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Precalculus : Pre-Calculus

Study concepts, example questions & explanations for Precalculus

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12 Diagnostic Tests 380 Practice Tests Question of the Day Flashcards Learn by Concept

Example Questions

Example Question #171 : Matrices And Vectors

Simplify: $\left \langle 2,-3,4\right \rangle+\left \langle-3,-3 \right \rangle-\left \langle5 \right \rangle$

Possible Answers:

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

$\left \langle -1,0,-1\right \rangle$

$\left \langle -1,-6,-1\right \rangle$

$\textup{ Cannot be solved.}$

$\left \langle -8\right \rangle$

Correct answer:

$\textup{ Cannot be solved.}$

Explanation:

The dimensions of the vectors are not the same. Placeholders cannot be added to a vector. Therefore, the values of the vectors cannot be added.

The correct answer is: $\textup{ Cannot be solved.}$

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Example Question #1591 : Pre Calculus

Find the norm of the vector $\vec{v}=[2,4,3]$ .

Possible Answers:

$\left | \vec{v}\right |= 2\sqrt{6}\approx 4.899$

$\left | \vec{v}\right |= \sqrt{29}\approx 5.385$

$\left | \vec{v}\right |= \sqrt{9} = 3$

$\left | \vec{v}\right |= 9$

Correct answer:

$\left | \vec{v}\right |= \sqrt{29}\approx 5.385$

Explanation:

We find the norm of a vector by finding the sum of each element squared and then taking the square root.

$\left | \vec{v} \right |= \sqrt{2^2+4^2+3^2}=\sqrt{29}$ .

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Example Question #1592 : Pre Calculus

Find the norm of the vector $\vec{r}=7\hat{i} +4 \hat{j}+5 \hat{k}$ .

Possible Answers:

$\left | \vec{r}\right | = 16$

$\left | \vec{r}\right | = 4$

$\left | \vec{r}\right | = 2 \sqrt{13}$

$\left | \vec{r} \ \right | = 3 \sqrt{10}$

Correct answer:

$\left | \vec{r} \ \right | = 3 \sqrt{10}$

Explanation:

We find the norm of a vector by finding the sum of each component squared and then taking the square root of that sum.

$\left | \vec{r}\right | = \sqrt{ 7^2+4^2+5^2 } = \sqrt{ 90 } = 3 \sqrt{10}$

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Example Question #23 : Evaluate Geometric Vectors

Find the norm of the vector: <3,-12,5.6,2,-9>

Possible Answers:

$\sqrt{269.36}\approx 16.41$

$\sqrt{31.6}\approx 5.62$

$\sqrt{184.36}\approx 13.58$

31.6

269.36

Correct answer:

$\sqrt{269.36}\approx 16.41$

Explanation:

The norm of a vector is also known as the length of the vector. The norm is given by the formula:

$\sqrt{\sum_{i=1}^nv_i^2}=\left \| v\right \|$ .

Here, we have

$\left \|v \right \|=\sqrt{3^2+(-12)^2+5.6^2+2^2+(-9)^2}=\sqrt{269.36}$ ,

the correct answer.

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Example Question #1593 : Pre Calculus

Find the norm of vector $\left \langle -2,2\right \rangle$ .

Possible Answers:

$\frac{1}{4}$

$2\sqrt2$

Correct answer:

$2\sqrt2$

Explanation:

Write the formula to find the norm, or the length the vector.

$\left \| v\right \|=\sqrt{a_{1}^2+b_{2}^2}$

Substitute the known values of the vector and solve.

$\left \| v\right \|=\sqrt{(-2)^2+2^2}= \sqrt{4+4}=\sqrt8=2\sqrt2$

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Example Question #25 : Evaluate Geometric Vectors

Find the norm (magnitude) of the following vector:

$\vec{b}=3i+5j-6k$

Possible Answers:

$\sqrt{72}$

$\sqrt{70}$

$-\sqrt{70}$

Correct answer:

$\sqrt{70}$

Explanation:

Use the following equation to find the magnitude of a vector:

$\left \| \vec{b}\right \|=\sqrt{i^2+j^2+k^2}$

In this case we have:

$\vec{b}=3i+5j-6k$

So plug in our values:

$\left \| \vec{b}\right \|=\sqrt{3^2+5^2+(-6)^2}=\sqrt{70}$

So:

$\left \| \vec{b} \right \|=\sqrt{70}$

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Example Question #1594 : Pre Calculus

Find the product of the vector $\vec{r}= 11\hat{i}+7\hat{j}$ and the scalar a=4 .

Possible Answers:

$a\vec{r}=44 \hat{i}+28\vec{j}$

$a\vec{r}=72$

$a\vec{r}=44 \hat{j}+28\vec{i}$

$a\vec{r}=72 \hat{i}$

Correct answer:

$a\vec{r}=44 \hat{i}+28\vec{j}$

Explanation:

When multiplying a vector by a scalar we multiply each component of the vector by the scalar and the result is a vector:

$a \vec{r}= 4 \cdot (11 \hat{i} + 7 \hat{j}) = 44 \hat{i}+28 \hat{j}$

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Example Question #1 : Polar Coordinates

Convert from polar form to rectangular form:

$r=4\cos\theta$

Possible Answers:

(x+y)^2=4

(x-2)^2+y^2=4

(x+2)^2+y^2=4

x^2+y^2+4=0

Correct answer:

(x-2)^2+y^2=4

Explanation:

Start by multiplying both sides by .

$r=4\cos\theta$

$r^2=4r\cos\theta$

Keep in mind that r^2=x^2+y^2

$x^2+y^2=4r\cos\theta$

Remember that $x=r\cos\theta$

So then,

x^2+y^2=4x

x^2-4x+y^2=0

Now, complete the square.

(x^2-4x+4)+y^2=4

(x-2)^2+y^2=4

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Example Question #2 : Convert Polar Equations To Rectangular Form And Vice Versa

Convert the polar equation to rectangular form:

$r=8\sin\theta$

Possible Answers:

x^2+(y-4)^2=16

(x-4)^2+y^2=15

x^2+(y+4)^2=16

(x+y-4)^2=16

Correct answer:

x^2+(y-4)^2=16

Explanation:

Start by multiplying both sides by .

$r=8\sin\theta$

$r^2=8r\sin\theta$

Remember that r^2=x^2+y^2

$x^2+y^2=8r\sin\theta$

Keep in mind that $y=r\sin\theta$

So then,

x^2+y^2=8y

x^2+y^2-8x=0

Now, complete the square.

x^2+(y^2-8x+16)=16

x^2+(y-4)^2=16

This is a graph of a circle with a radius of and a center at (0, 4)

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Example Question #1 : Convert Polar Equations To Rectangular Form And Vice Versa

Convert the polar equation into rectangular form:

$r=8\sec\theta$

Possible Answers:

x^2+y^2=8

y=8

x=8

(x-1)^2+y^2=64

Correct answer:

x=8

Explanation:

Remember that $\sec\theta=\frac{1}{\cos\theta}$

So then $r=8\sec\theta$ becomes

$r=\frac{8}{\cos\theta}$

Now, multiply both sides by $\cos\theta$ to get rid of the fraction.

$r\cos\theta=8$

Since $x=r\cos\theta,$ the rectangular form of this equation is

x=8

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Precalculus : Pre-Calculus

Study concepts, example questions & explanations for Precalculus

All Precalculus Resources

Example Questions

Example Question #171 : Matrices And Vectors

Example Question #1591 : Pre Calculus

Example Question #1592 : Pre Calculus

Example Question #23 : Evaluate Geometric Vectors

Example Question #1593 : Pre Calculus

Example Question #25 : Evaluate Geometric Vectors

Example Question #1594 : Pre Calculus

Example Question #1 : Polar Coordinates

Example Question #2 : Convert Polar Equations To Rectangular Form And Vice Versa

Example Question #1 : Convert Polar Equations To Rectangular Form And Vice Versa

All Precalculus Resources

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