Vectors and Vector Operations - AP Calculus BC

Card 0 of 3780

Question

Which of the following is true? (assume all vectors are -dimensional. $\theta$ is the acute angle between the two vectors.)

Answer

There are a few approaches to answering this.

One is to notice that the right hand side of each equation is some number times some number times some other number, whereas $\mathbf{a} \times \mathbf{b}$ is a vector. It is not possible for a vector to equal a number (or a "scalar" technically), so

$\mathbf{a} \times \mathbf{b} = |\mathbf{a}| |\mathbf{b}|\cos(\theta)$ , and

$\mathbf{a} \times \mathbf{b} = |\mathbf{a}| |\mathbf{b}|\sin(\theta)$ are out of the question.

$|\mathbf{a} \times \mathbf{b}| = |\mathbf{a}| |\mathbf{b}|\cos(\theta)$ is not correct since the right hand side of the equation is the definition of the dot product of two vectors, which is not represented by the left hand side.

$|\mathbf{a} \times \mathbf{b}| = |\mathbf{a}| |\mathbf{b}|\sin(\theta)$ is true. While it's not the definition of the cross product, it is a formula used to find the area of the parallelogram formed by vectors $\mathbf{a},\mathbf{b}$ .

Compare your answer with the correct one above

Question

What is the product of $\vec {a}=<-2,0,7>$ and $\vec {b}=<-3,4,1>$

Answer

Compare your answer with the correct one above

Question

Evaluate the dot product of the vectors .

Answer

To evaluate the dot product of any two vectors, we multiply them component-wise.

$23\times42 +78 \times 64 = 5958$ .

Compare your answer with the correct one above

Question

What is the angle between the vectors $\vec{a}= \hat{i}-3\hat{j} +7\hat{k}$ and $\vec{b}= -2\hat{i}+\hat{j} +4\hat{k}$ ?

Answer

To find the angle between vectors, we must use the dot product formula

$a\cdot b=\left | a\right |\left | b\right |cos\theta$

where $a\cdot b$ is the dot product of the vectors and , respectively.

$\left | a\right |$ and $\left | b\right |$ are the magnitudes of vectors and , respectively.

$cos\theta$ is the angle between the two vectors.

Let vector be represented as $\vec{a}= a_{1}\hat{i}+a_{2}\hat{j} +a_{3}\hat{k}$ and vector be represented as $\vec{b}= b_{1}\hat{i}+b_{2}\hat{j} +b_{3}\hat{k}$ .

The dot product of the vectors and is $a\cdot b=a_{1}( b_{1})+a_{2}( b_{2})+a_{3}( b_{3})$ .

The magnitude of vector is $\left | a\right |=\sqrt{(a_{1})^{2}+(a_{2})^{2}+(a_{3})^{2}}$ and vector is $\left | b\right |=\sqrt{(b_{1})^{2}+(b_{2})^{2}+(b_{3})^{2}}$ .

Rearranging the dot product formula to solve for $\theta$ gives us $\theta=cos^{-1}\left ( \frac{a\cdot b}{\left | a\right |\left | b\right |}\right )$

For this problem,

$a\cdot b=a_{1}( b_{1})+a_{2}( b_{2})+a_{3}( b_{3})=1(-2)+(-3)(1)+(7)(4)$

$a\cdot b=-2-3+28=23$

$\left | a\right |=\sqrt{(a_{1})^{2}+(a_{2})^{2}+(a_{3})^{2}}=\sqrt{(1)^{2}+(-3)^{2}+(7)^{2}}$

$\left | a\right |=\sqrt{1+9+49}=\sqrt{59}$

$\left | b\right |=\sqrt{(b_{1})^{2}+(b_{2})^{2}+(b_{3})^{2}}=\sqrt{(-2)^{2}+(1)^{2}+(4)^{2}}$

$\left | b\right |=\sqrt{4+1+16}=\sqrt{21}$

$\theta=cos^{-1}\left ( \frac{a\cdot b}{\left | a\right |\left | b\right |}\right )=cos^{-1}\left ( \frac{23}{\sqrt{59}\sqrt{21}}\right )=cos^{-1}\left ( \frac{23}{\sqrt{1239}}\right )$

$\theta\approx49.2^{\circ}$

Compare your answer with the correct one above

Question

Find the angle between these two vectors, , and .

Answer

Lets remember the formula for finding the angle between two vectors.

$\cos(\theta)=\frac{(u\cdot v)}{\left | u\right |\left | v\right |}$

$\cos(\theta)=\frac{((3\cdot 9)+(4\cdot (-2)))}{\sqrt{3^2+4^2}\sqrt{9^2+(-2)^2}}$

$\cos(\theta)=\frac{19}{\sqrt{25}\sqrt{85}}$

$\cos(\theta)=\frac{19}{5\sqrt{85}}$

$\theta=\cos^{-1}\Big(\frac{19}{5\sqrt{85}}\Big)\approx65.66^\circ$

Compare your answer with the correct one above

Question

Calculate the angle between , .

Answer

Lets recall the equation for finding the angle between vectors.

$\cos(\theta)=\frac{(u\cdot v)}{\left | u\right |\left | v\right |}$

$\cos(\theta)=\frac{((1\cdot 3)+(2\cdot 4))}{\sqrt{1^2+2^2}\sqrt{3^2+4^2}}$

$\cos(\theta)=\frac{11}{\sqrt{5}\sqrt{25}}$

$\cos(\theta)=\frac{11}{5\sqrt{5}}$

$\theta=\cos^{-1}\Big(\frac{11}{5\sqrt{5}}\Big)\approx 10.3^\circ$

Compare your answer with the correct one above

Question

What is the angle between the vectors $\vec{a}=2 \hat{i}-1\hat{j} +7\hat{k}$ and $\vec{b}= \hat{i}+2\hat{j}$ ?

Answer

To find the angle between vectors, we must use the dot product formula

$a\cdot b=\left | a\right |\left | b\right |cos\theta$

where $a\cdot b$ is the dot product of the vectors and , respectively.

$\left | a\right |$ and $\left | b\right |$ are the magnitudes of vectors and , respectively.

$cos\theta$ is the angle between the two vectors.

Let vector be represented as $\vec{a}= a_{1}\hat{i}+a_{2}\hat{j} +a_{3}\hat{k}$ and vector be represented as $\vec{b}= b_{1}\hat{i}+b_{2}\hat{j} +b_{3}\hat{k}$ .

The dot product of the vectors and is $a\cdot b=a_{1}( b_{1})+a_{2}( b_{2})+a_{3}( b_{3})$ .

The magnitude of vector is $\left | a\right |=\sqrt{(a_{1})^{2}+(a_{2})^{2}+(a_{3})^{2}}$ and vector is $\left | b\right |=\sqrt{(b_{1})^{2}+(b_{2})^{2}+(b_{3})^{2}}$ .

Rearranging the dot product formula to solve for $\theta$ gives us $\theta=cos^{-1}\left ( \frac{a\cdot b}{\left | a\right |\left | b\right |}\right )$

For this problem,

$a\cdot b=a_{1}( b_{1})+a_{2}( b_{2})+a_{3}( b_{3})=2(1)+(-1)(2)+(7)(0)$

$a\cdot b=2-2+0=0$

$\left | a\right |=\sqrt{(a_{1})^{2}+(a_{2})^{2}+(a_{3})^{2}}=\sqrt{(2)^{2}+(1)^{2}+(7)^{2}}$

$\left | a\right |=\sqrt{4+1+49}=\sqrt{54}$

$\left | b\right |=\sqrt{(b_{1})^{2}+(b_{2})^{2}+(b_{3})^{2}}=\sqrt{(1)^{2}+(2)^{2}+(0)^{2}}$

$\left | b\right |=\sqrt{1+4}=\sqrt{5}$

$\theta=cos^{-1}\left ( \frac{a\cdot b}{\left | a\right |\left | b\right |}\right )=cos^{-1}\left ( \frac{0}{\sqrt{54}\sqrt{5}}\right )=cos^{-1}\left (0\right )$

$\theta\approx90^{\circ}$

The vectors are perpendicular

Compare your answer with the correct one above

Question

What is the angle between the vectors $\vec{a}=-2 \hat{i}-3\hat{j} +\hat{k}$ and $\vec{b}= \hat{i}+2\hat{j}+5\hat{k}$ ?

Answer

To find the angle between vectors, we must use the dot product formula

$a\cdot b=\left | a\right |\left | b\right |cos\theta$

where $a\cdot b$ is the dot product of the vectors and , respectively.

$\left | a\right |$ and $\left | b\right |$ are the magnitudes of vectors and , respectively.

$cos\theta$ is the angle between the two vectors.

Let vector be represented as $\vec{a}= a_{1}\hat{i}+a_{2}\hat{j} +a_{3}\hat{k}$ and vector be represented as $\vec{b}= b_{1}\hat{i}+b_{2}\hat{j} +b_{3}\hat{k}$ .

The dot product of the vectors and is $a\cdot b=a_{1}( b_{1})+a_{2}( b_{2})+a_{3}( b_{3})$ .

The magnitude of vector is $\left | a\right |=\sqrt{(a_{1})^{2}+(a_{2})^{2}+(a_{3})^{2}}$ and vector is $\left | b\right |=\sqrt{(b_{1})^{2}+(b_{2})^{2}+(b_{3})^{2}}$ .

Rearranging the dot product formula to solve for $\theta$ gives us $\theta=cos^{-1}\left ( \frac{a\cdot b}{\left | a\right |\left | b\right |}\right )$

For this problem,

$a\cdot b=a_{1}( b_{1})+a_{2}( b_{2})+a_{3}( b_{3})=-2(1)+(-3)(2)+(1)(5)$

$a\cdot b=-2-6+5=-3$

$\left | a\right |=\sqrt{(a_{1})^{2}+(a_{2})^{2}+(a_{3})^{2}}=\sqrt{(-2)^{2}+(-3)^{2}+(1)^{2}}$

$\left | a\right |=\sqrt{4+9+1}=\sqrt{14}$

$\left | b\right |=\sqrt{(b_{1})^{2}+(b_{2})^{2}+(b_{3})^{2}}=\sqrt{(1)^{2}+(2)^{2}+(5)^{2}}$

$\left | b\right |=\sqrt{1+4+25}=\sqrt{30}$

$\theta=cos^{-1}\left ( \frac{a\cdot b}{\left | a\right |\left | b\right |}\right )=cos^{-1}\left ( \frac{-3}{\sqrt{14}\sqrt{30}}\right )=cos^{-1}\left (\frac{-3}{\sqrt{420}}\right )$

$\theta\approx98.42^{\circ}$

Compare your answer with the correct one above

Question

What is the angle between the vectors $\vec{a}= \hat{i}+3\hat{j} +5\hat{k}$ and $\vec{b}= 2\hat{i}+6\hat{j}+10\hat{k}$ ?

Answer

To find the angle between vectors, we must use the dot product formula

$a\cdot b=\left | a\right |\left | b\right |cos\theta$

where $a\cdot b$ is the dot product of the vectors and , respectively.

$\left | a\right |$ and $\left | b\right |$ are the magnitudes of vectors and , respectively.

$cos\theta$ is the angle between the two vectors.

Let vector be represented as $\vec{a}= a_{1}\hat{i}+a_{2}\hat{j} +a_{3}\hat{k}$ and vector be represented as $\vec{b}= b_{1}\hat{i}+b_{2}\hat{j} +b_{3}\hat{k}$ .

The dot product of the vectors and is $a\cdot b=a_{1}( b_{1})+a_{2}( b_{2})+a_{3}( b_{3})$ .

The magnitude of vector is $\left | a\right |=\sqrt{(a_{1})^{2}+(a_{2})^{2}+(a_{3})^{2}}$ and vector is $\left | b\right |=\sqrt{(b_{1})^{2}+(b_{2})^{2}+(b_{3})^{2}}$ .

Rearranging the dot product formula to solve for $\theta$ gives us $\theta=cos^{-1}\left ( \frac{a\cdot b}{\left | a\right |\left | b\right |}\right )$

For this problem,

$a\cdot b=a_{1}( b_{1})+a_{2}( b_{2})+a_{3}( b_{3})=1(2)+(3)(6)+(5)(10)$

$a\cdot b=2+18+50=70$

$\left | a\right |=\sqrt{(a_{1})^{2}+(a_{2})^{2}+(a_{3})^{2}}=\sqrt{(1)^{2}+(3)^{2}+(5)^{2}}$

$\left | a\right |=\sqrt{1+9+25}=\sqrt{35}$

$\left | b\right |=\sqrt{(b_{1})^{2}+(b_{2})^{2}+(b_{3})^{2}}=\sqrt{(2)^{2}+(6)^{2}+(10)^{2}}$

$\left | b\right |=\sqrt{4+36+100}=\sqrt{140}$ $\theta=cos^{-1}\left ( \frac{a\cdot b}{\left | a\right |\left | b\right |}\right )=cos^{-1}\left ( \frac{70}{\sqrt{35}\sqrt{140}}\right )=cos^{-1}\left (\frac{70}{\sqrt{4900}}\right )$

$\theta=cos^{-1}\left (\frac{70}{70}\right )=cos^{-1}\left (1\right )$

$\theta\approx0^{\circ}$

The two vectors are parallel.

Compare your answer with the correct one above

Question

Find the angle between the two vectors.

$v=<\sqrt{3}, 1>$

Answer

To find the angle between two vector we use the following formula

$cos , \theta = \frac{u\bullet v}{\left | u\right |\left | v\right |}$

and solve for $\theta$ .

Given

$v=<\sqrt{3}, 1>$ we find

$u\bullet v= 2\sqrt3+01=2\sqrt3$

$\left | u\right |=\sqrt{2^2+0^2}=\sqrt4=2$

$\left | v\right |=\sqrt{\sqrt{3}^2+1^2}=\sqrt4=2$

Plugging these values in we get

$cos, \theta = \frac{2\sqrt3}{22}=\frac{\sqrt3}{2}$

To find $\theta$ we calculate the $cos^{-1}$ of both sides

$cos^{-1}[cos(\theta)]=cos^{-1}(\frac{\sqrt3}{2})$

and find that

$\theta = \frac{\pi}{6}$

Compare your answer with the correct one above

Question

Find the approximate acute angle in degrees between the vectors .

Answer

To find the angle between two vectors, use the formula

$\theta = \cos^{-1}(\frac{\mathbf{a}\cdot\mathbf{b}}{|\mathbf{a}||\mathbf{b}|})$ .

$\theta = \cos^{-1}(\frac{<2,2,5>\cdot<0,5,-1>}{\sqrt{4+4+25}\sqrt{0+25+1}})$

$\theta = \cos^{-1}(\frac{5}{\sqrt{858}})$

$\theta \approx 80.22$

Compare your answer with the correct one above

Question

Find the angle between the following two vectors.

$\vec{A}=3\hat{i}+5\hat{j}-\hat{k}, \vec{B}=6\hat{i}-2\hat{j}+5\hat{k}$

Answer

In order to find the angle between two vectors, we need to take the quotient of their dot product and their magnitudes:

$\frac{\vec{A}\cdot \vec{B}}{\left|\vec{A} \right| \left| \vec{B} \right| } = \cos{\theta}\Rightarrow \theta = \cos^{-1} \left( \frac{\vec{A}\cdot \vec{B}}{\left|\vec{A} \right| \left| \vec{B} \right| }\right )$

Therefore, we find that

$\vec{A}\cdot\vec{B}=3(6)+5(-2)-1(5)=3,$

$\left| \vec{A}\right| = \sqrt{3^2+5^2+1^2}=\sqrt{35}, \left| \vec{B}\right| = \sqrt{6^2+2^2+5^2}=\sqrt{65}$

$\theta = \cos^{-1} \left( \frac{3}{\sqrt{35}\sqrt{65} }\right)=86.39 ^{\circ}$ .

Compare your answer with the correct one above

Question

Find the (acute) angle between the vectors $<1,0,\frac{1}{2}>, <\frac{3}{2},0, 12>$ in degrees.

Answer

To find the angle between vectors, we use the formula

$\theta = \cos^{-1}(\frac{\mathbf{a} \cdot \mathbf{b}}{|\mathbf{a}||\mathbf{b}|})$ .

Substituting in our values, we get

$\theta = \cos^{-1}(\frac{<1,0,\frac{1}{2}> \cdot <\frac{3}{2},0,12>}{\sqrt{1^2+0^2+\frac{1}{2}^2}\sqrt{\frac{3}{2}^2+0^2+12^2}}) \approx \cos^{-1}(\frac{7.5}{13.521}) = 56.310$

Compare your answer with the correct one above

Question

Find the angle between the two vectors.

Answer

To find the angle between two vector we use the following formula

$cos , \theta = \frac{u\bullet v}{\left | u\right |\left | v\right |}$

and solve for $\theta$ .

Given

we find

$u\bullet v= 1(-3)+13=0$

$\left | u\right |=\sqrt{1^2+1^2}=\sqrt2$

$\left | v\right |=\sqrt{(-3)^2+3^2}=\sqrt{18}=3\sqrt2$

Plugging these values in we get

$cos, \theta = \frac{0}{3\sqrt2\sqrt2}=0$

To find $\theta$ we calculate the $cos^{-1}$ of both sides

$cos^{-1}[cos(\theta)]=cos^{-1}(0)$

and find that

$\theta = \frac{\pi}{2}$

Compare your answer with the correct one above

Question

Find the approximate angle in degrees between the vectors $<1, e, e^2>, <e, \sqrt{e}, \sqrt[3]{e}>$ .

Answer

We can compute the (acute) angle between the two vectors using the formula

$\theta = \cos^{-1}(\frac{\mathbf{a}\cdot\mathbf{b}}{|\mathbf{a}||\mathbf{b}|})$

Hence we have

$\theta = \cos^{-1}(\frac{<1,e,e^2>\cdot<e,\sqrt{e},\sqrt[3]{e}>}{\sqrt{1+e^2+e^4}\sqrt{e^2+e+e^{3/2}}})$

$\approx \cos^{-1}(\frac{e+e\sqrt{e}+e^{7/3}}{30.314})$

$\approx \cos^{-1}(\frac{17.512}{30.314})$

$\approx 54.712$

Compare your answer with the correct one above

Question

Find the angle in degrees between the vectors $<1,1>, <\pi^e,e^{\pi}>$ .

Answer

The correct answer is approximately degrees.

To find the angle between two vectors, we use the equation $\theta = \cos^{-1}(\frac{\mathbf{a}\cdot \mathbf{b}}{|\mathbf{a}| |\mathbf{b}|})$ .

Hence we have

$\theta = \cos^{-1}(\frac{(1)(\pi^e)+(1)(e^{\pi})}{\sqrt{1^2+1^2}\sqrt{\pi^{2e}+e^{2\pi}}})$

(This answer is small due to the fact that the two vectors nearly point in the same direction, due to $e^\pi$ and $\pi^e$ being close in value.)

Compare your answer with the correct one above

Question

Find the acute angle in degrees between the vectors .

Answer

To find the angle between two vectors, we use the formula

$\theta = \cos^{-1}(\frac{\mathbf{a}\cdot\mathbf{b}}{|\mathbf{a}||\mathbf{b}|})$ .

So we have

$\theta = \cos^{-1}(\frac{<10,1,5>\cdot<1,-1,-1>}{\sqrt{10^2+1^2+5^2}\sqrt{1^2+(-1)^2+(-1)^2}})$

$\theta = \cos^{-1}(\frac{4}{\sqrt{126}\sqrt{3}})$

$\theta \approx 78.127$

Compare your answer with the correct one above

Question

Find the angle between the following vectors (to two decimal places):

$\mathbf{v}=[-2, 4, -1],;\mathbf{w}=[3, 1, -5]$

Answer

The dot product is defined as:

$\mathbf{v}\cdot \mathbf{w}=|\mathbf{v}||\mathbf{w}|\cos\theta$

Where theta is the angle between the two vectors. Solving for theta:

$\theta=\cos^{-1}(\frac{\mathbf{v}\cdot \mathbf{w}}{|\mathbf{v}||\mathbf{w}|})$

To solve each component:

$\mathbf{v}\cdot \mathbf{w}=(3)(-2)+(1)(4)+(-5)(-1)=-6+4+5=3$

$|\mathbf{v}|=\sqrt{3^2+1^2+(-5)^2}=\sqrt{9+1+25}=\sqrt{35}$

$|\mathbf{w}|=\sqrt{(-2)^2+4^2+(-1)^2}=\sqrt{4+16+1}=\sqrt{21}$

Putting it all together, we can solve for theta:

$\theta=\cos^{-1}(\frac{3}{\sqrt{35}\sqrt{21}})\approx 1.46$

Compare your answer with the correct one above

Question

Find the angle between vectors $u= \left \langle 1,-2,3\right \rangle$ and $v=\left \langle -3,-4,-6\right \rangle$ and round to the nearest degree.

Answer

Write the formula to find the angle between two vectors.

$\theta= cos^{-1}(\frac{u_1v_1+u_2v_2+u_3v_3}{\left | u\right |\left | v\right |})$

Evaluate each term.

$\left | u\right | = \sqrt{1^2+(-2)^2+(3)^2} = \sqrt{1+4+9} = \sqrt{14}$

$\left | v\right | = \sqrt{(-3)^2+(-4)^2+(-6)^2} = \sqrt{9+16+36}= \sqrt{61}$

$\left | u\right |\left | v\right | = \sqrt{14}\sqrt{61} = \sqrt{ 854}$

Substitute the values into the equation.

$\theta= cos^{-1}(\frac{-13}{\sqrt{854}})= 116.41379$

The correct answer is:

Compare your answer with the correct one above

Question

Find the angle between the two vectors

Answer

In order to find the angle between the two vectors, we follow the formula

$cos(\theta)=\frac{u\bullet v}{||u||||v||}$

and solve for $\theta$ .

Using the vectors in the problem, we get

$cos(\theta)=\frac{<3,7>\bullet <-7,10>}{\sqrt{3^2+7^2}\sqrt{(-7)^2+10^2}}=\frac{3(-7)+7(10)}{\sqrt{58}*\sqrt{149}}$

Simplifying we get

$cos(\theta)=\frac{49}{\sqrt{8642}}$

To solve for $\theta$ we find the $cos^{-1}$ of both sides and get

$cos^{-1}(cos(\theta))=cos^{-1}(\frac{49}{\sqrt{8642}})$

and find that

$\theta=58^o$

Compare your answer with the correct one above

Tap the card to reveal the answer