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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:

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

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

V_n=(3, 3, 4 )

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

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

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 #2 : Normal Vectors

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

Possible Answers:

Does not exist

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

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

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

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:

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

N(t)=1

Does not exist

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

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:

DNE

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

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

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

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 #2 : Normal Vectors

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

Possible Answers:

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

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

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

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

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 #5 : 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 #5 : Normal Vectors

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

Possible Answers:

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

It is orthogonal to the plane.

All the other answers are true.

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

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

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 #2 : 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 orthogonal.

The two vectors are not 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 #3 : 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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Example Questions

Example Question #1 : Normal Vectors

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