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Calculus 3 : Normal Vectors

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Example Questions

Example Question #91 : Normal Vectors

Find the normal vector to the plane containing the vectors $\left \langle0,-5,-4 \right \rangle$ and $\left \langle2,-4,7\right \rangle$

Possible Answers:

$\left \langle -51,-8,10\right \rangle$

$\left \langle -51,-5,10\right \rangle$

$\left \langle -41,-8,10\right \rangle$

$\left \langle -50,-8,9\right \rangle$

Correct answer:

$\left \langle -51,-8,10\right \rangle$

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

$determinant=\hat{i}(-35-16)-\hat{j}(0+8)+\hat{k}(0+10)=\left \langle -51,-8,10\right \rangle$

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Example Question #91 : Normal Vectors

Find the normal vector to the plane containing the vectors $\left \langle -2,0,6\right \rangle$ and $\left \langle 2,-3,-1\right \rangle$

Possible Answers:

$\left \langle 18,11,5\right \rangle$

$\left \langle 18,-14,6\right \rangle$

$\left \langle 10,13,6\right \rangle$

$\left \langle 18,14,6\right \rangle$

Correct answer:

$\left \langle 18,14,6\right \rangle$

Explanation:

To obtain the normal vector to a plane containing 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 compute 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 this formula, we evaluate using the vectors from the problem statement:

$\hat{i}(0+18)-\hat{j}(-2-12)+\hat{k}(6-0)=\left \langle 18,14,6\right \rangle$

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Example Question #2581 : Calculus 3

Find the normal vector to the plane that contains the vectors $\left \langle 0,4,9\right \rangle$ and $\left \langle 0,7,5\right \rangle$

Possible Answers:

$\left \langle -25,0,0\right \rangle$

$\left \langle -27,0,0\right \rangle$

$\left \langle -27,3,6\right \rangle$

$\left \langle -27,1,1\right \rangle$

Correct answer:

$\left \langle -27,0,0\right \rangle$

Explanation:

$determinant=\hat{i}(a_2b_3-b_2a_3)-\hat{j}(a_1b_3-b_1a_3)+\hat{k}(a_1b_2-b_1a_2)$

Using this formula, we evaluate using the vectors from the problem statement:

$\hat{i}(36-63)-\hat{j}(0-0)+\hat{k}(0-0)=\left \langle -27,0,0\right \rangle$

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Example Question #92 : Normal Vectors

Find the normal vector to the plane that contains the vectors $\left \langle 5,0,-6\right \rangle$ and $\left \langle 7,0,2\right \rangle$

Possible Answers:

$\left \langle -3,12,0\right \rangle$

$\left \langle 0,51,0\right \rangle$

$\left \langle 0,52,0\right \rangle$

$\left \langle -3,16,3\right \rangle$

Correct answer:

$\left \langle 0,52,0\right \rangle$

Explanation:

$determinant=\hat{i}(a_2b_3-b_2a_3)-\hat{j}(a_1b_3-b_1a_3)+\hat{k}(a_1b_2-b_1a_2)$

Using this formula, we evaluate using the vectors from the problem statement:

$\hat{i}(0-0)-\hat{j}(10+42)+\hat{k}(0-0)=\left \langle 0,52,0\right \rangle$

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Example Question #2591 : Calculus 3

Find the normal vector to the plane that contains the vectors $\left \langle 2,0,9\right \rangle$ and $\left \langle -3,0,12\right \rangle$

Possible Answers:

$\left \langle 0,51,0\right \rangle$

$\left \langle 27,33,3\right \rangle$

$\left \langle -2,48,0\right \rangle$

$\left \langle 3,5,18\right \rangle$

Correct answer:

$\left \langle 0,51,0\right \rangle$

Explanation:

$determinant=\hat{i}(a_2b_3-b_2a_3)-\hat{j}(a_1b_3-b_1a_3)+\hat{k}(a_1b_2-b_1a_2)$

Using this formula, we evaluate using the vectors from the problem statement:

$\hat{i}(0-0)-\hat{j}(24+27)+\hat{k}(0-0)=\left \langle 0,51,0\right \rangle$

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Example Question #2591 : Calculus 3

Find the Unit Normal Vector to the given plane.

3x+3y+4z=10 .

Possible Answers:

V_n=(3, 3, 4 )

$V_n=(\frac{1}{\sqrt{34}}, \frac{1}{\sqrt{34}}, \frac{1}{\sqrt{34}} )$

$V_n=(\frac{3}{\sqrt{34}}, \frac{3}{\sqrt{34}}, \frac{4}{\sqrt{34}} )$

$V_n=(\frac{3}{{34}}, \frac{3}{{34}}, \frac{4}{{34}} )$

$V_n=(\frac{1}{{34}}, \frac{1}{{34}}, \frac{1}{{34}} )$

Correct answer:

$V_n=(\frac{3}{\sqrt{34}}, \frac{3}{\sqrt{34}}, \frac{4}{\sqrt{34}} )$

Explanation:

Recall the definition of the Unit Normal Vector.

$V_n=\frac{V}{||V||}$

Let V=(3,3,4)

$||V||=\sqrt{3^2+3^2+4^2}=\sqrt{34}$

$V_n=(\frac{3}{\sqrt{34}}, \frac{3}{\sqrt{34}}, \frac{4}{\sqrt{34}} )$

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Example Question #2591 : Calculus 3

Find the Unit Normal Vector to the given plane.

3x+3y+4z=10 .

Possible Answers:

$V_n=(\frac{1}{\sqrt{34}}, \frac{1}{\sqrt{34}}, \frac{1}{\sqrt{34}} )$

$V_n=(\frac{1}{{34}}, \frac{1}{{34}}, \frac{1}{{34}} )$

$V_n=(\frac{3}{\sqrt{34}}, \frac{3}{\sqrt{34}}, \frac{4}{\sqrt{34}} )$

V_n=(3, 3, 4 )

$V_n=(\frac{3}{{34}}, \frac{3}{{34}}, \frac{4}{{34}} )$

Correct answer:

$V_n=(\frac{3}{\sqrt{34}}, \frac{3}{\sqrt{34}}, \frac{4}{\sqrt{34}} )$

Explanation:

Recall the definition of the Unit Normal Vector.

$V_n=\frac{V}{||V||}$

Let V=(3,3,4)

$||V||=\sqrt{3^2+3^2+4^2}=\sqrt{34}$

$V_n=(\frac{3}{\sqrt{34}}, \frac{3}{\sqrt{34}}, \frac{4}{\sqrt{34}} )$

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Calculus 3 : Normal Vectors

Study concepts, example questions & explanations for Calculus 3

All Calculus 3 Resources

Example Questions

Example Question #91 : Normal Vectors

Example Question #91 : Normal Vectors

Example Question #2581 : Calculus 3

Example Question #92 : Normal Vectors

Example Question #2591 : Calculus 3

Example Question #2591 : Calculus 3

Example Question #2591 : Calculus 3

All Calculus 3 Resources

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