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Calculus 2 : Vector Calculations

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Example Questions

Example Question #1 : Vector Calculations

Find the cross product of the following two vectors:

$\vec{a}=\left \langle -4,3,-1 \right \rangle$

$\vec{b}=\left \langle 7,-2,3 \right \rangle$

Possible Answers:

$\left \langle -6,8,11 \right \rangle$

$\left \langle 4,-2,17 \right \rangle$

$\left \langle 12,4,7 \right \rangle$

$\left \langle 7,5,-13 \right \rangle$

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

Correct answer:

$\left \langle 7,5,-13 \right \rangle$

Explanation:

We obtain the cross product of the two vectors by setting up the following matrix:

$\begin{bmatrix} \vec{i} &\vec{j} &\vec{k} \\-4 &3 &-1 \\7 &-2 &3 \end{bmatrix}$

Where our first row represents the unit vectors, the second row represents vector a, and the third row represents vector b. The first component of our cross product is obtained by taking the determinant of the matrix left by crossing out the row and column in which $\vec{i}$ is located. Accordingly, our second and third components are found by taking the determinant of the matrix left by crossing out the rows and columns in which $\vec{j}$ and $\vec{k}$ are located, respectively. This process gives us the following simple equation for expressing the cross product of two vectors, into which we can plug in the components of our vectors to find the cross product:

$\vec{a}\times \vec{b}=\left \langle a_2b_3-a_3b_2,a_3b_1-a_1b_3,a_1b_2-a_2b_1 \right \rangle$

$=\left \langle (3)(3)-(-1)(-2),(-1)(7)-(-4)(3),(-4)(-2)-(3)(7) \right \rangle$

$=\left \langle 7,5,-13 \right \rangle$

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Example Question #1 : Vector Calculations

Evaluate the dot product: $\left \langle 3,2\right \rangle\cdot\left \langle -6,9\right \rangle$

Possible Answers:

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

$\left \langle -3,11\right \rangle$

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

Correct answer:

Explanation:

Let vector $\vec{a}=\left \langle a1,a2\right \rangle$ and $\vec{b}=\left \langle b1,b2\right \rangle$ .

The dot product is equal to:

$\vec{a}\cdot\vec{b} = a1\cdot b1+ a2\cdot b2$

Following this rule for the current problem, simplify.

$\left \langle 3,2\right \rangle\cdot\left \langle -6,9\right \rangle = (3)(-6)+(2)(9)=0$

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Example Question #3 : Vector Calculations

Evaluate: $\left \langle3,-1,2 \right \rangle \cdot \left \langle -2,4,5\right \rangle$

Possible Answers:

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

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

Correct answer:

Explanation:

Let vector $\vec{a}=\left \langle a1,a2,a3\right \rangle$ and $\vec{b}=\left \langle b1,b2,b3\right \rangle$ .

Use the following formula to solve this dot product:

$\vec{a}\cdot\vec{b} = a1\cdot b1+ a2\cdot b2+a3\cdot b3$

Substitute and solve.

$\left \langle3,-1,2 \right \rangle \cdot \left \langle -2,4,5\right \rangle=(3)(-2)+(-1)(4)+(2)(5 )=0$

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Example Question #1 : Vector Calculations

Calculate the cross product: $(i-j)\times-k$

Possible Answers:

i-j

-j+k

i-j-k

i-k

i+j

Correct answer:

i+j

Explanation:

The vectors can be rewritten in the following form:

$\begin{vmatrix} i& j& k\\ 1&-1 &0 \\ 0& 0& -1 \end{vmatrix}$

One method of solving the 3 by 3 determinant is to break this down into 2 by 2 determinants. The determinant of 2 by 2 matrices are computed by ad-bc , such that:

$D=\begin{vmatrix} a&b \\ c& d \end{vmatrix}$

Rewrite the 3 by 3 determinant.

$=i\begin{vmatrix} -1& 0\\ 0&-1 \end{vmatrix}- j\begin{vmatrix} 1&0 \\ 0&-1 \end{vmatrix}+k\begin{bmatrix} 1&-1 \\ 0&0 \end{bmatrix}$

=i[(-1)(-1)-0] -j[(1)(-1)-0]+k[0]

= i[1]-j[-1]

= i+j

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Example Question #1 : Vector Calculations

A sling shoots a rock feet per second at an elevation angle of degrees. What are the horizontal and vertical components in vector form?

Possible Answers:

$\left \langle -9.848,-1.736\right \rangle$

$\left \langle -9.848,1.736\right \rangle$

$\left \langle 9.848,1.736\right \rangle$

$\left \langle -8.391,-5.440\right \rangle$

$\left \langle 8.391,5.440\right \rangle$

Correct answer:

$\left \langle 9.848,1.736\right \rangle$

Explanation:

The horizontal and vertical components are shown below:

$Vx=Vcos\Theta$

$Vy=Vsin\Theta$

Plug in the velocity and the given angle to the equations.

$Vx=Vcos\Theta=10cos(10) = 9.848$

$Vy=Vsin\Theta = 10sin(10)=1.736$

Therefore, the components in vector form is $\left \langle 9.848,1.736\right \rangle$ .

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Example Question #2 : Vector Calculations

Evaluate the dot product of $\left \langle 1,6\right \rangle$ and $\left \langle 1,6\right \rangle$ .

Possible Answers:

$\infty$

Correct answer:

Explanation:

Let vectors $\vec{a}=\left \langle a1,a2\right \rangle$ and $\vec{b}=\left \langle b1,b2\right \rangle$ .

The formula for the dot product is:

$\vec{a}\cdot\vec{b} = a1\cdot b1+ a2\cdot b2$

Follow this formula and simplify.

$\left \langle 1,6\right \rangle\cdot\left \langle 1,6\right \rangle = (1)(1)+(6)(6)=37$

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Example Question #2 : Vector Calculations

Solve: $\left \langle 3,4,5\right \rangle\cdot\left \langle1,2,3 \right \rangle$

Possible Answers:

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

$\left \langle 3,8,15\right \rangle$

$\left \langle 4,6,8\right \rangle$

Correct answer:

Explanation:

The problem $\left \langle 3,4,5\right \rangle\cdot\left \langle1,2,3 \right \rangle$ is in the form of a dot product. The final answer must be an integer, and not in vector form.

Write the formula for the dot product.

$\left \langle a_1,a_2,a_3\right \rangle\cdot\left \langle b_1,b_2,b_3\right \rangle=a_1b_1+a_2b_2+a_3b_3$

Substitute the givens and solve.

(3)(1)+(4)(2)+(5)(3)=3+8+15=26

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Example Question #8 : Vector Calculations

Suppose $\vec{a}= \left \langle 2,4,-3\right \rangle$ . Find the magnitude of $2\vec{a}$ .

Possible Answers:

$\sqrt{69}$

$2\sqrt{29}$

$\sqrt{29}$

Correct answer:

$2\sqrt{29}$

Explanation:

Calculate $2\vec{a}$ .

$\vec{a}= \left \langle 2,4,-3\right \rangle$

$2\vec{a}= \left \langle 4,8,-6\right \rangle$

Find the magnitude.

$|2\vec{a}|=\sqrt{4^2+8^2+(-6)^2}=\sqrt{16+64+36}=\sqrt{116}=2\sqrt{29}$

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Example Question #161 : Vector

Two particles move freely in two dimensional space. The first particle's location as a function of time is $\small \left<3t^{2}-t,4\right>$ , and the second particle's location is $\small \small \left<6,t+1/t\right>$ . Will the particles ever collide for $\small t>0$ ?

Possible Answers:

Impossible to determine

No, because the particles' $\small x$ and $\small y$ coordinates are never the same simultaneously at any instant in time.

Yes, because the particles have the same $\small x$ or $\small y$ component (not necessarily simultaneously).

No, because the particles never have the same $\small x$ or $\small y$ component (not necessarily simultaneously).

Yes, because the particles' $\small x$ and $\small y$ coordinates are the same simultaneously at a certain instant in time.

Correct answer:

No, because the particles' $\small x$ and $\small y$ coordinates are never the same simultaneously at any instant in time.

Explanation:

In order for the particles to collide, their $\small x$ and $\small y$ coordinates must be equal simultaneously. In order to check if this happens, we can set the particles' $\small x$ -coordinates and $\small y$ -coordinates equal to each other.

Let's start with the $\small x$ -coordinate:

$\small \small \small 3t^{2}-t=6\rightarrow 3t^2-t-6=0$ .

Using the quadratic formula

$\small \small t=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$ , we get $\small \small t=\frac{1+\sqrt{73}}{6}\approx1.59$

the other root is negative, so it can be discarded since $\small t>0$ .

Now let's do the $\small y$ -coordinate: $\small \small t+1/t=4\rightarrow t^2+1=4t\rightarrow t^2-4t+1=0$ . Use the quadratic formula again to solve for $\small t$ , and you'll get $\small 2\pm\sqrt{3}$ (these roots are approximately $\small 0.27$ and $\small 3.73$ ). The particles never have the same $\small x$ and $\small y$ coordinate simultaneously, so they do not collide.

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Example Question #1 : Vector Calculations

$\vec{a}=<3,2,1>$

$\vec{b}=<1,0,10>$

Calculate $\vec{a}\cdot\vec{b}$

Possible Answers:

Correct answer:

Explanation:

$\vec{a}=<3,2,1>$

$\vec{b}=<1,0,10>$

$\vec{a}\cdot\vec{b}$ is simply the dot product of these two vectors. Mathematically, this is calculated as follows.

$(3\cdot 1)+(2\cdot 0)+(1\cdot 10)$ =13

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Calculus 2 : Vector Calculations

Study concepts, example questions & explanations for Calculus 2

All Calculus 2 Resources

Example Questions

Example Question #1 : Vector Calculations

Example Question #1 : Vector Calculations

Example Question #3 : Vector Calculations

Example Question #1 : Vector Calculations

Example Question #1 : Vector Calculations

Example Question #2 : Vector Calculations

Example Question #2 : Vector Calculations

Example Question #8 : Vector Calculations

Example Question #161 : Vector

Example Question #1 : Vector Calculations

All Calculus 2 Resources

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