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Calculus 3 : Normal Vectors

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Example Questions

Example Question #81 : Normal Vectors

Find the normal vector of the plane that is parallel to the plane given by the equation 2x+5y-6z=13

Possible Answers:

$\left \langle 2,5,-6\right \rangle$

$\left \langle 5,-6,13\right \rangle$

$\left \langle 4,-5,8\right \rangle$

$\left \langle 3,-2,9\right \rangle$

Correct answer:

$\left \langle 2,5,-6\right \rangle$

Explanation:

To solve the problem, we use the fact that two parallel planes have the same normal vector. The equation of on of the planes is given, and from that we know its normal vector is $\left \langle 2,5,-6\right \rangle$ , which is the normal vector of the plane in question

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Example Question #82 : Normal Vectors

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

$\left \langle 3,0,1\right \rangle$ and $\left \langle 4,5,2\right \rangle$

Possible Answers:

$\left \langle -5,2,13\right \rangle$

$\left \langle -5,-2,15\right \rangle$

$\left \langle -5,-3,15\right \rangle$

$\left \langle -5,2,15\right \rangle$

Correct answer:

$\left \langle -5,-2,15\right \rangle$

Explanation:

To find the normal vector to the plane containing vectors $\left \langle x_1,y_1,z_1\right \rangle$ and $\left \langle x_2,y_2,z_2\right \rangle$ , we find the determinant of the 3x3 matrix $\begin{bmatrix} i&j &k \\ x_1&y_1 & z_1\\ x_2&y_2 &z_2 \end{bmatrix}$

Plugging in the vectors and solving, we get

$i(0-5)-j(6-4)+k(15-0)=-5i-2j+15k=\left \langle -5,-2,15\right \rangle$

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Example Question #81 : Normal Vectors

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

$\left \langle -4,4,1\right \rangle$ and $\left \langle 2,1,0\right \rangle$

Possible Answers:

$\left \langle -1,2,-12\right \rangle$

$\left \langle -3,-2,-12\right \rangle$

$\left \langle -1,2,-10\right \rangle$

$\left \langle -1,0,-12\right \rangle$

Correct answer:

$\left \langle -1,2,-12\right \rangle$

Explanation:

Plugging in the vectors and solving, we get

$i(0-1)-j(0-2)+k(-4-8)=-i+2j+-12k=\left \langle-1,2,-12 \right \rangle$

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Example Question #84 : Normal Vectors

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

$\left \langle 3, 0, 3\right \rangle, \left \langle 8, 0, 8\right \rangle$

Possible Answers:

$-24\hat{j}$

$24\hat{j}$

$\vec{0}$

Correct answer:

$\vec{0}$

Explanation:

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

$\begin{vmatrix} \hat{i}& \hat{j}&\hat{k} \\ 3&0 &3 \\ 8&0 &8 \end{vmatrix}$

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:

$0\hat{i}+0\hat{k}+24\hat{j}-(0\hat{k}+0\hat{i}+24\hat{j})=\vec{0}$

The difference between zero and the zero vector is an important one, because the result of a cross product is always a vector (the dot product of two vectors gives a scalar).

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Example Question #81 : Normal Vectors

Find the normal vector to the plane containing the vectors $\left \langle -1,-3,-7\right \rangle$ and $\left \langle 2,5,0\right \rangle$

Possible Answers:

$\left \langle 35,14,1\right \rangle$

$\left \langle 39,-13,2\right \rangle$

$\left \langle 35,-14,1\right \rangle$

$\left \langle 30,-12,1\right \rangle$

Correct answer:

$\left \langle 35,-14,1\right \rangle$

Explanation:

To find the normal vector to the plane containing vectors and , we take the cross product of the two.

To find the cross product between the vectors $a=\left \langle a_1,a_2,a_3\right \rangle$ and $b=\left \langle b_1,b_2,b_3\right \rangle$ , we find the determinant of the 3x3 matrix $\begin{bmatrix} i&j &k \\ a_1&a_2 &a_3 \\ b_1&b_2 &b_3 \end{bmatrix}$ which follows the formula determinant= i(a_2b_3-b_2a_3)-j(a_1b_3-b_1a_3)+k(a_1b_2-b_1a_2)

Applying to the vectors from the problem statement, we get

$n=i(0+35)-j(0+14)+k(-5+6)=\left \langle35,-14,1\right \rangle$

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Example Question #81 : Normal Vectors

Find the normal vector to the plane containing the vectors $\left \langle 0,-3,1\right \rangle$ and $\left \langle 0,5,3\right \rangle$

Possible Answers:

$\left \langle -14,14,0\right \rangle$

$\left \langle -12,0,0\right \rangle$

$\left \langle -14,0,0\right \rangle$

$\left \langle -14,2,0\right \rangle$

Correct answer:

$\left \langle -14,0,0\right \rangle$

Explanation:

To find the normal vector to the plane containing vectors and , we take the cross product of the two.

Applying to the vectors from the problem statement, we get

$n=i(-9-5)-j(0-0)+k(0-0)=\left \langle-14,0,0\right \rangle$

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Example Question #81 : Normal Vectors

Find the normal vector to the following vectors:

$\left \langle 8, 1, -4\right \rangle, \left \langle -1, 3, 4\right \rangle$

Possible Answers:

$\left \langle 16, 28, 23\right \rangle$

$\left \langle 16, -28, 23\right \rangle$

$\left \langle -8, 36, 25\right \rangle$

Correct answer:

$\left \langle 16, -28, 23\right \rangle$

Explanation:

To find the normal vector, we must take the cross product of the two vectors.

Now, we can write the determinant in order to take the cross product of the two vectors:
$\begin{vmatrix} \hat{i}& \hat{j}&\hat{k} \\ 8&1 &-4 \\ -1&3 &4 \end{vmatrix}$

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

$4\hat{i}+24\hat{k}+4\hat{j}-(\hat{k}-12\hat{i}+32\hat{j})=16\hat{i}-28\hat{j}+23\hat{k}=\left \langle 16, -28, 23\right \rangle$

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Example Question #88 : Normal Vectors

Determine the vector normal to the plane created by the following two vectors:

$\left \langle 10, 2, 4\right \rangle, \left \langle 11, 2, 6\right \rangle$

Possible Answers:

$\left \langle 20, 104, 42\right \rangle$

$\left \langle -16, 4, -2\right \rangle$

$\left \langle 4, 16, 2\right \rangle$

$\left \langle 4, -16, -2\right \rangle$

Correct answer:

$\left \langle 4, -16, -2\right \rangle$

Explanation:

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

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

$\begin{vmatrix} \hat{i}& \hat{j}&\hat{k} \\ 10&2 &4 \\ 11&2 &6 \end{vmatrix}$

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

$12\hat{i}+20\hat{k}+44\hat{j}-(22\hat{k}+8\hat{i}+60\hat{j})=4\hat{i}-16\hat{j}-2\hat{k}=\left \langle 4, -16, -2\right \rangle$

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Example Question #89 : Normal Vectors

Find the normal vector to the plane containing the vectors $\left \langle3,4,-6 \right \rangle$ and $\left \langle 0,5,3\right \rangle$

Possible Answers:

$\left \langle 20,-9,15\right \rangle$

$\left \langle 42,-9,15\right \rangle$

$\left \langle 23,9,15\right \rangle$

$\left \langle 23,-9,11\right \rangle$

Correct answer:

$\left \langle 42,-9,15\right \rangle$

Explanation:

To find the normal vector to a plane containing vectors $a=\left \langle a_1,a_2,a_3\right \rangle$ and $b=\left \langle b_1,b_2,b_3\right \rangle$ , you take the cross product of the two vectors.

To find the cross product between the two vectors $a=\left \langle a_1,a_2,a_3\right \rangle$ and $b=\left \langle b_1,b_2,b_3\right \rangle$ , you take the determinant of the 3x3 matrix

$\begin{bmatrix} \hat{i}&\hat{j} &\hat{k} \\ a_1& a_2&a_3 \\ b_1&b_2 &b_3 \end{bmatrix}$ . $determinant=\hat{i}(a_2b_3-b_2a_3)-\hat{j}(a_1b_3-b_1a_3)+\hat{k}(a_1b_2-b_1a_2)$

Using the vectors from the problem statement, we get

$determinant=\hat{i}(12+30)-\hat{j}(9-0)+\hat{k}(15-0)=\left \langle 42,-9,15\right \rangle$

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Example Question #90 : Normal Vectors

Find the normal vector to the plane containing the vectors $\left \langle0,7,5 \right \rangle$ and $\left \langle -9,-1,-4\right \rangle$

Possible Answers:

$\left \langle -20,-45,61\right \rangle$

$\left \langle -23,-45,63\right \rangle$

$\left \langle -23,45,63\right \rangle$

$\left \langle -23,-44,63\right \rangle$

Correct answer:

$\left \langle -23,-45,63\right \rangle$

Explanation:

To find the normal vector to a plane containing vectors $a=\left \langle a_1,a_2,a_3\right \rangle$ and $b=\left \langle b_1,b_2,b_3\right \rangle$ , you take the cross product of the two vectors.

To find the cross product between the two vectors $a=\left \langle a_1,a_2,a_3\right \rangle$ and $b=\left \langle b_1,b_2,b_3\right \rangle$ , you take the determinant of the 3x3 matrix

$\begin{bmatrix} \hat{i}&\hat{j} &\hat{k} \\ a_1& a_2&a_3 \\ b_1&b_2 &b_3 \end{bmatrix}$ . $determinant=\hat{i}(a_2b_3-b_2a_3)-\hat{j}(a_1b_3-b_1a_3)+\hat{k}(a_1b_2-b_1a_2)$

Using the vectors from the problem statement, we get

$determinant=\hat{i}(-28+5)-\hat{j}(0+45)+\hat{k}(0+63)=\left \langle -23,-45,63\right \rangle$

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Calculus 3 : Normal Vectors

Study concepts, example questions & explanations for Calculus 3

All Calculus 3 Resources

Example Questions

Example Question #81 : Normal Vectors

Example Question #82 : Normal Vectors

Example Question #81 : Normal Vectors

Example Question #84 : Normal Vectors

Example Question #81 : Normal Vectors

Example Question #81 : Normal Vectors

Example Question #81 : Normal Vectors

Example Question #88 : Normal Vectors

Example Question #89 : Normal Vectors

Example Question #90 : Normal Vectors

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