystyle \left \| T'(t)\right \|=\sqrt{\left (0\right )^{2}}=0$

$\displaystyle N(t)=\frac{T'(t)}{\left \| T'(t)\right \|}=\frac{0}{0}=DNE$

There is no unit normal vector of $\displaystyle v(t)=t^{2}\hat{i}+3t^{2}\hat{j}$.

To find the unit normal vector, you must first find the unit tangent vector.  The equation for the unit tangent vector, $\displaystyle T(t)$,  is

$\displaystyle T(t)=\frac{r(t)}{\left \| r(t)\right \|}$

where $\displaystyle r(t)$ is the vector and $\displaystyle \left \| r(t)\right \|$ is the magnitude of the vector.

The equation for the unit normal vector,$\displaystyle N(t)$,  is 

$\displaystyle N(t)=\frac{T'(t)}{\left \| T'(t)\right \|}$

where $\displaystyle T'(t)$ is the derivative of the unit tangent vector and $\displaystyle \left \| T'(t)\right \|$ is the magnitude of the derivative of the unit vector.

For this problem

$\displaystyle v(t)=sin(t)\hat{i}+cos(t)\hat{j}$

$\displaystyle \left \| v(t)\right \|=\sqrt{\left (sin(t) \right )^{2}+\left (cos(t) \right )^{2}}=\sqrt{1}=1$

$\displaystyle T(t)=\frac{v(t)}{\left \| v(t)\right \|}=\frac{sin(t)\hat{i}+cos(t)\hat{j}}{1}=sin(t)\hat{i}+cos(t)\hat{j}$

$\displaystyle T'(t)=\frac{d}{dt}\left (sin(t)\hat{i}+cos(t)\hat{j} \right )=cos(t)\hat{i}-sin(t)\hat{j}$

$\displaystyle \left \| T'(t)\right \|=\sqrt{\left (cos(t) \right )^{2}+\left (sin(t) \right )^{2}}=\sqrt{1}=1$

$\displaystyle N(t)=\frac{T'(t)}{\left \| T'(t)\right \|}=\frac{cos(t)\hat{i}-sin(t)\hat{j}}{1}=cos(t)\hat{i}-sin(t)\hat{j}$

To find the unit normal vector, you must first find the unit tangent vector.  The equation for the unit tangent vector, $\displaystyle T(t)$,  is

$\displaystyle T(t)=\frac{r(t)}{\left \| r(t)\right \|}$

where $\displaystyle r(t)$ is the vector and $\displaystyle \left \| r(t)\right \|$ is the magnitude of the vector.

The equation for the unit normal vector,$\displaystyle N(t)$,  is 

$\displaystyle N(t)=\frac{T'(t)}{\left \| T'(t)\right \|}$

where $\displaystyle T'(t)$ is the derivative of the unit tangent vector and $\displaystyle \left \| T'(t)\right \|$ is the magnitude of the derivative of the unit vector.

For this problem

$\displaystyle v(t)=e^{t}\hat{i}+2e^{t}\hat{j}$

$\displaystyle \left \| v(t)\right \|=\sqrt{\left (e^{t} \right )^{2}+\left (2e^{t} \right )^{2}}=\sqrt{e^{2t}+4e^{2t}}=\sqrt{5e^{2t}}=e^{t}\sqrt{5}$

$\displaystyle T(t)=\frac{v(t)}{\left \| v(t)\right \|}=\frac{e^{t}\hat{i}+2e^{t}\hat{j}}{e^{t}\sqrt{5}}=\frac{1}{\sqrt{5}}\hat{i}+\frac{2}{\sqrt{5}}\hat{j}$

$\displaystyle T'(t)=\frac{d}{dt}\left (\frac{1}{\sqrt{5}}\hat{i}+\frac{2}{\sqrt{5}}\hat{j} \right )=0$

$\displaystyle \left \| T'(t)\right \|=0$

$\displaystyle N(t)=\frac{T'(t)}{\left \| T'(t)\right \|}=\frac{0}{0}=DNE$

The normal vector of $\displaystyle v(t)=e^{t}\hat{i}+2e^{t}\hat{j}$ does not exist.

To find the unit normal vector, you must first find the unit tangent vector.  The equation for the unit tangent vector, $\displaystyle T(t)$,  is

$\displaystyle T(t)=\frac{r(t)}{\left \| r(t)\right \|}$

where $\displaystyle r(t)$ is the vector and $\displaystyle \left \| r(t)\right \|$ is the magnitude of the vector.

The equation for the unit normal vector,$\displaystyle N(t)$,  is 

$\displaystyle N(t)=\frac{T'(t)}{\left \| T'(t)\right \|}$

where $\displaystyle T'(t)$ is the derivative of the unit tangent vector and $\displaystyle \left \| T'(t)\right \|$ is the magnitude of the derivative of the unit vector.

For this problem

$\displaystyle v(t)=5cos(t)\hat{i}+5sin(t)\hat{j}+2\hat{k}$

$\displaystyle \left \| v(t)\right \|=\sqrt{\left (5cos(t) \right )^{2}+\left (5sin(t) \right )^{2}+(2)^{2}}=\sqrt{25+4}=\sqrt{29}$

$\displaystyle T(t)=\frac{v(t)}{\left \| v(t)\right \|}=\frac{5cos(t)\hat{i}+5sin(t)\hat{j}+2\hat{k}}{\sqrt{29}}$

$\displaystyle T(t)=\frac{v(t)}{\left \| v(t)\right \|}=\frac{5}{\sqrt{29}}cos(t)\hat{i}+\frac{5}{\sqrt{29}}sin(t)\hat{j}+\frac{2}{\sqrt{29}}\hat{k}$

$\displaystyle T'(t)=\frac{d}{dt}\left (\frac{5}{\sqrt{29}}cos(t)\hat{i}+\frac{5}{\sqrt{29}}sin(t)\hat{j}+\frac{2}{\sqrt{29}}\hat{k} \right )$

$\displaystyle T'(t)=\frac{-5}{\sqrt{29}}sin(t)\hat{i}+\frac{5}{\sqrt{29}}cos(t)\hat{j}+0\hat{k}$

$\displaystyle T'(t)=\frac{-5}{\sqrt{29}}sin(t)\hat{i}+\frac{5}{\sqrt{29}}cos(t)\hat{j}$

$\displaystyle \left \| T'(t)\right \|=\sqrt{\left (\frac{-5}{\sqrt{29}}sin(t) \right )^{2}+\left (\frac{5}{\sqrt{29}}cos(t) \right )^{2}}$

$\displaystyle \left \| T'(t)\right \|=\sqrt{\frac{25}{29}sin^2(t)+\frac{25}{29}cos^2(t)}=\sqrt{\frac{25}{29}}=\frac{5}{\sqrt{29}}$

$\displaystyle N(t)=\frac{T'(t)}{\left \| T'(t)\right \|}=\frac{\frac{-5}{\sqrt{29}}sin(t)\hat{i}+\frac{5}{\sqrt{29}}cos(t)\hat{j}}{\frac{5}{\sqrt{29}}}=-sin(t)\hat{i}+cos(t)\hat{j}$

Derived from properties of plane equations, one can simply pick off the coefficients of the cartesian coordinate variable to give a normal vector $\displaystyle \vec{n}$ that is perpendicular to that plane.  For a given plane, we can write

$\displaystyle ax+by+cz=C, \vec{n}= a \hat{x}+b\hat{y}+c\hat{z}$.

From this result, we find that for our case, 

$\displaystyle \vec{n}= -3 \hat{x}+5\hat{y}-\hat{z}$.

These are all true facts about normal vectors to a plane. (If the surface is not a plane, then a few of these no longer hold.)

Vectors can be said to be orthogonal, that is to say perpendicular or normal, if their dot product amounts to zero:

$\displaystyle \overrightarrow{a}\cdot\overrightarrow{b}=0\rightarrow \overrightarrow{a}\perp\overrightarrow{b}$

To find the dot product of two vectors given the notation

$\displaystyle \overrightarrow{a}\begin{bmatrix} a_1\\a_2 \\ a_3 \end{bmatrix};\overrightarrow{b}\begin{bmatrix} b_1\\b_2 \\ b_3 \end{bmatrix}$

Simply multiply terms across rows:

$\displaystyle \overrightarrow{a}\cdot\overrightarrow{b}=a_1b_1+a_2b_2+a_3b_3$

For our vectors, $\displaystyle \overrightarrow{a}=\begin{bmatrix} 5\\0 \\7 \end{bmatrix}$ and $\displaystyle \overrightarrow{b}=\begin{bmatrix} -14\\11 \\10 \end{bmatrix}$

$\displaystyle \overrightarrow{a}\cdot\overrightarrow{b}=-70+0+70=0$

The two vectors are orthogonal.

Vectors can be said to be orthogonal, that is to say perpendicular or normal, if their dot product amounts to zero:

$\displaystyle \overrightarrow{a}\cdot\overrightarrow{b}=0\rightarrow \overrightarrow{a}\perp\overrightarrow{b}$

To find the dot product of two vectors given the notation

$\displaystyle \overrightarrow{a}\begin{bmatrix} a_1\\a_2 \\ a_3 \end{bmatrix};\overrightarrow{b}\begin{bmatrix} b_1\\b_2 \\ b_3 \end{bmatrix}$

Simply multiply terms across rows:

$\displaystyle \overrightarrow{a}\cdot\overrightarrow{b}=a_1b_1+a_2b_2+a_3b_3$

For our vectors, $\displaystyle \overrightarrow{a}=\begin{bmatrix} 7\\7 \\3 \end{bmatrix}$ and $\displaystyle \overrightarrow{b}=\begin{bmatrix}1 \\-1 \\2 \end{bmatrix}$

$\displaystyle \overrightarrow{a}\cdot\overrightarrow{b}=7-7+6=6$

The two vectors are not orthogonal.

Vectors can be said to be orthogonal, that is to say perpendicular or normal, if their dot product amounts to zero:

$\displaystyle \overrightarrow{a}\cdot\overrightarrow{b}=0\rightarrow \overrightarrow{a}\perp\overrightarrow{b}$

To find the dot product of two vectors given the notation

$\displaystyle \overrightarrow{a}\begin{bmatrix} a_1\\a_2 \\ a_3 \end{bmatrix};\overrightarrow{b}\begin{bmatrix} b_1\\b_2 \\ b_3 \end{bmatrix}$

Simply multiply terms across rows:

$\displaystyle \overrightarrow{a}\cdot\overrightarrow{b}=a_1b_1+a_2b_2+a_3b_3$

For our vectors, $\displaystyle \overrightarrow{a}=\begin{bmatrix} 11\\8 \\-10 \end{bmatrix}$ and $\displaystyle \overrightarrow{b}=\begin{bmatrix} 2\\3 \\4 \end{bmatrix}$

$\displaystyle \overrightarrow{a}\cdot\overrightarrow{b}=22+24-40=6$

The two vectors are not orthogonal.