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Calculus 3 : Normal Vectors

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Example Questions

Example Question #1 : Normal Vectors

Find the Unit Normal Vector to the given plane.

3x+3y+4z=10 .

Possible Answers:

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

V_n=(3, 3, 4 )

$V_n=(\frac{1}{\sqrt{34}}, \frac{1}{\sqrt{34}}, \frac{1}{\sqrt{34}} )$

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

$V_n=(\frac{1}{{34}}, \frac{1}{{34}}, \frac{1}{{34}} )$

Correct answer:

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

Explanation:

Recall the definition of the Unit Normal Vector.

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

Let V=(3,3,4)

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

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

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Example Question #1 : Normal Vectors

Find the unit normal vector of $v(t)=t^{2}\hat{i}+3t^{2}\hat{j}$ .

Possible Answers:

$N(t)=2t\hat{i}+6t\hat{j}$

$N(t)=\frac{1}{\sqrt{10}}\hat{i}+\frac{3}{\sqrt{10}}\hat{j}$

Does not exist

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

Correct answer:

Does not exist

Explanation:

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

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

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

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

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

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

For this problem

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

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

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

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

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

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

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

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Example Question #3 : Normal Vectors

Find the unit normal vector of $v(t)=sin(t)\hat{i}+cos(t)\hat{j}$ .

Possible Answers:

$N(t)=cos(t)\hat{i}-sin(t)\hat{j}$

Does not exist

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

N(t)=1

Correct answer:

$N(t)=cos(t)\hat{i}-sin(t)\hat{j}$

Explanation:

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

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

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

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

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

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

For this problem

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

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

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

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

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

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

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Example Question #4 : Normal Vectors

Find the unit normal vector of $v(t)=e^{t}\hat{i}+2e^{t}\hat{j}$ .

Possible Answers:

$N(t)=\frac{1}{\sqrt{5}}\hat{i}+\frac{2}{\sqrt{5}}\hat{j}$

$N(t)=e^{t}\sqrt{5}$

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

DNE

Correct answer:

DNE

Explanation:

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

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

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

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

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

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

For this problem

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

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

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

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

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

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

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

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Example Question #5 : Normal Vectors

Find the unit normal vector of $v(t)=5cos(t)\hat{i}+5sin(t)\hat{j}+2\hat{k}$ .

Possible Answers:

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

$N(t)=-sin(t)\hat{i}+cos(t)\hat{j}$

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

$N(t)=\frac{5}{\sqrt{29}}cos(t)\hat{i}+\frac{5}{\sqrt{29}}sin(t)\hat{j}+\frac{2}{\sqrt{29}}\hat{k}$

Correct answer:

$N(t)=-sin(t)\hat{i}+cos(t)\hat{j}$

Explanation:

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

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

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

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

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

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

For this problem

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

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

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

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

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

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

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

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

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

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

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Example Question #6 : Normal Vectors

Find a normal vector $\vec{n}$ that is perpendicular to the plane given below.

-3x + 5y -z = 25

Possible Answers:

$\vec{n}= \hat{x}+\hat{y}+\hat{z}$

No such vector exists.

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

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

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

Correct answer:

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

Explanation:

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

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

From this result, we find that for our case,

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

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Example Question #1 : Normal Vectors

Which of the following is FALSE concerning a vector normal to a plane (in -dimensional space)?

Possible Answers:

The cross product of any two normal vectors to the plane is <0,0,0> .

It is parallel to any other normal vector to the plane.

Multiplying it by a scalar gives another normal vector to the plane.

It is orthogonal to the plane.

All the other answers are true.

Correct answer:

All the other answers are true.

Explanation:

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.)

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Example Question #8 : Normal Vectors

Determine whether the two vectors, $\overrightarrow{a}=\begin{bmatrix} 5\\0 \\7 \end{bmatrix}$ and $\overrightarrow{b}=\begin{bmatrix} -14\\11 \\10 \end{bmatrix}$ , are orthogonal or not.

Possible Answers:

The two vectors are not orthogonal.

The two vectors are orthogonal.

Correct answer:

The two vectors are orthogonal.

Explanation:

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

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

To find the dot product of two vectors given the notation

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

Simply multiply terms across rows:

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

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

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

The two vectors are orthogonal.

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Example Question #9 : Normal Vectors

Determine whether the two vectors, $\overrightarrow{a}=\begin{bmatrix} 7\\7 \\3 \end{bmatrix}$ and $\overrightarrow{b}=\begin{bmatrix}1 \\-1 \\2 \end{bmatrix}$ , are orthogonal or not.

Possible Answers:

The two vectors are not orthogonal.

The two vectors are orthogonal.

Correct answer:

The two vectors are not orthogonal.

Explanation:

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

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

To find the dot product of two vectors given the notation

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

Simply multiply terms across rows:

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

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

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

The two vectors are not orthogonal.

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Example Question #10 : Normal Vectors

Determine whether the two vectors, $\overrightarrow{a}=\begin{bmatrix} 11\\8 \\-10 \end{bmatrix}$ and $\overrightarrow{b}=\begin{bmatrix} 2\\3 \\4 \end{bmatrix}$ , are orthogonal or not.

Possible Answers:

The two vectors are not orthogonal.

The two vectors are orthogonal.

Correct answer:

The two vectors are not orthogonal.

Explanation:

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

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

To find the dot product of two vectors given the notation

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

Simply multiply terms across rows:

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

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

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

The two vectors are not orthogonal.

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Calculus 3 : Normal Vectors

Study concepts, example questions & explanations for Calculus 3

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