Precalculus : Pre-Calculus

Study concepts, example questions & explanations for Precalculus

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Example Questions

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

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

Possible Answers:

\(\displaystyle < -4, -1>\)

\(\displaystyle < 1, -4>\)

\(\displaystyle < 4,1>\)

\(\displaystyle < 4, -1>\)

Correct answer:

\(\displaystyle < 4,1>\)

Explanation:

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

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

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

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

Possible Answers:

\(\displaystyle < -1, -4>\)

\(\displaystyle < -4, 1>\)

\(\displaystyle < -4, -1>\)

\(\displaystyle < 1, 4>\)

Correct answer:

\(\displaystyle < -1, -4>\)

Explanation:

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

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

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

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

Possible Answers:

\(\displaystyle < 9, 2>\)

\(\displaystyle < -9, -2>\)

\(\displaystyle < 2, 9>\)

\(\displaystyle < -2, -9>\)

Correct answer:

\(\displaystyle < -9, -2>\)

Explanation:

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

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

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 \(\displaystyle S(4, 9)\).

Possible Answers:

\(\displaystyle < 5,5>\)

\(\displaystyle < 5, -5>\)

\(\displaystyle < -5, 5>\)

\(\displaystyle < -5, -5>\)

Correct answer:

\(\displaystyle < 5,5>\)

Explanation:

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

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

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

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

Possible Answers:

\(\displaystyle < -3, 7>\)

\(\displaystyle < 3, 7>\)

\(\displaystyle < -3, -7>\)

\(\displaystyle < 7, 3>\)

Correct answer:

\(\displaystyle < 3, 7>\)

Explanation:

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

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

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:

\(\displaystyle < -8, 6>\)

\(\displaystyle < -8, 6>\)

\(\displaystyle < -6, 8>\)

\(\displaystyle < 8, -6>\)

Correct answer:

\(\displaystyle < 8, -6>\)

Explanation:

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

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

Example Question #101 : Matrices And Vectors

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

Possible Answers:

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

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

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

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

Correct answer:

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

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 \(\displaystyle D\)

\(\displaystyle -2-x=5\)

\(\displaystyle x=-7\)

To find the y-coordinate of \(\displaystyle D\),

\(\displaystyle -1-y=-1\)

\(\displaystyle y=0\)

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

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

Possible Answers:

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

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

\(\displaystyle (3, 3)\)

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

Correct answer:

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

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 \(\displaystyle F\),

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

\(\displaystyle x=1\)

To find the y-coordinate of \(\displaystyle F\),

\(\displaystyle y-5=-8\)

\(\displaystyle y=-3\)

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

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

Possible Answers:

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

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

\(\displaystyle (6, 8)\)

\(\displaystyle (8, 6)\)

Correct answer:

\(\displaystyle (6, 8)\)

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 \(\displaystyle P\),

\(\displaystyle x-2=4\)

\(\displaystyle x=6\)

To find the y-coordinate of \(\displaystyle P\),

\(\displaystyle y-10=-2\)

\(\displaystyle y=8\)

Example Question #1 : Express A Vector In Polar Form

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

 

Possible Answers:

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

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

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

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

Correct answer:

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

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 \(\displaystyle \theta\). This is shown in the figure below.

Vecfssec

We find the angle using trigonometric identities:

\(\displaystyle \tan{\theta}= \frac{5}{3}\)

Using a calculator,

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

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

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

So, our vector written in polar form is

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

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