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Calculus 3 : Normal Vectors

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Example Questions

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 #71 : Normal Vectors

Calculate the norm of the vector:

2i+j-2k

Possible Answers:

None of the Above.

$\frac{1}{2}$

$\sqrt{5}$

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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Example Question #571 : Vectors And Vector Operations

Find the normal vector to the plane containing the following vectors:

$\left \langle 5, 5, 6\right \rangle, \left \langle 0, 1, 2\right \rangle$

Possible Answers:

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

$\left \langle 16,10, 5\right \rangle$

$\left \langle 4, -10, 5\right \rangle$

Correct answer:

$\left \langle 4, -10, 5\right \rangle$

Explanation:

The normal vector to the plane is given by the cross product of the two vectors in the plane.

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

$\begin{vmatrix} \hat{i}&\hat{j} &\hat{k} \\ 5&5 &6 \\ 0&1 &2 \end{vmatrix}$

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

$10\hat{i}+5\hat{k}-(6\hat{i}+10\hat{j})=4\hat{i}-10\hat{j}+5\hat{k}=\left \langle 4, -10, 5\right \rangle$

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Example Question #72 : Normal Vectors

Find the normal vector to the plane containing the following vectors:

$\left \langle 5, -10, 5\right \rangle, \left \langle 3, 4, 6\right \rangle$

Possible Answers:

$\left \langle 80, 15, 50\right \rangle$

$\left \langle -80, -15, 50\right \rangle$

$\left \langle -40, 45, -10\right \rangle$

$\left \langle -80, -15, -50\right \rangle$

Correct answer:

$\left \langle -80, -15, 50\right \rangle$

Explanation:

The normal vector to the plane is given by the cross product of the two vectors in the plane.

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

$\begin{vmatrix} \hat{i}&\hat{j} &\hat{k} \\ 5&-10 &5 \\ 3&4 &6 \end{vmatrix}$

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

$-60\hat{i}+20\hat{k}+15\hat{j}-(-30\hat{k}+20\hat{i}+30\hat{j})=-80\hat{i}-15\hat{j}+50\hat{k}=\left \langle -80, -15, 50\right \rangle$

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Example Question #2571 : Calculus 3

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

$\left \langle -3, -2, 4\right \rangle, \left \langle 1, 5,5\right \rangle$

Possible Answers:

$\left \langle 30, 19, 13\right \rangle$

$\left \langle -30, 19, -13\right \rangle$

$\left \langle -30, 19, 13\right \rangle$

$\left \langle 10, -11, -17\right \rangle$

Correct answer:

$\left \langle -30, 19, -13\right \rangle$

Explanation:

The normal vector to the plane is given by the cross product of two vectors in the plane.

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

$\begin{vmatrix} \hat{i}&\hat{j} &\hat{k} \\ -3&-2 &4 \\ 1& 5& 5 \end{vmatrix}$

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

$-10\hat{i}+-15\hat{k}+4\hat{j}-(-2\hat{k}+20\hat{i}-15\hat{j})=-30\hat{i}+19\hat{j}-13\hat{k}=\left \langle -30, 19, -13\right \rangle$

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Example Question #2572 : Calculus 3

Find the normal vector to the plane given the vectors in the plane

$\left \langle 3, 3, 3\right \rangle$ and $\left \langle 5, 0, 2\right \rangle$

Possible Answers:

$\left \langle 6, 9, 15\right \rangle$

$\left \langle 15, 0, 6\right \rangle$

$\left \langle 5, 3, 5\right \rangle$

$\left \langle 6, 9, -15\right \rangle$

Correct answer:

$\left \langle 6, 9, -15\right \rangle$

Explanation:

To determine the normal vector to the plane, we must take the cross product of the two vectors in the plane.

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

$\begin{vmatrix} \hat{i}&\hat{j} &\hat{k} \\ 3&3 &3 \\ 5&0 &2 \end{vmatrix}$

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

$6\hat{i}+15\hat{j}-(15\hat{k}+6\hat{j})=6\hat{i}+9\hat{j} -15\hat{k}=\left \langle 6, 9, -15\right \rangle$

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Calculus 3 : Normal Vectors

Study concepts, example questions & explanations for Calculus 3

All Calculus 3 Resources

Example Questions

Example Question #71 : Normal Vectors

Example Question #71 : Normal Vectors

Example Question #73 : Normal Vectors

Example Question #74 : Normal Vectors

Example Question #75 : Normal Vectors

Example Question #76 : Normal Vectors

Example Question #571 : Vectors And Vector Operations

Example Question #72 : Normal Vectors

Example Question #2571 : Calculus 3

Example Question #2572 : Calculus 3

All Calculus 3 Resources

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