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Calculus 3 : Vectors and Vector Operations

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Example Questions

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 #561 : Vectors And Vector Operations

Calculate the norm of the vector:

2i+j-2k

Possible Answers:

None of the Above.

$\sqrt{5}$

$\frac{1}{2}$

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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Example Question #75 : Normal Vectors

Find the normal vector to the plane given by the following vectors:

$\left \langle 2, 0, 1\right \rangle, \left \langle 3, 3, 3\right \rangle$

Possible Answers:

$\left \langle -3, -3, 6\right \rangle$

$\left \langle -3, 6, -3 \right \rangle$

$\left \langle 3, 9, 6\right \rangle$

$\left \langle 0, 0, 0\right \rangle$

Correct answer:

$\left \langle -3, -3, 6\right \rangle$

Explanation:

The normal vector is given by the cross product of the vectors.

First, we must write the determinant in order to take the cross product of the two vectors:

$\begin{vmatrix} \hat{i}&\hat{j} &\hat{k} \\ 2& 0&1 \\ 3&3 &3 \end{vmatrix}$

where i, j, and k are the unit vectors corresponding to the x, y, and z direction respectively.

Next, we take the cross product. One can do this by multiplying across from the top left to the lower right, and continuing downward, and then subtracting the terms multiplied from top right to the bottom left:

$6\hat{k}+3\hat{j-(3\hat{i}+6\hat{j}})=6\hat{k}-3\hat{i}-3\hat{j}=\left \langle -3, -3, 6\right \rangle$

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Example Question #76 : Normal Vectors

Find the normal vector to the plane that contains $a=\left \langle 3,0,1\right \rangle$ and $b=\left \langle 5,2,2\right \rangle$

Possible Answers:

$\left \langle -2,-1,6\right \rangle$

$\left \langle 1,3,7\right \rangle$

$\left \langle -2,1,6\right \rangle$

$\left \langle 0,1,5\right \rangle$

Correct answer:

$\left \langle -2,1,6\right \rangle$

Explanation:

The normal vector to the plane is found by taking the cross product of and . Using the formula for taking the cross product of two vectors, where a=(a,b,c) and b=(x,y,z) , we get $a\times b=i(bz-yc)-j(az-xc)+k(ay-xb)$ . Using the vectors from the problem statement, we then get $a\times b=i(0-2)-j(6-5)+k(6-0)=-2i+j+6k$ . In vector notation, this becomes $\left \langle -2,1,6\right \rangle$ .

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Calculus 3 : Vectors and Vector Operations

Study concepts, example questions & explanations for Calculus 3

All Calculus 3 Resources

Example Questions

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 #561 : Vectors And Vector Operations

Example Question #73 : Normal Vectors

Example Question #74 : Normal Vectors

Example Question #75 : Normal Vectors

Example Question #76 : Normal Vectors

All Calculus 3 Resources

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