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Calculus 3 : Normal Vectors

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6 Diagnostic Tests 373 Practice Tests Question of the Day Flashcards Learn by Concept

Example Questions

Example Question #61 : Normal Vectors

Determine whether the two vectors, $\overrightarrow{a}=\begin{bmatrix} -4\\-6 \\-7 \end{bmatrix}$ and $\overrightarrow{b}=\begin{bmatrix} 1\\-5 \\4 \end{bmatrix}$ , 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, $\overrightarrow{a}=\begin{bmatrix} -4\\-6 \\-7 \end{bmatrix}$ and $\overrightarrow{b}=\begin{bmatrix} 1\\-5 \\4 \end{bmatrix}$

$\overrightarrow{a}\cdot\overrightarrow{b}=-4+30-28=-2$

The two vectors are not orthogonal.

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Example Question #551 : Vectors And Vector Operations

Determine whether the two vectors, $\overrightarrow{a}=\begin{bmatrix} 6\\ -6\\5 \end{bmatrix}$ and $\overrightarrow{b}=\begin{bmatrix} 13\\ -2\\ -18\end{bmatrix}$ , 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, $\overrightarrow{a}=\begin{bmatrix} 6\\ -6\\5 \end{bmatrix}$ and $\overrightarrow{b}=\begin{bmatrix} 13\\ -2\\ -18\end{bmatrix}$

$\overrightarrow{a}\cdot\overrightarrow{b}=78+12-90=0$

The two vectors are orthogonal.

Report an Error

Example Question #552 : Vectors And Vector Operations

Determine whether the two vectors, $\overrightarrow{a}=\begin{bmatrix} -1\\5 \\8 \end{bmatrix}$ and $\overrightarrow{b}=\begin{bmatrix} 3\\4 \\ -3\end{bmatrix}$ , 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, $\overrightarrow{a}=\begin{bmatrix} -1\\5 \\8 \end{bmatrix}$ and $\overrightarrow{b}=\begin{bmatrix} 3\\4 \\ -3\end{bmatrix}$

$\overrightarrow{a}\cdot\overrightarrow{b}=-3+20-24=-7$

The two vectors are not orthogonal.

Report an Error

Example Question #61 : Normal Vectors

Determine whether the two vectors, $\overrightarrow{a}=\begin{bmatrix}1 \\ -9\\1 \end{bmatrix}$ and $\overrightarrow{b}=\begin{bmatrix} 13\\1 \\-4 \end{bmatrix}$ , are orthogonal or not.

Possible Answers:

The two vectors are not orthogonal.

The two vectors are orthogonal.

Correct answer:

The two vectors are 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, $\overrightarrow{a}=\begin{bmatrix}1 \\ -9\\1 \end{bmatrix}$ and $\overrightarrow{b}=\begin{bmatrix} 13\\1 \\-4 \end{bmatrix}$

$\overrightarrow{a}\cdot\overrightarrow{b}=13-9-4=0$

The two vectors are orthogonal.

Report an Error

Example Question #554 : Vectors And Vector Operations

Determine whether the two vectors, $\overrightarrow{a}=\begin{bmatrix} 5\\ 7\\-1 \end{bmatrix}$ and $\overrightarrow{b}=\begin{bmatrix} 2\\1 \\3 \end{bmatrix}$ , 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, $\overrightarrow{a}=\begin{bmatrix} 5\\ 7\\-1 \end{bmatrix}$ and $\overrightarrow{b}=\begin{bmatrix} 2\\1 \\3 \end{bmatrix}$

$\overrightarrow{a}\cdot\overrightarrow{b}=10+7-3=14$

The two vectors are not orthogonal.

Report an Error

Example Question #555 : Vectors And Vector Operations

Determine whether the two vectors, $\overrightarrow{a}=\begin{bmatrix} 3\\-4 \\-4 \end{bmatrix}$ and $\overrightarrow{b}=\begin{bmatrix} 12\\ 3\\6 \end{bmatrix}$ , are orthogonal or not.

Possible Answers:

The two vectors are orthogonal.

The two vectors are not orthogonal.

Correct answer:

The two vectors are 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, $\overrightarrow{a}=\begin{bmatrix} 3\\-4 \\-4 \end{bmatrix}$ and $\overrightarrow{b}=\begin{bmatrix} 12\\ 3\\6 \end{bmatrix}$

$\overrightarrow{a}\cdot\overrightarrow{b}=36-12-24=0$

The two vectors are orthogonal.

Report an Error

Example Question #64 : Normal Vectors

Determine whether the two vectors, $\overrightarrow{a}=\begin{bmatrix} 9\\ -6\\4 \end{bmatrix}$ and $\overrightarrow{b}=\begin{bmatrix} 3\\5 \\2 \end{bmatrix}$ , 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, $\overrightarrow{a}=\begin{bmatrix} 9\\ -6\\4 \end{bmatrix}$ and $\overrightarrow{b}=\begin{bmatrix} 3\\5 \\2 \end{bmatrix}$

$\overrightarrow{a}\cdot\overrightarrow{b}=27-30+8=5$

The two vectors are not orthogonal.

Report an Error

Example Question #65 : Normal Vectors

Which of the following vectors is perpendicular to the plane given by the following equation:

3x-7y+2z=5

Possible Answers:

$6\mathbf{i}-14\mathbf{j}+4\mathbf{k}$

$3\mathbf{i}-5\mathbf{j}-4\mathbf{k}$

$-6\mathbf{i}+8\mathbf{j}+\mathbf{k}$

$8\mathbf{i}+2\mathbf{j}-12\mathbf{k}$

Correct answer:

$6\mathbf{i}-14\mathbf{j}+4\mathbf{k}$

Explanation:

A normal vector to a plane of the form

f(x, y, z)=ax+by+cz+d=0

is given by the gradient of f. First, we have to put the equation into a form where it equals zero:

f(x, y, z)=3x-7y+2z-5=0

The gradient is given by:

$\bigtriangledown f=\frac{\partial f}{\partial x}\mathbf{i}+\frac{\partial f}{\partial y}\mathbf{j}+\frac{\partial f}{\partial z}\mathbf{k}=3\mathbf{i}-7\mathbf{i}+2\mathbf{i}$

A vector multiplied by a constant is parallel to the original vector, so the above vector multiplied by a constant is perpendicular to the plane. The correct answer is the above vector multiplied by two.

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Example Question #66 : Normal Vectors

Find the tangent vector for $\vec{r}(t)=<t,t^2,t^3>$

Possible Answers:

<t^4,-2t^3,t^2>

$\frac{1}{\sqrt{1+4t^2+9t^4}}$

<1,2t,3t^2>

$\frac{1}{\sqrt{1+4t^2+9t^4}}<1,2t,3t^2>$

Correct answer:

$\frac{1}{\sqrt{1+4t^2+9t^4}}<1,2t,3t^2>$

Explanation:

$\vec{r}(t)=<t,t^2,t^3>$

$\vec{r'}(t)=<1,2t,3t^2>$

$||\vec{r'}(t)||=\sqrt{1+4t^2+9t^4}$

$\vec{T}(t)=\frac{\vec{r'}(t)}{||\vec{r}'(t)||}=\frac{1}{\sqrt{1+4t^2+9t^4}}<1,2t,3t^2>$

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Example Question #67 : Normal Vectors

Find the normal vector (in standard notation) to the plane:

-2x+4y=-2

Possible Answers:

$-2\hat{i}+4\hat{j}+\hat{k}$

$-2\hat{i}+4\hat{j}+2\hat{k}$

$-2\hat{i}+4\hat{j}$

$\left \langle -2, 4, 0 \right \rangle$

Correct answer:

$-2\hat{i}+4\hat{j}$

Explanation:

To determine the normal vector to a plane, we simply report the coefficients of the x, y, and z terms, as the equation of a plane is given by

a(x-x_0)+b(y-y_0)+c(z-z_0)=0

where $\left \langle a, b, c \right \rangle$ is the normal vector.

So, our normal vector is

$\left \langle -2, 4, 0 \right \rangle$

We were asked to write this in standard notation, which gives us

$-2\hat{i}+4\hat{j}$

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Calculus 3 : Normal Vectors

Study concepts, example questions & explanations for Calculus 3

All Calculus 3 Resources

Example Questions

Example Question #61 : Normal Vectors

Example Question #551 : Vectors And Vector Operations

Example Question #552 : Vectors And Vector Operations

Example Question #61 : Normal Vectors

Example Question #554 : Vectors And Vector Operations

Example Question #555 : Vectors And Vector Operations

Example Question #64 : Normal Vectors

Example Question #65 : Normal Vectors

Example Question #66 : Normal Vectors

Example Question #67 : Normal Vectors

All Calculus 3 Resources

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