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SAT II Math II : 3-Dimensional Axes and Coordinates

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Example Questions

Example Question #1 : 3 Dimensional Axes And Coordinates

Which of the following numbers comes closest to the length of line segment in three-dimensional coordinate space whose endpoints are the origin and the point (3,4,5) ?

Possible Answers:

8.5

6.5

7.5

Correct answer:

Explanation:

Use the three-dimensional version of the distance formula:

$d= \sqrt{(x_{1}-x_{2})^{2} +(y_{1}-y_{2})^{2}+(z_{1}-z_{2})^{2}}$

$d= \sqrt{(3-0)^{2} +(4-0)^{2}+(5-0)^{2}}$

$d= \sqrt{3^{2} +4^{2}+5^{2}}$

$d= \sqrt{9+16+25}$

$d= \sqrt{50}$

$d \approx 7.1$

The closest of the five choices is 7.

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Example Question #1 : 3 Dimensional Axes And Coordinates

A line segment $\overline{AB}$ in three-dimensional space has midpoint ; $\overline{AM}$ has midpoint .

has Cartesian coordinates $\left ( 100, \frac{1}{2}, 100\right )$ ; has Cartesian coordinates $\left ( 150, - \frac{1}{3}, -100\right )$ . Give the -coordinate of .

Possible Answers:

$-\frac{5}{6}$

$2\frac{1}{6}$

$-1\frac{5}{6}$

$- 2\frac{5}{6}$

Correct answer:

$- 2\frac{5}{6}$

Explanation:

The midpoint formula for the -coordinate

$\frac{y_{1}+y_{2}}{2}=y_{M}$

will be applied twice, once to find the -coordinate of , then again to find that of .

First, set $y_{2}=\frac{ 1 }{2}$ , the -coordinate of , and $y_{M}= - \frac{1}{3}$ , the -coordinate of , and solve for $y_{1}$ , the -coordinate of :

$\frac{y_{1}+\frac{ 1 }{2}}{2}=- \frac{1}{3}$

$y_{1}+\frac{ 1 }{2}=- \frac{2}{3}$

$y_{1} =- 1\frac{1}{6}$

Now, set $y_{2}=\frac{ 1 }{2}$ , the -coordinate of , and $y_{M}=- 1\frac{1}{6}$ , the -coordinate of , and solve for $y_{1}$ , the -coordinate of :

$\frac{y_{1}+\frac{ 1 }{2}}{2}= - 1\frac{1}{6}$

$y_{1}+\frac{ 1 }{2} = - 2\frac{1}{3}$

$y_{1} = - 2\frac{5}{6}$

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Example Question #3 : 3 Dimensional Axes And Coordinates

A line segment in three-dimensional space has endpoints with Cartesian coordinates (-4, 1, 8) and $\left (5, -2, 7 \right )$ . To the nearest tenth, give the length of the segment.

Possible Answers:

1.7

3.0

9.5

9.1

15.1

Correct answer:

9.5

Explanation:

Use the three-dimensional version of the distance formula:

$d= \sqrt{(x_{1}-x_{2})^{2} +(y_{1}-y_{2})^{2}+(z_{1}-z_{2})^{2}}$

$d= \sqrt{(-4-5)^{2} +(1-(-2))^{2}+(8-7)^{2}}$

$d= \sqrt{(-9)^{2} +3^{2}+1^{2}}$

$d= \sqrt{81+9+1}$

$d= \sqrt{91}$

$d \approx 9.5$

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Example Question #4 : 3 Dimensional Axes And Coordinates

A pyramid is positioned in three-dimensional space so that its four vertices are located at the points with coordinates (6,0,0) ,(0,9,0), (0,0,12) , and the origin. Give the volume of this pyramid.

Possible Answers:

162

108

648

Correct answer:

108

Explanation:

The three segments that connect the origin to the other points are all contained in one of the -, -, and - axes. Thus, this figure can be seen as a pyramid with, as its base, a right triangle in the -plane with vertices (6,0,0) ,(0,9,0) , and the origin, and, as its altitude, the segment with the origin and (0,0, 12) as its endpoints.

The segment connecting the origin and (6,0,0) is one leg of the base and has length 6; the segment connecting the origin and (0,9,0) is the other leg of the base and has length 9; the area of the base is therefore

$B = \frac{1}{2} \cdot 6 \cdot 9 = 27$

The segment connecting the origin and (0,0, 12) is the altitude; its length - the height of the pyramid - is 12.

The volume of the pyramid is

$V = \frac{1}{3} Bh = \frac{1}{3} \cdot 27 \cdot 12 = 108$

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Example Question #5 : 3 Dimensional Axes And Coordinates

A pyramid is positioned in three-dimensional space so that its four vertices are located at the points with coordinates (X,0,0) ,(0,Y,0), (0,0,Z) , and the origin. Give the volume of this pyramid.

Possible Answers:

$\frac{1}{3}XYZ$

$\frac{1}{6}XYZ$

$\frac{1}{2}XYZ$

$\frac{2}{3}XYZ$

XYZ

Correct answer:

$\frac{1}{6}XYZ$

Explanation:

The three segments that connect the origin to the other points are all contained in one of the -, -, and - axes. Thus, this figure can be seen as a pyramid with, as its base, a right triangle in the -plane with vertices (X,0,0) ,(0,Y,0) , and the origin, and, as its altitude, the segment with the origin and (0,0, Z) as its endpoints.

The segment connecting the origin and (X,0,0) is one leg of the base and has length ; the segment connecting the origin and (0,Y,0) is the other leg of the base and has length ; the area of the base is therefore

$B = \frac{1}{2} XY$

The segment connecting the origin and (0,0, Z) is the altitude; its length - the height of the pyramid - is .

The volume of the pyramid is

$V = \frac{1}{3} Bh = \frac{1}{3} \cdot \frac{1}{2}XY \cdot Z= \frac{1}{6}XYZ$

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Example Question #121 : Geometry

A line segment $\overline{AB}$ in three-dimensional space has midpoint ; $\overline{AM}$ has midpoint .

has Cartesian coordinates $\left ( 100, \frac{1}{2}, 100\right )$ ; has Cartesian coordinates $\left ( 150, - \frac{1}{3}, -100\right )$ . Give the -coordinate of .

Possible Answers:

Correct answer:

Explanation:

The midpoint formula for the -coordinate

$\frac{z_{1}+z_{2}}{2}=z_{M}$

will be applied twice, once to find the -coordinate of , then again to find that of .

First, set $z_{2}= 100$ , the -coordinate of , and $z_{M}= -100$ , the -coordinate of , and solve for $z_{1}$ , the -coordinate of :

$\frac{z_{1}+100}{2}=-100$

$z_{1}+100 = -200$

$z_{1}=-300$

Now, set $z_{2}= 100$ , the -coordinate of , and $z_{M}= -300$ , the -coordinate of , and solve for $z_{1}$ , the -coordinate of :

$\frac{z_{1}+100}{2}= -300$

$z_{1}+100 = -600$

$z_{1}=-700$

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Example Question #1 : 3 Dimensional Axes And Coordinates

A line segment $\overline{AB}$ in three-dimensional space has midpoint ; $\overline{AM}$ has midpoint .

has Cartesian coordinates $\left ( 100, \frac{1}{2}, 100\right )$ ; has Cartesian coordinates $\left ( 150, - \frac{1}{3}, -100\right )$ . Give the -coordinate of .

Possible Answers:

250

500

300

Correct answer:

300

Explanation:

The midpoint formula for the -coordinate

$\frac{x_{1}+x_{2}}{2}= x_{M}$

will be applied twice, once to find the -coordinate of , then again to find that of .

First, set $x_{2}= 100$ , the -coordinate of , and $x_{M}= 150$ , the -coordinate of , and solve for $x_{1}$ , the -coordinate of :

$\frac{x_{1}+100}{2}= 150$

$x_{1}+100 = 300$

$x_{1}=200$

Now, set $x_{2}= 100$ , the -coordinate of , and $x_{M}= 200$ , the -coordinate of , and solve for $x_{1}$ , the -coordinate of :

$\frac{x_{1}+100}{2}= 200$

$x_{1}+100 = 400$

$x_{1}=300$

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SAT II Math II : 3-Dimensional Axes and Coordinates

Study concepts, example questions & explanations for SAT II Math II

All SAT II Math II Resources

Example Questions

Example Question #1 : 3 Dimensional Axes And Coordinates

Example Question #1 : 3 Dimensional Axes And Coordinates

Example Question #3 : 3 Dimensional Axes And Coordinates

Example Question #4 : 3 Dimensional Axes And Coordinates

Example Question #5 : 3 Dimensional Axes And Coordinates

Example Question #121 : Geometry

Example Question #1 : 3 Dimensional Axes And Coordinates

All SAT II Math II Resources

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