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Precalculus : Pre-Calculus

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

Example Questions

Example Question #1 : Find A Direction Vector When Given Two Points

Find $\overrightarrow{MN}$ $\displaystyle \overrightarrow{MN}$ if $\displaystyle N=(2, 2)$ and $\displaystyle M=(-2, 1)$.

Possible Answers:

<-4, -1> $\displaystyle < -4, -1>$

<1, -4> $\displaystyle < 1, -4>$

<4,1> $\displaystyle < 4,1>$

<4, -1> $\displaystyle < 4, -1>$

Correct answer:

<4,1> $\displaystyle < 4,1>$

Explanation:

To find the direction vector from $\displaystyle M$ to $\displaystyle N$, subtract the x- and y-coordinates of $\displaystyle M$ from $\displaystyle N$.

$\overrightarrow{MN}=<2-(-2), 2-1>=<4, 1>$ $\displaystyle \overrightarrow{MN}=< 2-(-2), 2-1>=< 4, 1>$

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Example Question #1 : Find A Direction Vector When Given Two Points

Find the direction vector that has an initial point at C(1,5) $\displaystyle C(1,5)$ and a terminal point of D(0, 1) $\displaystyle D(0, 1)$.

Possible Answers:

<-1, -4> $\displaystyle < -1, -4>$

<-4, 1> $\displaystyle < -4, 1>$

<-4, -1> $\displaystyle < -4, -1>$

<1, 4> $\displaystyle < 1, 4>$

Correct answer:

<-1, -4> $\displaystyle < -1, -4>$

Explanation:

To find the directional vector, subtract the coordinates of the initial point from the coordinates of the terminal point.

$\overrightarrow{CD}=<0-1, 1-5>=<-1, -4>$ $\displaystyle \overrightarrow{CD}=< 0-1, 1-5>=< -1, -4>$

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Example Question #1 : Find A Direction Vector When Given Two Points

Find the direction vector that has an initial point at E(4, 1) $\displaystyle E(4, 1)$ and a terminal point at $\displaystyle F(-5, -1)$.

Possible Answers:

<9, 2> $\displaystyle < 9, 2>$

<-9, -2> $\displaystyle < -9, -2>$

<2, 9> $\displaystyle < 2, 9>$

<-2, -9> $\displaystyle < -2, -9>$

Correct answer:

<-9, -2> $\displaystyle < -9, -2>$

Explanation:

To find the directional vector, subtract the coordinates of the initial point from the coordinates of the terminal point.

$\overrightarrow{EF}=<-5-4, -1-1>=<-9, -2>$ $\displaystyle \overrightarrow{EF}=< -5-4, -1-1>=< -9, -2>$

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Example Question #1 : Find A Direction Vector When Given Two Points

Find the direction vector with an initial point of $\displaystyle R(-1, 4)$ and an terminal point of S(4, 9) $\displaystyle S(4, 9)$.

Possible Answers:

<5,5> $\displaystyle < 5,5>$

<5, -5> $\displaystyle < 5, -5>$

<-5, 5> $\displaystyle < -5, 5>$

<-5, -5> $\displaystyle < -5, -5>$

Correct answer:

<5,5> $\displaystyle < 5,5>$

Explanation:

To find the directional vector, subtract the coordinates of the initial point from the coordinates of the terminal point.

$\overrightarrow{RS}=<4-(-1), 9-4>=<5, 5>$ $\displaystyle \overrightarrow{RS}=< 4-(-1), 9-4>=< 5, 5>$

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Example Question #1 : Find A Direction Vector When Given Two Points

Find the direction vector that has an initial point at Z(1, 1) $\displaystyle Z(1, 1)$ and a terminal point at A(4, 8) $\displaystyle A(4, 8)$.

Possible Answers:

<-3, 7> $\displaystyle < -3, 7>$

<3, 7> $\displaystyle < 3, 7>$

<-3, -7> $\displaystyle < -3, -7>$

<7, 3> $\displaystyle < 7, 3>$

Correct answer:

<3, 7> $\displaystyle < 3, 7>$

Explanation:

To find the directional vector, subtract the coordinates of the initial point from the coordinates of the terminal point.

$\overrightarrow{ZA}=<4-1, 8-1>=<3, 7>$ $\displaystyle \overrightarrow{ZA}=< 4-1, 8-1>=< 3, 7>$

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Example Question #101 : Matrices And Vectors

Find the direction vector with an initial point of $\displaystyle T(-4, 5)$ and a terminal point $\displaystyle B(4, -1)$.

Possible Answers:

<-8, 6> $\displaystyle < -8, 6>$

<-6, 8> $\displaystyle < -6, 8>$

<8, -6> $\displaystyle < 8, -6>$

Correct answer:

<8, -6> $\displaystyle < 8, -6>$

Explanation:

To find the directional vector, subtract the coordinates of the initial point from the coordinates of the terminal point.

$\overrightarrow{TB}=<4-(-4), -1-5>=<8, -6>$ $\displaystyle \overrightarrow{TB}=< 4-(-4), -1-5>=< 8, -6>$

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Example Question #101 : Matrices And Vectors

If $\overrightarrow{DE}=<5, -1>$ $\displaystyle \overrightarrow{DE}=< 5, -1>$ and $\displaystyle E=(-2, -1)$, which of the following is a possible point for $\displaystyle D$?

Possible Answers:

$\displaystyle (-3, -2)$

(3, -2) $\displaystyle (3, -2)$

(0, -3) $\displaystyle (0, -3)$

(-7,0) $\displaystyle (-7,0)$

Correct answer:

(-7,0) $\displaystyle (-7,0)$

Explanation:

To find the directional vector, subtract the coordinates of the initial point from the coordinates of the terminal point.

To find the x-coordinate of $\displaystyle D$,

-2-x=5 $\displaystyle -2-x=5$

x=-7 $\displaystyle x=-7$

To find the y-coordinate of $\displaystyle D$,

-1-y=-1 $\displaystyle -1-y=-1$

y=0 $\displaystyle y=0$

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Example Question #11 : Find A Direction Vector When Given Two Points

If $\overrightarrow{EF}=<2, -8>$ $\displaystyle \overrightarrow{EF}=< 2, -8>$ and $\displaystyle E=(-1, 5)$, what is one possible point for $\displaystyle F$?

Possible Answers:

(1, -3) $\displaystyle (1, -3)$

$\displaystyle (-3, -3)$

(3, 3) $\displaystyle (3, 3)$

(-3, 3) $\displaystyle (-3, 3)$

Correct answer:

(1, -3) $\displaystyle (1, -3)$

Explanation:

To find the directional vector, subtract the coordinates of the initial point from the coordinates of the terminal point.

To find the x-coordinate of $\displaystyle F$,

$\displaystyle x-(-1)=2$

x=1 $\displaystyle x=1$

To find the y-coordinate of $\displaystyle F$,

y-5=-8 $\displaystyle y-5=-8$

y=-3 $\displaystyle y=-3$

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Example Question #12 : Find A Direction Vector When Given Two Points

If $\overrightarrow{OP}=<4, -2>$ $\displaystyle \overrightarrow{OP}=< 4, -2>$ and $\displaystyle O=(2, 10)$, what is a possible value of $\displaystyle P$?

Possible Answers:

(-6, 8) $\displaystyle (-6, 8)$

(6, -2) $\displaystyle (6, -2)$

(6, 8) $\displaystyle (6, 8)$

(8, 6) $\displaystyle (8, 6)$

Correct answer:

(6, 8) $\displaystyle (6, 8)$

Explanation:

To find the directional vector, subtract the coordinates of the initial point from the coordinates of the terminal point.

To find the x-coordinate of $\displaystyle P$,

x-2=4 $\displaystyle x-2=4$

x=6 $\displaystyle x=6$

To find the y-coordinate of $\displaystyle P$,

y-10=-2 $\displaystyle y-10=-2$

y=8 $\displaystyle y=8$

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Example Question #1 : Express A Vector In Polar Form

Rewrite the vector $\vec{v}= 3 \hat{i} + 5 \hat{j}$ $\displaystyle \vec{v}= 3 \hat{i} + 5 \hat{j}$ from Cartesian coordinates to polar coordinates $\vec{v} = \left \langle r \angle \theta \right \rangle$ $\displaystyle \vec{v} = \left \langle r \angle \theta \right \rangle$.

Possible Answers:

$\vec{v} = \left \langle \sqrt{21} \angle 32.64^\circ \right \rangle$ $\displaystyle \vec{v} = \left \langle \sqrt{21} \angle 32.64^\circ \right \rangle$

$\vec{v} = \left \langle \sqrt{8} \angle 60^\circ \right \rangle$ $\displaystyle \vec{v} = \left \langle \sqrt{8} \angle 60^\circ \right \rangle$

$\vec{v} = \left \langle 16 \angle 60^\circ \right \rangle$ $\displaystyle \vec{v} = \left \langle 16 \angle 60^\circ \right \rangle$

$\vec{v} = \left \langle \sqrt{34} \angle 59.04^\circ \right \rangle$ $\displaystyle \vec{v} = \left \langle \sqrt{34} \angle 59.04^\circ \right \rangle$

Correct answer:

$\vec{v} = \left \langle \sqrt{34} \angle 59.04^\circ \right \rangle$ $\displaystyle \vec{v} = \left \langle \sqrt{34} \angle 59.04^\circ \right \rangle$

Explanation:

To convert to polar form, we need to find the magnitude of the vector, $\displaystyle r$, and the angle it forms with the positive $\displaystyle x$-axis going counterclockwise, or $\theta$ $\displaystyle \theta$. This is shown in the figure below.

Vecfssec

We find the angle using trigonometric identities:

$\tan{\theta}= \frac{5}{3}$ $\displaystyle \tan{\theta}= \frac{5}{3}$

Using a calculator,

$\theta = \arctan{\frac{5}{3}} \approx 59.04^\circ$ $\displaystyle \theta = \arctan{\frac{5}{3}} \approx 59.04^\circ$

To find the magnitude of a vector, we add up the squares of each component and take the square root:

$r= \left | \vec{v}\right | = \sqrt{3^2+5^2}=\sqrt{34}$ $\displaystyle r= \left | \vec{v}\right | = \sqrt{3^2+5^2}=\sqrt{34}$.

So, our vector written in polar form is

$\vec{v} = \left \langle \sqrt{34} \angle 59.04^\circ \right \rangle$ $\displaystyle \vec{v} = \left \langle \sqrt{34} \angle 59.04^\circ \right \rangle$

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Precalculus : Pre-Calculus

Study concepts, example questions & explanations for Precalculus

All Precalculus Resources

Example Questions

Example Question #1 : Find A Direction Vector When Given Two Points

Example Question #1 : Find A Direction Vector When Given Two Points

Example Question #1 : Find A Direction Vector When Given Two Points

Example Question #1 : Find A Direction Vector When Given Two Points

Example Question #1 : Find A Direction Vector When Given Two Points

Example Question #101 : Matrices And Vectors

Example Question #101 : Matrices And Vectors

Example Question #11 : Find A Direction Vector When Given Two Points

Example Question #12 : Find A Direction Vector When Given Two Points

Example Question #1 : Express A Vector In Polar Form

All Precalculus Resources

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