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Calculus 3 : Vectors and Vector Operations

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Example Questions

Example Question #591 : Vectors And Vector Operations

Find the Unit Normal Vector to the given plane.

3x+3y+4z=10 .

Possible Answers:

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

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

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

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

V_n=(3, 3, 4 )

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 #1 : Binormal Vectors

Find the binormal vector of $v(t)=sin(t)\hat{i}+cos(t)\hat{j}$ .

Possible Answers:

Does not exist.

$B(t)=sin(t)\hat{i}+cos(t)\hat{j}$

$B(t)=-\hat{k}$

$B(t)=cos(t)\hat{i}-sin(t)\hat{j}$

Correct answer:

$B(t)=-\hat{k}$

Explanation:

To find the binormal vector, you must first find the unit tangent vector, then the unit normal vector.

The equation for the unit tangent vector, T(t) , is

$T(t)=\frac{r(t)}{\left \| r(t)\right \|}$

where r(t) is the vector and $\left \| r(t)\right \|$ is the magnitude of the vector.

The equation for the unit normal vector, N(t) , is

$N(t)=\frac{T'(t)}{\left \| T'(t)\right \|}$

where T'(t) is the derivative of the unit tangent vector and $\left \| T'(t)\right \|$ is the magnitude of the derivative of the unit vector.

The binormal vector is the cross product of unit tangent and unit normal vectors, or

$B(t)=T(t)\times N(t)$

For this problem

$v(t)=sin(t)\hat{i}+cos(t)\hat{j}$

$\left \| v(t)\right \|=\sqrt{\left (sin(t) \right )^{2}+\left (cos(t) \right )^{2}}=\sqrt{1}=1$

$T(t)=\frac{v(t)}{\left \| v(t)\right \|}=\frac{sin(t)\hat{i}+cos(t)\hat{j}}{1}=sin(t)\hat{i}+cos(t)\hat{j}$

$T'(t)=\frac{d}{dt}\left (sin(t)\hat{i}+cos(t)\hat{j} \right )=cos(t)\hat{i}-sin(t)\hat{j}$

$\left \| T'(t)\right \|=\sqrt{\left (cos(t) \right )^{2}+\left (sin(t) \right )^{2}}=\sqrt{1}=1$

$N(t)=\frac{T'(t)}{\left \| T'(t)\right \|}=\frac{cos(t)\hat{i}-sin(t)\hat{j}}{1}=cos(t)\hat{i}-sin(t)\hat{j}$

$B(t)=T(t)\times N(t)=\begin{vmatrix} \hat{i}& \hat{j} &\hat{k} \\ sin(t)& cos(t)& 0\\ cos(t)&-sin(t) & 0 \end{vmatrix}$

$B(t)=\hat{i}\left [ cos(t)*0+sin(t)*0\right ] -\hat{j}\left [ sin(t)*0-cos(t)*0\right ] +\hat{k}\left [ sin(t)*(-sin(t))-cos(t)*cos(t)\right ]$

$B(t)=0\hat{i}-0\hat{j} -\hat{k}\left [ sin^{2}(t)+cos^{2}(t)\right ]=-\hat{k}$

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Example Question #2 : Binormal Vectors

Find the binormal vector of $v(t)=5cos(t)\hat{i}+5sin(t)\hat{j}+2\hat{k}$ .

Possible Answers:

$B(t)=-\frac{2}{\sqrt{29}}cos(t)\hat{i} -\frac{2}{\sqrt{29}}sin(t)\hat{j} + \frac{5}{\sqrt{29}}\hat{k}$

$N(t)=\frac{-5}{\sqrt{29}}sin(t)\hat{i}+\frac{5}{\sqrt{29}}cos(t)\hat{j}+0\hat{k}$

$B(t)=5cos(t)\hat{i}+5sin(t)\hat{j}+2\hat{k}$

Does not exist

Correct answer:

$B(t)=-\frac{2}{\sqrt{29}}cos(t)\hat{i} -\frac{2}{\sqrt{29}}sin(t)\hat{j} + \frac{5}{\sqrt{29}}\hat{k}$

Explanation:

To find the binormal vector, you must first find the unit tangent vector, then the unit normal vector.

The equation for the unit tangent vector, T(t) , is

$T(t)=\frac{r(t)}{\left \| r(t)\right \|}$

where r(t) is the vector and $\left \| r(t)\right \|$ is the magnitude of the vector.

The equation for the unit normal vector, N(t) , is

$N(t)=\frac{T'(t)}{\left \| T'(t)\right \|}$

where T'(t) is the derivative of the unit tangent vector and $\left \| T'(t)\right \|$ is the magnitude of the derivative of the unit vector.

For this problem

$v(t)=5cos(t)\hat{i}+5sin(t)\hat{j}+2\hat{k}$

$\left \| v(t)\right \|=\sqrt{\left (5cos(t) \right )^{2}+\left (5sin(t) \right )^{2}+(2)^{2}}=\sqrt{25+4}=\sqrt{29}$

$T(t)=\frac{v(t)}{\left \| v(t)\right \|}=\frac{5cos(t)\hat{i}+5sin(t)\hat{j}+2\hat{k}}{\sqrt{29}}$

$T(t)=\frac{v(t)}{\left \| v(t)\right \|}=\frac{5}{\sqrt{29}}cos(t)\hat{i}+\frac{5}{\sqrt{29}}sin(t)\hat{j}+\frac{2}{\sqrt{29}}\hat{k}$

$T'(t)=\frac{d}{dt}\left (\frac{5}{\sqrt{29}}cos(t)\hat{i}+\frac{5}{\sqrt{29}}sin(t)\hat{j}+\frac{2}{\sqrt{29}}\hat{k} \right )$

$T'(t)=\frac{-5}{\sqrt{29}}sin(t)\hat{i}+\frac{5}{\sqrt{29}}cos(t)\hat{j}+0\hat{k}$

$T'(t)=\frac{-5}{\sqrt{29}}sin(t)\hat{i}+\frac{5}{\sqrt{29}}cos(t)\hat{j}$

$\left \| T'(t)\right \|=\sqrt{\left (\frac{-5}{\sqrt{29}}sin(t) \right )^{2}+\left (\frac{5}{\sqrt{29}}cos(t) \right )^{2}}$

$\left \| T'(t)\right \|=\sqrt{\frac{25}{29}sin^2(t)+\frac{25}{29}cos^2(t)}=\sqrt{\frac{25}{29}}=\frac{5}{\sqrt{29}}$

$N(t)=\frac{T'(t)}{\left \| T'(t)\right \|}=\frac{\frac{-5}{\sqrt{29}}sin(t)\hat{i}+\frac{5}{\sqrt{29}}cos(t)\hat{j}}{\frac{5}{\sqrt{29}}}=-sin(t)\hat{i}+cos(t)\hat{j}$

$B(t)=T(t)\times N(t)=\begin{vmatrix} \hat{i}& \hat{j} &\hat{k} \\ \frac{5}{\sqrt{29}}cos(t)& \frac{5}{\sqrt{29}}sin(t)& \frac{2}{\sqrt{29}}\\ -sin(t)& cos(t) & 0 \end{vmatrix}$

$B(t)=\hat{i}\left [ \frac{5}{\sqrt{29}}sin(t)*0-cos(t)*\frac{2}{\sqrt{29}}\right ] -\hat{j}\left [ \frac{5}{\sqrt{29}}cos(t)*0+sin(t)*\frac{2}{\sqrt{29}}\right ] +\hat{k}\left [ \frac{5}{\sqrt{29}}cos(t)*cos(t)+\frac{5}{\sqrt{29}}sin(t)*(sin(t))\right ]$

$B(t)=\hat{i}\left [ -\frac{2}{\sqrt{29}}cos(t)\right ] -\hat{j}\left [ \frac{2}{\sqrt{29}}sin(t)\right ] +\hat{k}\left [ \frac{5}{\sqrt{29}}cos^{2}(t)+\frac{5}{\sqrt{29}}sin^{2}(t)\right ]$

$B(t)=-\frac{2}{\sqrt{29}}cos(t)\hat{i} -\frac{2}{\sqrt{29}}sin(t)\hat{j} + \frac{5}{\sqrt{29}}\hat{k}$

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Example Question #3 : Binormal Vectors

Find the binormal vector for:

$\mathbf{r}(t)=[2\sin(t), t, 2\cos(t)]$

Possible Answers:

$\left [ \frac{\cos(t)}{\sqrt{5}}, \frac{-2}{\sqrt{5}}, \frac{-\sin(t)}{\sqrt{5}}\right ]$

$\left [ \frac{-\sin(t)}{\sqrt{5}}, \frac{2}{\sqrt{5}}, \frac{\cos(t)}{\sqrt{5}}\right ]$

$\left [ \frac{2}{\sqrt{5}}, \frac{-\sin(t)}{\sqrt{5}},\frac{\cos(t)}{\sqrt{5}}\right ]$

$\left [ \frac{\sin(t)}{\sqrt{5}}, \frac{-2}{\sqrt{5}}, \frac{-\cos(t)}{\sqrt{5}}\right ]$

$\left [ \frac{-\cos(t)}{\sqrt{5}}, \frac{2}{\sqrt{5}}, \frac{\sin(t)}{\sqrt{5}}\right ]$

Correct answer:

$\left [ \frac{-\cos(t)}{\sqrt{5}}, \frac{2}{\sqrt{5}}, \frac{\sin(t)}{\sqrt{5}}\right ]$

Explanation:

The binormal vector is defined as

$\mathbf{B}(t)=\mathbf{T}(t)\times\mathbf{N}(t)$ $\mathbf{B}(t)=\mathbf{T}(t)\times\mathbf{N}(t)$

Where T(t) (the tangent vector) and N(t) (the normal vector) are:

$\mathbf{T}(t)=\frac{\mathbf{r}'(t)}{|\mathbf{r}'(t)|}=\frac{1}{\sqrt{5}}[2\cos (t), 1, -2\sin (t)]$

and

$\mathbf{N}(t)=\frac{\mathbf{T}'(t)}{|\mathbf{T}'(t)|}=[-\sin (t), 0, -\cos (t)]$

The binormal vector is:

$\mathbf{B}(t)=\left [ \frac{-\cos(t)}{\sqrt{5}}, \frac{2}{\sqrt{5}}, \frac{\sin(t)}{\sqrt{5}}\right ]$

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Example Question #1 : Matrices

Calculate the determinant of Matrix .

$A=\begin{bmatrix} 1 & 2&3 \\ 0& 2& 1\\ 3& 4& 5 \end{bmatrix}$

Possible Answers:

$\det(A)=-2$

$\det(A)=1$

$\det(A)=0$

$\det(A)=-6$

$\det(A)=6$

Correct answer:

$\det(A)=-6$

Explanation:

In order to find the determinant of , we first need to copy down the first two columns into columns 4 and 5.

$A=\begin{bmatrix} 1 & 2&3 & 1 &2\\ 0& 2& 1&0 &2\\ 3& 4& 5 &3 &4\end{bmatrix}$

The next step is to multiply the down diagonals.

$A_d=1\cdot2\cdot5+2\cdot1\cdot 3+3\cdot0\cdot4=10+6+0=16$

The next step is to multiply the up diagonals.

$A_u=3\cdot2\cdot3+4\cdot1\cdot1+5\cdot0\cdot2=18+4=22$

The last step is to substract A_u from A_d .

$\det(A)=A_d-A_u=16-22=-6$

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Example Question #1 : Matrices

Calculate the determinant of Matrix .

$A=\begin{bmatrix} 0 & 1&5 \\ 10& -3& 2\\ 1& 2& -5 \end{bmatrix}$

Possible Answers:

$\det(A)=38$

$\det(A)=67$

$\det(A)=-38$

$\det(A)=167$

$\det(A)=0$

Correct answer:

$\det(A)=167$

Explanation:

In order to find the determinant of , we first need to copy down the first two columns into columns 4 and 5.

$A=\begin{bmatrix} 0 & 1&5 & 0 &1\\ 10& -3& 2&10 &-3\\ 1& 2& -5 &1 &2\end{bmatrix}$

The next step is to multiply the down diagonals.

$A_d=0\cdot-3\cdot-5+1\cdot2\cdot 1+5\cdot10\cdot2=0+2+100=102$

The next step is to multiply the up diagonals.

$A_u=1\cdot-3\cdot5+2\cdot2\cdot0+(-5)\cdot10\cdot1=-15+0-50=-65$

The last step is to substract A_u from A_d .

$\det(A)=A_d-A_u=102-(-65)=167$

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Example Question #1 : Matrices

Calculate the determinant of .

$A=\begin{bmatrix} 3 & 4\\ -2 & 10 \end{bmatrix}$

Possible Answers:

Correct answer:

Explanation:

All we need to do is multiply the main diagonal and substract it from the off diagonal.

$\det(A)=3\cdot10-4(-2)=30+8=38$

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Example Question #2 : Matrices

Which of the following is a way to represent the computation $<2,6,1> \cdot <5,2,4>$ using matrices?

Possible Answers:

$\begin{bmatrix} 2 \\ 6 \\ 1 \end{bmatrix} \begin{bmatrix} 5 \\ 2 \\ 4 \end{bmatrix}$

None of the other answers

$\begin{bmatrix} 2 & 6 & 1 \end{bmatrix} \begin{bmatrix} 5& 2 & 4 \end{bmatrix}$

$\begin{bmatrix} 2 & 6 & 1 \end{bmatrix} \begin{bmatrix} 5\\ 2\\ 4\\ \end{bmatrix}$

All of the above answers

Correct answer:

$\begin{bmatrix} 2 & 6 & 1 \end{bmatrix} \begin{bmatrix} 5\\ 2\\ 4\\ \end{bmatrix}$

Explanation:

This choice is the only pair with a well-defined operation; the others are meaningless in terms of matrix multiplication.

Computing this we get

$\begin{bmatrix} 2 & 6 & 1 \end{bmatrix} \begin{bmatrix} 5\\ 2\\ 4\\ \end{bmatrix} = (2)(5)+(6)(2)+(1)(4) =26$ .

Which is the same as

$<2,6,1> \cdot <5,2,4> = (2)(5)+(6)(2)+(1)(4) =26$ .

This operation is true for (real) -dimensional vectors in general;

$\begin{bmatrix} a & b & c \end{bmatrix} \begin{bmatrix} d\\ e\\ f\\ \end{bmatrix} = ad+be+cf$

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Example Question #5 : Matrices

Find the determinant of the matrix $A=\begin{vmatrix} 1&2 \\ 5& 9\end{vmatrix}$

Possible Answers:

Correct answer:

Explanation:

The determinant of a 2x2 matrix can be found by cross multiplying terms as follows:

$det\begin{vmatrix} a&b \\ c&d \end{vmatrix}=ad-bc$

For the matrix $A=\begin{vmatrix} 1&2 \\ 5& 9\end{vmatrix}$

The determinant is thus:

|A|=9-10=-1

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Example Question #1 : Matrices

Find the determinant of the matrix $A=\begin{vmatrix} 4&2 \\-1 &3 \end{vmatrix}$

Possible Answers:

Correct answer:

Explanation:

The determinant of a 2x2 matrix can be found by cross multiplying terms as follows:

$det\begin{vmatrix} a&b \\ c&d \end{vmatrix}=ad-bc$

For the matrix $A=\begin{vmatrix} 4&2 \\-1 &3 \end{vmatrix}$

The determinant is thus:

|A|=12-(-2)=14

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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 #591 : Vectors And Vector Operations

Example Question #1 : Binormal Vectors

Example Question #2 : Binormal Vectors

Example Question #3 : Binormal Vectors

Example Question #1 : Matrices

Example Question #1 : Matrices

Example Question #1 : Matrices

Example Question #2 : Matrices

Example Question #5 : Matrices

Example Question #1 : Matrices

All Calculus 3 Resources

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