3-Dimensional Axes and Coordinates - SAT Subject Test in Math II

Card 0 of 7

Question

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 ?

Answer

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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Question

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 .

Answer

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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Question

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

Answer

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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Question

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

Answer

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 , and the origin, and, as its altitude, the segment with the origin and as its endpoints.

The segment connecting the origin and is one leg of the base and has length 6; the segment connecting the origin and 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 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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Question

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

Answer

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 , and the origin, and, as its altitude, the segment with the origin and as its endpoints.

The segment connecting the origin and is one leg of the base and has length ; the segment connecting the origin and 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 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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Question

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 .

Answer

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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Question

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 .

Answer

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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