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Calculus 3 : Calculus 3

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

Example Questions

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.

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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.

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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.

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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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Example Question #71 : Normal Vectors

Find the normal vector to plane given by the equation of two vectors on the plane: $\left \langle 2,2,5\right \rangle$ and $\left \langle 3,-1,0\right \rangle$ .

Possible Answers:

$n=\left \langle 5,15,-8\right \rangle$

$n=\left \langle -5,-15,-8\right \rangle$

$n=\left \langle 5,15,8\right \rangle$

$n=\left \langle 5,5,0\right \rangle$

Correct answer:

$n=\left \langle 5,15,-8\right \rangle$

Explanation:

To find the normal vector, you must take the cross product of the two vectors. Once you take the cross product, you get 5i+15j-8k . In vector notation, this is $\left \langle 5,15,-8\right \rangle$ .

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Example Question #2561 : Calculus 3

Calculate the norm of the vector:

2i+j-2k

Possible Answers:

$\sqrt{5}$

$\frac{1}{2}$

None of the Above.

Correct answer:

Explanation:

Norm of the Vector is = $\sqrt{2^2+1^2+2^2}=\sqrt{9}=3$

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Example Question #73 : Normal Vectors

Two vectors $\left \langle 1,3,5\right \rangle$ and $\left \langle 2,0,5\right \rangle$ are parallel to a plane. Find the normal vector to the plane.

Possible Answers:

$\left \langle -15,-5,-6\right \rangle$

$\left \langle 15,5,-6\right \rangle$

$\left \langle 15,-5,6\right \rangle$

$\left \langle 15,-5,-6\right \rangle$

Correct answer:

$\left \langle 15,5,-6\right \rangle$

Explanation:

To find the normal vector to the plane, we must take the cross product of the two vectors. Using the 3x3 matrix $\begin{bmatrix} i&j &k \\ 1&3 &5 \\2 & 0&5 \end{bmatrix}$ , we perform the cross product.

Using the formula for the determinant of a 3x3 matrix

$A=\begin{bmatrix} a&b &c \\ d &e & f\\ g&h &i \end{bmatrix}$

$\left | A\right |=a(ei-hf)-b(di-gf)+c(dh-ge)$ ,

we get

$i(15-0)-j(5-10)+k(0-6)=\left \langle 15,5,-6\right \rangle$

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Example Question #74 : Normal Vectors

Find the normal vector of the plane that contains the lines $\left \langle 2,5,1\right \rangle$ and $\left \langle 0,2,6\right \rangle$

Possible Answers:

$\left \langle 24,8,16\right \rangle$

$\left \langle -28,-12,4\right \rangle$

$\left \langle 28,-12,4\right \rangle$

$\left \langle 12,2,8\right \rangle$

Correct answer:

$\left \langle 28,-12,4\right \rangle$

Explanation:

To find the normal vector to the plane, you must the the cross (determinant) between the vectors .The formula for the determinant of a 3x3 matrix $A=\begin{bmatrix} a&b &c \\ d& e&f \\ g&h &i \end{bmatrix}$ is $\left | A\right |=a(ei-hf)-b(di-gf)+c(dh-ge)$ . Using the matrix in the problem statement, we get $i(30-2)-j(12-0)+k(4-0)=28i-12j+4k=\left \langle 28,-12,4\right \rangle$

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Calculus 3 : Calculus 3

Study concepts, example questions & explanations for Calculus 3

All Calculus 3 Resources

Example Questions

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

Example Question #71 : Normal Vectors

Example Question #2561 : Calculus 3

Example Question #73 : Normal Vectors

Example Question #74 : Normal Vectors

All Calculus 3 Resources

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