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High School Math : Calculus II — Integrals

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Example Question #1 :Understanding Vector Calculations

Let $\vec{a}, \vec{b},\vec{c}$ be vectors. All of the following are defined EXCEPT:

Possible Answers:

$(\vec{a}\times \vec{b})\cdot(\vec{a}\times \vec{c})$

$(\vec{a}\times \vec{b})\times\vec{c}$

$(\vec{a}\cdot \vec{b})\times\vec{c}$

$(\vec{a}\cdot \vec{b})+(\vec{a}\cdot \vec{c})$

$(\vec{a}\times \vec{b})\cdot \vec{c}$

Correct answer:

$(\vec{a}\cdot \vec{b})\times\vec{c}$

Explanation:

The cross product of two vectors (represented by "x") requires two vectors and results in another vector. By contrast, the dot product (represented by " $\cdot$ ") between two vectors requires two vectors and results in a scalar, not a vector.

If we were to evaluate $(\vec{a}\cdot \vec{b})\times\vec{c}$ , we would first have to evaluate $\vec{a}\cdot\vec{b}$ , which would result in a scalar, because it is a dot product.

However, once we have a scalar value, we cannot calculate a cross product with another vector, because a cross product requires two vectors. For example, we cannot find the cross product between 4 and the vector <1, 2, 3>; the cross product is only defined for two vectors, not scalars.

The answer is $(\vec{a}\cdot \vec{b})\times\vec{c}$ .

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Example Question #2 :Understanding Vector Calculations

Find the magnitude of vector v=<3,-2> :

Possible Answers:

$\sqrt{5}$

$\sqrt{13}$

$\sqrt{11}$

$\sqrt{7}$

Correct answer:

$\sqrt{13}$

Explanation:

To solve for the magnitude of a vector, we use the following formula:

$\left \| v \right \|=\sqrt{x^2+y^2}$

v=<3,-2>

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

$\left \| v \right \|=\sqrt{9+}4$

$\left \| v \right \|=\sqrt{13}$

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Example Question #3 :Understanding Vector Calculations

Given vector u=<0,0> and v=<2,1> , solve for u+v .

Possible Answers:

<0,0>

<0,1>

<2,1>

<2,0>

Correct answer:

<2,1>

Explanation:

u=<0,0> v=<2,1>

To solve for u+v , we need to add thecomponents in the vector and thecomponents together:

u+v=<0+2,0+1>

u+v=<2,1>

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Example Question #11 :Calculus Ii — Integrals

Given vector u=<0,0> and v=<2,1> , solve for u-v .

Possible Answers:

<-2,-1>

<0,-1>

<2,0>

<2,1>

Correct answer:

<-2,-1>

Explanation:

u=<0,0> v=<2,1>

To solve for u-v , we need to subtract thecomponents in the vector and thecomponents together:

u-v= <0-2,0-1>

u-v= <-2,-1>

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Example Question #5 :Understanding Vector Calculations

Given vector u=<0,0> and v=<2,1> , solve for 2u-3v .

Possible Answers:

<2,-3>

<-6,-3>

<6,3>

<-6,1>

Correct answer:

<-6,-3>

Explanation:

u=<0,0> v=<2,1>

To solve for 2u-3v ,我们首先需要相乘into vectorto findand multiplyinto vectorto find; then we need to subtract thecomponents in the vector and thecomponents together:

2u=<2*0,2*0> = <0,0>

3u=<3*2,3*1> = <6,3>

2u-3v= <0-6,0-3>

2u-3v= <-6,-3>

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Example Question #6 :Understanding Vector Calculations

Find the unit vector of v=<5,-12> .

Possible Answers:

$<\frac{5}{13},\frac{12}{13}>$

$<\frac{-5}{13},\frac{12}{13}>$

$<\frac{-5}{13},\frac{-12}{13}>$

$<\frac{5}{13},\frac{-12}{13}>$

Correct answer:

$<\frac{5}{13},\frac{-12}{13}>$

Explanation:

To solve for the unit vector, the following formula must be used:

$\frac{v}{||v||}$

v=<5,-12>

$\left \| v \right \|=\sqrt{(5)^2+(-12)^2}$

$\left \| v \right \|=\sqrt{25+144}$

$\left \| v \right \|=\sqrt{169}$

$\left \| v \right \|=13$

unit vector:

$\frac{<5,-12>}{13}$

$=\frac{1}{13}*<5,-12>$

$=<\frac{5}{13},\frac{-12}{13}>$

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Example Question #7 :Understanding Vector Calculations

Is $\langle \frac{5}{13},\frac{-12}{13}\rangle$ a unit vector?

Possible Answers:

yes, because magnitude is equal to

not enough information given

no, because magnitude is not equal to

Correct answer:

yes, because magnitude is equal to

Explanation:

To verify where a vector is a unit vector, we must solve for its magnitude. If the magnitude is equal to 1 then the vector is a unit vector:

$<\frac{5}{13},\frac{-12}{13}>$

$\sqrt{\left(\frac{5}{13}\right)^2+\left(\frac{-12}{13}\right)^2}$

$\sqrt{\left(\frac{25}{169}\right)+\left(\frac{144}{169}\right)$

$\sqrt{\frac{169}{169}}$

$\sqrt{1}$

$<\frac{5}{13},\frac{-12}{13}>$ is a unit vector because magnitude is equal to.

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Example Question #8 :Understanding Vector Calculations

Given vector u=<3,3> . Solve for the direction (angle) of the vector:

Possible Answers:

$60^{\circ}$

$15^{\circ}$

$90^{\circ}$

$45^{\circ}$

Correct answer:

$45^{\circ}$

Explanation:

To solve for the direction of a vector, we use the following formula:

$tan\alpha$ = $\frac{b}{a}$

with the vector being $u=\left \langle a,b \right \rangle$

$u=\left \langle 3,3 \right \rangle$

a=3

b=3

$\tan\alpha=\frac{3}{3}=1$

$\alpha = \arctan (1)$

$\alpha =45^{\circ}$

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Example Question #1 :Understanding Vector Calculations

Solve for vectorgiven direction of $45^{\circ}$ and magnitude of.

Possible Answers:

$\left \langle 2,2} \right \rangle$

$\left \langle 4,4} \right \rangle$

$\left \langle 2\sqrt{2},2\sqrt{2}} \right \rangle$

$\left \langle 4\sqrt{2},4\sqrt{2}} \right \rangle$

Correct answer:

$\left \langle 4\sqrt{2},4\sqrt{2}} \right \rangle$

Explanation:

To solve for a vector with the magnitude and direction given, we use the following formula:

v=(||v||cos) i + (||v||sin)j

$v=(8\cos 45^{\circ})i +(8\sin45^{\circ})j$

$v=4\sqrt{2}i+4\sqrt{2}j$

$v=\left \langle 4\sqrt{2},4\sqrt{2}} \right \rangle$

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Example Question #10 :Understanding Vector Calculations

Given vector u=<2,-1> and v=<1,2> , solve for u-2v .

Possible Answers:

<0,-3>

<0,-5>

<1,-3>

<0,3>

Correct answer:

<0,-5>

Explanation:

u=<2,-1> v=<1,2>

To solve for u-2v , We need to multiplyinto vectorto find; then we need to subtract thecomponents in the vector and thecomponents together:

2v= <2*1,2*2> = <2,4>

u-2v= <2-2,-1-4>

u-2v= <0,-5>

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High School Math : Calculus II — Integrals

Study concepts, example questions & explanations for High School Math

All High School Math Resources

Example Questions

Example Question #1 :Understanding Vector Calculations

Example Question #2 :Understanding Vector Calculations

Example Question #3 :Understanding Vector Calculations

Example Question #11 :Calculus Ii — Integrals

Example Question #5 :Understanding Vector Calculations

Example Question #6 :Understanding Vector Calculations

Example Question #7 :Understanding Vector Calculations

Example Question #8 :Understanding Vector Calculations

Example Question #1 :Understanding Vector Calculations

Example Question #10 :Understanding Vector Calculations

All High School Math Resources

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