Chapter 11 Introduction To Three Dimensional Geometry

Given:

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To Find:

The coordinates of point F.

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Solution:

From Figure 11.3 (which is not displayed here, but the description implies the geometry), point P is located in three-dimensional space with coordinates $(x, y, z) = (2, 4, 5)$.

Observing typical diagrams in 3D geometry showing projections onto coordinate planes, the point labelled F is usually the foot of the perpendicular from P onto the XZ-plane.

When a point $(x, y, z)$ is projected onto the XZ-plane, the y-coordinate becomes $0$. The x-coordinate and the z-coordinate remain unchanged.

Given the coordinates of P are $(2, 4, 5)$, its projection onto the XZ-plane (point F) will have:

x-coordinate = 2 (same as P's x-coordinate)

y-coordinate = 0 (because it's on the XZ-plane)

z-coordinate = 5 (same as P's z-coordinate)

Therefore, the coordinates of F are $(2, 0, 5)$.

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Answer:

The coordinates of F are $\mathbf{(2, 0, 5)}$.

Given:

The two given points are $(-3, 1, 2)$ and $(-3, 1, -2)$.

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To Find:

The octant in which each of the given points lies.

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Solution:

The three coordinate planes divide the space into eight regions, called octants. The position of a point $(x, y, z)$ in space is determined by the signs of its coordinates.

The signs of the coordinates $(x, y, z)$ for each octant are as follows:

Octant I: $(+, +, +)$

Octant II: $(-, +, +)$

Octant III: $(-, -, +)$

Octant IV: $(+, -, +)$

Octant V: $(+, +, -)$

Octant VI: $(-, +, -)$

Octant VII: $(-, -, -)$

Octant VIII: $(+, -, -)$

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Let's consider the first point $(-3, 1, 2)$.

The coordinates are $x = -3$, $y = 1$, and $z = 2$.

The signs of the coordinates are: x is negative, y is positive, and z is positive.

The combination of signs is $(-, +, +)$.

Comparing this with the list of octants, the point $(-3, 1, 2)$ lies in the octant with signs $(-, +, +)$, which is Octant II.

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Now let's consider the second point $(-3, 1, -2)$.

The coordinates are $x = -3$, $y = 1$, and $z = -2$.

The signs of the coordinates are: x is negative, y is positive, and z is negative.

The combination of signs is $(-, +, -)$.

Comparing this with the list of octants, the point $(-3, 1, -2)$ lies in the octant with signs $(-, +, -)$, which is Octant VI.

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Therefore, the point $(-3, 1, 2)$ lies in Octant II and the point $(-3, 1, -2)$ lies in Octant VI.

Given:

A point lies on the x-axis.

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To Find:

The y-coordinate and z-coordinate of the point.

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Solution:

In a three-dimensional coordinate system, the axes are mutually perpendicular.

The x-axis is defined as the line where the y-coordinate and the z-coordinate are both zero.

Any point on the x-axis has the form $(x, 0, 0)$, where x is its distance from the origin along the x-axis.

Therefore, if a point is on the x-axis, its distance from the YZ-plane (where $x=0$) is its x-coordinate, and its distances from the XY-plane (where $z=0$) and XZ-plane (where $y=0$) are zero.

Thus, for a point on the x-axis, the y-coordinate and the z-coordinate must be $0$.

The y-coordinate is $\mathbf{0}$.

The z-coordinate is $\mathbf{0}$.

Given:

A point lies in the XZ-plane.

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To Find:

The y-coordinate of the point.

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Solution:

In a three-dimensional coordinate system, the XZ-plane is one of the three principal coordinate planes.

The XZ-plane is defined as the set of all points where the y-coordinate is equal to $0$.

Any point in the XZ-plane has coordinates of the form $(x, 0, z)$, where x and z can be any real numbers.

This is because the distance from the point to the XZ-plane is represented by the absolute value of its y-coordinate. If the point lies in the XZ-plane, its distance from the plane must be zero, which implies the y-coordinate must be zero.

Therefore, if a point is in the XZ-plane, its y-coordinate is always $0$.

The y-coordinate is $\mathbf{0}$.

Given:

The following points are given: (1, 2, 3), (4, –2, 3), (4, –2, –5), (4, 2, –5), (– 4, 2, –5), (– 4, 2, 5), (–3, –1, 6), (– 2, – 4, –7).

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To Find:

The octant in which each point lies.

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Solution:

The position of a point $(x, y, z)$ in three-dimensional space determines the octant it lies in based on the signs of its coordinates. The eight octants are defined by the combinations of signs of the x, y, and z coordinates.

The sign conventions for the octants are:

Octant I: $(+, +, +)$

Octant II: $(-, +, +)$

Octant III: $(-, -, +)$

Octant IV: $(+, -, +)$

Octant V: $(+, +, -)$

Octant VI: $(-, +, -)$

Octant VII: $(-, -, -)$

Octant VIII: $(+, -, -)$

Let's determine the octant for each given point:

1. Point: (1, 2, 3)

Signs of coordinates: $(+, +, +)$. This corresponds to Octant I.

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2. Point: (4, –2, 3)

Signs of coordinates: $(+, -, +)$. This corresponds to Octant IV.

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3. Point: (4, –2, –5)

Signs of coordinates: $(+, -, -)$. This corresponds to Octant VIII.

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4. Point: (4, 2, –5)

Signs of coordinates: $(+, +, -)$. This corresponds to Octant V.

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5. Point: (– 4, 2, –5)

Signs of coordinates: $(-, +, -)$. This corresponds to Octant VI.

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6. Point: (– 4, 2, 5)

Signs of coordinates: $(-, +, +)$. This corresponds to Octant II.

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7. Point: (–3, –1, 6)

Signs of coordinates: $(-, -, +)$. This corresponds to Octant III.

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8. Point: (– 2, – 4, –7)

Signs of coordinates: $(-, -, -)$. This corresponds to Octant VII.

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Summary:

(1, 2, 3) lies in Octant I.

(4, –2, 3) lies in Octant IV.

(4, –2, –5) lies in Octant VIII.

(4, 2, –5) lies in Octant V.

(– 4, 2, –5) lies in Octant VI.

(– 4, 2, 5) lies in Octant II.

(–3, –1, 6) lies in Octant III.

(– 2, – 4, –7) lies in Octant VII.

Given:

Three statements with blanks to be filled.

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To Fill:

The correct terms or forms in the blanks.

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Solution:

(i) The x-axis and y-axis lie in a specific coordinate plane. This plane is where the z-coordinate is always zero. This plane is called the XY-plane.

The x-axis and y-axis taken together determine a plane known as the XY-plane.

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(ii) A point in the XY-plane has no displacement along the z-axis from the origin. Therefore, its z-coordinate is always zero, while its x and y coordinates can be any real numbers.

The coordinates of points in the XY-plane are of the form (x, y, 0).

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(iii) The three coordinate planes (XY-plane, YZ-plane, and XZ-plane) intersect at the origin and divide the entire three-dimensional space into eight regions.

Coordinate planes divide the space into eight octants.

Given:

Two points P and Q with coordinates P(1, –3, 4) and Q (– 4, 1, 2).

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To Find:

The distance between the points P and Q.

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Solution:

The distance between two points $P(x_1, y_1, z_1)$ and $Q(x_2, y_2, z_2)$ in three-dimensional space is given by the distance formula:

$\text{Distance PQ} = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2 + (z_2 - z_1)^2}$

Given points are P(1, –3, 4) and Q (– 4, 1, 2).

Here, $x_1 = 1$, $y_1 = -3$, $z_1 = 4$

And $x_2 = -4$, $y_2 = 1$, $z_2 = 2$

Substitute these values into the distance formula:

$\text{Distance PQ} = \sqrt{(-4 - 1)^2 + (1 - (-3))^2 + (2 - 4)^2}$

$\text{Distance PQ} = \sqrt{(-5)^2 + (1 + 3)^2 + (-2)^2}$

$\text{Distance PQ} = \sqrt{(-5)^2 + (4)^2 + (-2)^2}$

$\text{Distance PQ} = \sqrt{25 + 16 + 4}$

$\text{Distance PQ} = \sqrt{45}$

We can simplify the square root of 45:

$\sqrt{45} = \sqrt{9 \times 5} = \sqrt{9} \times \sqrt{5} = 3\sqrt{5}$

Therefore, the distance between points P and Q is $3\sqrt{5}$ units.

The distance between the points P and Q is $\mathbf{3\sqrt{5}}$.

Given:

Three points P, Q, and R with coordinates P(–2, 3, 5), Q(1, 2, 3), and R(7, 0, –1).

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To Show:

The points P, Q, and R are collinear.

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Solution:

To show that three points are collinear, we can calculate the distances between each pair of points. If the sum of the lengths of the two shorter segments equals the length of the longest segment, then the points are collinear.

The distance between two points $(x_1, y_1, z_1)$ and $(x_2, y_2, z_2)$ is given by the formula:

$\text{Distance} = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2 + (z_2 - z_1)^2}$

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Let's calculate the distance between P(–2, 3, 5) and Q(1, 2, 3).

$x_1 = -2, y_1 = 3, z_1 = 5$

$x_2 = 1, y_2 = 2, z_2 = 3$

$\text{PQ} = \sqrt{(1 - (-2))^2 + (2 - 3)^2 + (3 - 5)^2}$

$\text{PQ} = \sqrt{(1 + 2)^2 + (-1)^2 + (-2)^2}$

$\text{PQ} = \sqrt{(3)^2 + (-1)^2 + (-2)^2}$

$\text{PQ} = \sqrt{9 + 1 + 4}$

$\text{PQ} = \sqrt{14}$

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Let's calculate the distance between Q(1, 2, 3) and R(7, 0, –1).

$x_1 = 1, y_1 = 2, z_1 = 3$

$x_2 = 7, y_2 = 0, z_2 = -1$

$\text{QR} = \sqrt{(7 - 1)^2 + (0 - 2)^2 + (-1 - 3)^2}$

$\text{QR} = \sqrt{(6)^2 + (-2)^2 + (-4)^2}$

$\text{QR} = \sqrt{36 + 4 + 16}$

$\text{QR} = \sqrt{56}$

$\text{QR} = \sqrt{4 \times 14} = 2\sqrt{14}$

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Let's calculate the distance between P(–2, 3, 5) and R(7, 0, –1).

$x_1 = -2, y_1 = 3, z_1 = 5$

$x_2 = 7, y_2 = 0, z_2 = -1$

$\text{PR} = \sqrt{(7 - (-2))^2 + (0 - 3)^2 + (-1 - 5)^2}$

$\text{PR} = \sqrt{(7 + 2)^2 + (-3)^2 + (-6)^2}$

$\text{PR} = \sqrt{(9)^2 + (-3)^2 + (-6)^2}$

$\text{PR} = \sqrt{81 + 9 + 36}$

$\text{PR} = \sqrt{126}$

$\text{PR} = \sqrt{9 \times 14} = 3\sqrt{14}$

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Now, let's check the sum of the distances:

$\text{PQ} = \sqrt{14}$

$\text{QR} = 2\sqrt{14}$

$\text{PR} = 3\sqrt{14}$

Consider $\text{PQ} + \text{QR}$: $\sqrt{14} + 2\sqrt{14} = (1+2)\sqrt{14} = 3\sqrt{14}$

We observe that $\text{PQ} + \text{QR} = 3\sqrt{14}$, which is equal to $\text{PR}$.

Since the sum of the distances between two pairs of points is equal to the distance of the third pair, the points P, Q, and R are collinear, and point Q lies between P and R.

Therefore, the points P(–2, 3, 5), Q(1, 2, 3), and R(7, 0, –1) are collinear.

Given:

Three points A, B, and C with coordinates A(3, 6, 9), B(10, 20, 30), and C(25, –41, 5).

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To Determine:

If the given points are the vertices of a right-angled triangle.

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Solution:

To determine if the points form a right-angled triangle, we can calculate the square of the length of each side and check if they satisfy the Pythagorean theorem ($a^2 + b^2 = c^2$).

The square of the distance between two points $(x_1, y_1, z_1)$ and $(x_2, y_2, z_2)$ is given by the formula:

$\text{Distance}^2 = (x_2 - x_1)^2 + (y_2 - y_1)^2 + (z_2 - z_1)^2$

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Let's calculate the square of the distance between points A(3, 6, 9) and B(10, 20, 30):

$AB^2 = (10 - 3)^2 + (20 - 6)^2 + (30 - 9)^2$

$AB^2 = (7)^2 + (14)^2 + (21)^2$

$AB^2 = 49 + 196 + 441$

$AB^2 = 686$

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Let's calculate the square of the distance between points B(10, 20, 30) and C(25, –41, 5):

$BC^2 = (25 - 10)^2 + (-41 - 20)^2 + (5 - 30)^2$

$BC^2 = (15)^2 + (-61)^2 + (-25)^2$

$BC^2 = 225 + 3721 + 625$

$BC^2 = 4571$

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Let's calculate the square of the distance between points A(3, 6, 9) and C(25, –41, 5):

$AC^2 = (25 - 3)^2 + (-41 - 6)^2 + (5 - 9)^2$

$AC^2 = (22)^2 + (-47)^2 + (-4)^2$

$AC^2 = 484 + 2209 + 16$

$AC^2 = 2709$

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Now, let's check if the Pythagorean theorem holds by comparing the sum of the squares of two sides with the square of the third side.

The squared lengths are $AB^2 = 686$, $BC^2 = 4571$, and $AC^2 = 2709$.

The largest squared length is $BC^2 = 4571$. For the triangle to be right-angled, the sum of the other two squared lengths must equal the largest one.

Check if $AB^2 + AC^2 = BC^2$:

$686 + 2709 = 3395$

Since $3395 \neq 4571$, the condition $AB^2 + AC^2 = BC^2$ is not satisfied.

We can also check other combinations just to be thorough, though with the largest side identified, this step is technically not required to confirm it's *not* a right triangle.

Check if $AB^2 + BC^2 = AC^2$:

$686 + 4571 = 5257$

Since $5257 \neq 2709$, the condition $AB^2 + BC^2 = AC^2$ is not satisfied.

Check if $AC^2 + BC^2 = AB^2$:

$2709 + 4571 = 7280$

Since $7280 \neq 686$, the condition $AC^2 + BC^2 = AB^2$ is not satisfied.

Since the square of the longest side is not equal to the sum of the squares of the other two sides, the given points do not form a right-angled triangle.

Therefore, the points A, B, and C are not the vertices of a right-angled triangle.

Given:

Point A has coordinates $(3, 4, 5)$.

Point B has coordinates $(-1, 3, -7)$.

The condition $PA^2 + PB^2 = 2k^2$, where P is a variable point and $k$ is a constant.

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To Find:

The equation of the set of points P.

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Solution:

Let the coordinates of the variable point P be $(x, y, z)$.

The distance formula in three dimensions states that the distance between two points $(x_1, y_1, z_1)$ and $(x_2, y_2, z_2)$ is $\sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2 + (z_2 - z_1)^2}$.

The square of the distance between two points $(x_1, y_1, z_1)$ and $(x_2, y_2, z_2)$ is $(x_2 - x_1)^2 + (y_2 - y_1)^2 + (z_2 - z_1)^2$.

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First, let's calculate $PA^2$ using the coordinates of P$(x, y, z)$ and A$(3, 4, 5)$.

$PA^2 = (x - 3)^2 + (y - 4)^2 + (z - 5)^2$

Expanding the terms:

$PA^2 = (x^2 - 6x + 9) + (y^2 - 8y + 16) + (z^2 - 10z + 25)$

$PA^2 = x^2 + y^2 + z^2 - 6x - 8y - 10z + 9 + 16 + 25$

$PA^2 = x^2 + y^2 + z^2 - 6x - 8y - 10z + 50$

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Next, let's calculate $PB^2$ using the coordinates of P$(x, y, z)$ and B$(-1, 3, -7)$.

$PB^2 = (x - (-1))^2 + (y - 3)^2 + (z - (-7))^2$

$PB^2 = (x + 1)^2 + (y - 3)^2 + (z + 7)^2$

Expanding the terms:

$PB^2 = (x^2 + 2x + 1) + (y^2 - 6y + 9) + (z^2 + 14z + 49)$

$PB^2 = x^2 + y^2 + z^2 + 2x - 6y + 14z + 1 + 9 + 49$

$PB^2 = x^2 + y^2 + z^2 + 2x - 6y + 14z + 59$

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Now, substitute the expressions for $PA^2$ and $PB^2$ into the given condition $PA^2 + PB^2 = 2k^2$.

$(x^2 + y^2 + z^2 - 6x - 8y - 10z + 50) + (x^2 + y^2 + z^2 + 2x - 6y + 14z + 59) = 2k^2$

Combine like terms:

$(x^2 + x^2) + (y^2 + y^2) + (z^2 + z^2) + (-6x + 2x) + (-8y - 6y) + (-10z + 14z) + (50 + 59) = 2k^2$

$2x^2 + 2y^2 + 2z^2 - 4x - 14y + 4z + 109 = 2k^2$

Rearrange the equation to find the standard form:

$2x^2 + 2y^2 + 2z^2 - 4x - 14y + 4z + 109 - 2k^2 = 0$

We can divide the entire equation by 2 to simplify it:

$x^2 + y^2 + z^2 - 2x - 7y + 2z + \frac{109 - 2k^2}{2} = 0$

This is the required equation of the set of points P such that $PA^2 + PB^2 = 2k^2$. This equation represents a sphere (provided the constant term allows for a positive radius squared).

The equation of the set of points P is $\mathbf{x^2 + y^2 + z^2 - 2x - 7y + 2z + \frac{109 - 2k^2}{2} = 0}$.

Given:

Four pairs of points in three-dimensional space.

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To Find:

The distance between the points in each pair.

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Solution:

The distance between two points $P_1(x_1, y_1, z_1)$ and $P_2(x_2, y_2, z_2)$ in three-dimensional space is given by the distance formula:

$\text{Distance} = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2 + (z_2 - z_1)^2}$

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(i) Distance between (2, 3, 5) and (4, 3, 1)

Let $P_1 = (2, 3, 5)$ and $P_2 = (4, 3, 1)$.

$x_1 = 2, y_1 = 3, z_1 = 5$

$x_2 = 4, y_2 = 3, z_2 = 1$

Distance = $\sqrt{(4 - 2)^2 + (3 - 3)^2 + (1 - 5)^2}$

Distance = $\sqrt{(2)^2 + (0)^2 + (-4)^2}$

Distance = $\sqrt{4 + 0 + 16}$

Distance = $\sqrt{20}$

Distance = $\sqrt{4 \times 5} = 2\sqrt{5}$

The distance between (2, 3, 5) and (4, 3, 1) is $\mathbf{2\sqrt{5}}$.

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(ii) Distance between (–3, 7, 2) and (2, 4, –1)

Let $P_1 = (-3, 7, 2)$ and $P_2 = (2, 4, -1)$.

$x_1 = -3, y_1 = 7, z_1 = 2$

$x_2 = 2, y_2 = 4, z_2 = -1$

Distance = $\sqrt{(2 - (-3))^2 + (4 - 7)^2 + (-1 - 2)^2}$

Distance = $\sqrt{(2 + 3)^2 + (-3)^2 + (-3)^2}$

Distance = $\sqrt{(5)^2 + (-3)^2 + (-3)^2}$

Distance = $\sqrt{25 + 9 + 9}$

Distance = $\sqrt{43}$

The distance between (–3, 7, 2) and (2, 4, –1) is $\mathbf{\sqrt{43}}$.

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(iii) Distance between (–1, 3, – 4) and (1, –3, 4)

Let $P_1 = (-1, 3, -4)$ and $P_2 = (1, -3, 4)$.

$x_1 = -1, y_1 = 3, z_1 = -4$

$x_2 = 1, y_2 = -3, z_2 = 4$

Distance = $\sqrt{(1 - (-1))^2 + (-3 - 3)^2 + (4 - (-4))^2}$

Distance = $\sqrt{(1 + 1)^2 + (-6)^2 + (4 + 4)^2}$

Distance = $\sqrt{(2)^2 + (-6)^2 + (8)^2}$

Distance = $\sqrt{4 + 36 + 64}$

Distance = $\sqrt{104}$

Distance = $\sqrt{4 \times 26} = 2\sqrt{26}$

The distance between (–1, 3, –4) and (1, –3, 4) is $\mathbf{2\sqrt{26}}$.

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(iv) Distance between (2, –1, 3) and (–2, 1, 3)

Let $P_1 = (2, -1, 3)$ and $P_2 = (-2, 1, 3)$.

$x_1 = 2, y_1 = -1, z_1 = 3$

$x_2 = -2, y_2 = 1, z_2 = 3$

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

Distance = $\sqrt{(-4)^2 + (1 + 1)^2 + (0)^2}$

Distance = $\sqrt{(-4)^2 + (2)^2 + (0)^2}$

Distance = $\sqrt{16 + 4 + 0}$

Distance = $\sqrt{20}$

Distance = $\sqrt{4 \times 5} = 2\sqrt{5}$

The distance between (2, –1, 3) and (–2, 1, 3) is $\mathbf{2\sqrt{5}}$.

Given:

Three points P, Q, and R with coordinates P(–2, 3, 5), Q(1, 2, 3), and R(7, 0, –1).

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To Show:

The points P, Q, and R are collinear.

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Solution:

Three points are collinear if the sum of the distances between two pairs of points is equal to the distance of the third pair of points. We will use the distance formula in three dimensions.

The distance between two points $(x_1, y_1, z_1)$ and $(x_2, y_2, z_2)$ is given by:

$\text{Distance} = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2 + (z_2 - z_1)^2}$

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Let's calculate the distance between P(–2, 3, 5) and Q(1, 2, 3):

$x_1 = -2, y_1 = 3, z_1 = 5$

$x_2 = 1, y_2 = 2, z_2 = 3$

$\text{PQ} = \sqrt{(1 - (-2))^2 + (2 - 3)^2 + (3 - 5)^2}$

$\text{PQ} = \sqrt{(1 + 2)^2 + (-1)^2 + (-2)^2}$

$\text{PQ} = \sqrt{(3)^2 + (-1)^2 + (-2)^2}$

$\text{PQ} = \sqrt{9 + 1 + 4}$

$\text{PQ} = \sqrt{14}$

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Let's calculate the distance between Q(1, 2, 3) and R(7, 0, –1):

$x_1 = 1, y_1 = 2, z_1 = 3$

$x_2 = 7, y_2 = 0, z_2 = -1$

$\text{QR} = \sqrt{(7 - 1)^2 + (0 - 2)^2 + (-1 - 3)^2}$

$\text{QR} = \sqrt{(6)^2 + (-2)^2 + (-4)^2}$

$\text{QR} = \sqrt{36 + 4 + 16}$

$\text{QR} = \sqrt{56}$

We can simplify $\sqrt{56}$: $\sqrt{56} = \sqrt{4 \times 14} = \sqrt{4} \times \sqrt{14} = 2\sqrt{14}$

$\text{QR} = 2\sqrt{14}$

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Let's calculate the distance between P(–2, 3, 5) and R(7, 0, –1):

$x_1 = -2, y_1 = 3, z_1 = 5$

$x_2 = 7, y_2 = 0, z_2 = -1$

$\text{PR} = \sqrt{(7 - (-2))^2 + (0 - 3)^2 + (-1 - 5)^2}$

$\text{PR} = \sqrt{(7 + 2)^2 + (-3)^2 + (-6)^2}$

$\text{PR} = \sqrt{(9)^2 + (-3)^2 + (-6)^2}$

$\text{PR} = \sqrt{81 + 9 + 36}$

$\text{PR} = \sqrt{126}$

We can simplify $\sqrt{126}$: $\sqrt{126} = \sqrt{9 \times 14} = \sqrt{9} \times \sqrt{14} = 3\sqrt{14}$

$\text{PR} = 3\sqrt{14}$

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Now we compare the distances:

$\text{PQ} = \sqrt{14}$

$\text{QR} = 2\sqrt{14}$

$\text{PR} = 3\sqrt{14}$

Let's check if the sum of two distances equals the third distance. We can see that $\text{PQ} + \text{QR} = \sqrt{14} + 2\sqrt{14} = (1+2)\sqrt{14} = 3\sqrt{14}$.

So, $\text{PQ} + \text{QR} = 3\sqrt{14}$ and $\text{PR} = 3\sqrt{14}$.

Since $\text{PQ} + \text{QR} = \text{PR}$, the points P, Q, and R lie on the same straight line, with Q located between P and R.

Therefore, the points P(–2, 3, 5), Q(1, 2, 3) and R(7, 0, –1) are collinear.

We will use the distance formula in three dimensions to find the lengths of the sides between the given points.

The distance between two points $(x_1, y_1, z_1)$ and $(x_2, y_2, z_2)$ is given by:

$\text{Distance} = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2 + (z_2 - z_1)^2}$

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(i) Verify if (0, 7, –10), (1, 6, – 6) and (4, 9, – 6) are the vertices of an isosceles triangle.

Let the points be A(0, 7, –10), B(1, 6, –6), and C(4, 9, –6).

An isosceles triangle has at least two sides of equal length.

Calculate the length of side AB:

$AB = \sqrt{(1 - 0)^2 + (6 - 7)^2 + (-6 - (-10))^2}$

$AB = \sqrt{(1)^2 + (-1)^2 + (-6 + 10)^2}$

$AB = \sqrt{1^2 + (-1)^2 + (4)^2}$

$AB = \sqrt{1 + 1 + 16} = \sqrt{18}$

Calculate the length of side BC:

$BC = \sqrt{(4 - 1)^2 + (9 - 6)^2 + (-6 - (-6))^2}$

$BC = \sqrt{(3)^2 + (3)^2 + (-6 + 6)^2}$

$BC = \sqrt{3^2 + 3^2 + 0^2}$

$BC = \sqrt{9 + 9 + 0} = \sqrt{18}$

Calculate the length of side AC:

$AC = \sqrt{(4 - 0)^2 + (9 - 7)^2 + (-6 - (-10))^2}$

$AC = \sqrt{(4)^2 + (2)^2 + (-6 + 10)^2}$

$AC = \sqrt{4^2 + 2^2 + 4^2}$

$AC = \sqrt{16 + 4 + 16} = \sqrt{36} = 6$

Comparing the side lengths, we have $AB = \sqrt{18}$, $BC = \sqrt{18}$, and $AC = 6$.

Since $AB = BC$, two sides of the triangle ABC are equal in length.

Therefore, the points (0, 7, –10), (1, 6, –6) and (4, 9, –6) are indeed the vertices of an isosceles triangle.

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(ii) Verify if (0, 7, 10), (–1, 6, 6) and (– 4, 9, 6) are the vertices of a right angled triangle.

Let the points be A(0, 7, 10), B(–1, 6, 6), and C(– 4, 9, 6).

A triangle is right-angled if the square of the length of the longest side is equal to the sum of the squares of the lengths of the other two sides (Pythagorean theorem).

Calculate the square of the length of side AB:

$AB^2 = (-1 - 0)^2 + (6 - 7)^2 + (6 - 10)^2$

$AB^2 = (-1)^2 + (-1)^2 + (-4)^2$

$AB^2 = 1 + 1 + 16 = 18$

Calculate the square of the length of side BC:

$BC^2 = (-4 - (-1))^2 + (9 - 6)^2 + (6 - 6)^2$

$BC^2 = (-4 + 1)^2 + (3)^2 + (0)^2$

$BC^2 = (-3)^2 + 3^2 + 0^2$

$BC^2 = 9 + 9 + 0 = 18$

Calculate the square of the length of side AC:

$AC^2 = (-4 - 0)^2 + (9 - 7)^2 + (6 - 10)^2$

$AC^2 = (-4)^2 + (2)^2 + (-4)^2$

$AC^2 = 16 + 4 + 16 = 36$

We have $AB^2 = 18$, $BC^2 = 18$, and $AC^2 = 36$.

Check if the Pythagorean theorem holds for the largest squared length $AC^2 = 36$:

$AB^2 + BC^2 = 18 + 18 = 36$

Since $AB^2 + BC^2 = AC^2$, the Pythagorean theorem is satisfied.

Therefore, the points (0, 7, 10), (–1, 6, 6) and (–4, 9, 6) are indeed the vertices of a right-angled triangle, with the right angle at vertex B.

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(iii) Verify if (–1, 2, 1), (1, –2, 5), (4, –7, 8) and (2, –3, 4) are the vertices of a parallelogram.

Let the points be A(–1, 2, 1), B(1, –2, 5), C(4, –7, 8), and D(2, –3, 4) in order.

A quadrilateral is a parallelogram if its opposite sides are equal in length.

Calculate the length of side AB:

$AB = \sqrt{(1 - (-1))^2 + (-2 - 2)^2 + (5 - 1)^2}$

$AB = \sqrt{(1 + 1)^2 + (-4)^2 + (4)^2}$

$AB = \sqrt{2^2 + (-4)^2 + 4^2}$

$AB = \sqrt{4 + 16 + 16} = \sqrt{36} = 6$

Calculate the length of side BC:

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

$BC = \sqrt{(3)^2 + (-7 + 2)^2 + (3)^2}$

$BC = \sqrt{3^2 + (-5)^2 + 3^2}$

$BC = \sqrt{9 + 25 + 9} = \sqrt{43}$

Calculate the length of side CD:

$CD = \sqrt{(2 - 4)^2 + (-3 - (-7))^2 + (4 - 8)^2}$

$CD = \sqrt{(-2)^2 + (-3 + 7)^2 + (-4)^2}$

$CD = \sqrt{(-2)^2 + (4)^2 + (-4)^2}$

$CD = \sqrt{4 + 16 + 16} = \sqrt{36} = 6$

Calculate the length of side DA:

$DA = \sqrt{(-1 - 2)^2 + (2 - (-3))^2 + (1 - 4)^2}$

$DA = \sqrt{(-3)^2 + (2 + 3)^2 + (-3)^2}$

$DA = \sqrt{(-3)^2 + (5)^2 + (-3)^2}$

$DA = \sqrt{9 + 25 + 9} = \sqrt{43}$

Comparing the lengths of the sides:

$AB = 6$ and $CD = 6$. Thus, $AB = CD$.

$BC = \sqrt{43}$ and $DA = \sqrt{43}$. Thus, $BC = DA$.

Since both pairs of opposite sides are equal in length, the points A, B, C, and D taken in order form a parallelogram.

Therefore, the points (–1, 2, 1), (1, –2, 5), (4, –7, 8) and (2, –3, 4) are indeed the vertices of a parallelogram.

Given:

Point A with coordinates $(1, 2, 3)$.

Point B with coordinates $(3, 2, -1)$.

We are looking for the set of points P that are equidistant from A and B.

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To Find:

The equation of the set of points P.

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Solution:

Let P be a point with coordinates $(x, y, z)$ such that it is equidistant from points A and B.

This means the distance from P to A is equal to the distance from P to B.

$PA = PB$

Squaring both sides, we get:

$PA^2 = PB^2$

Using the square of the distance formula between two points $(x_1, y_1, z_1)$ and $(x_2, y_2, z_2)$, which is $(x_2 - x_1)^2 + (y_2 - y_1)^2 + (z_2 - z_1)^2$:

For $PA^2$, using P$(x, y, z)$ and A$(1, 2, 3)$: $PA^2 = (x - 1)^2 + (y - 2)^2 + (z - 3)^2$

For $PB^2$, using P$(x, y, z)$ and B$(3, 2, -1)$: $PB^2 = (x - 3)^2 + (y - 2)^2 + (z - (-1))^2$ $PB^2 = (x - 3)^2 + (y - 2)^2 + (z + 1)^2$

Set $PA^2 = PB^2$:

$(x - 1)^2 + (y - 2)^2 + (z - 3)^2 = (x - 3)^2 + (y - 2)^2 + (z + 1)^2$

Expand the squared terms:

$(x^2 - 2x + 1) + (y^2 - 4y + 4) + (z^2 - 6z + 9) = (x^2 - 6x + 9) + (y^2 - 4y + 4) + (z^2 + 2z + 1)$

Cancel the terms that are common on both sides ($x^2$, $y^2$, $z^2$, $-4y$, $4$, $9$):

$-2x + 1 - 6z = -6x + 2z + 1$

Move all terms involving x, y, and z to one side and constant terms to the other side:

$-2x + 6x - 6z - 2z + 1 - 1 = 0$

$4x - 8z = 0$

Divide the equation by 4:

$x - 2z = 0$

This equation represents a plane in three-dimensional space. The set of points equidistant from two fixed points forms the perpendicular bisector plane of the line segment joining the two points.

The equation of the set of points is $\mathbf{x - 2z = 0}$.

Given:

Point A has coordinates $(4, 0, 0)$.

Point B has coordinates $(-4, 0, 0)$.

The sum of the distances from a point P to A and B is 10, i.e., $PA + PB = 10$.

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To Find:

The equation of the set of points P.

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Solution:

Let the coordinates of the variable point P be $(x, y, z)$.

The distance between two points $(x_1, y_1, z_1)$ and $(x_2, y_2, z_2)$ is given by the distance formula:

$\text{Distance} = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2 + (z_2 - z_1)^2}$

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Using the distance formula for PA and PB:

$PA = \sqrt{(x - 4)^2 + (y - 0)^2 + (z - 0)^2} = \sqrt{(x - 4)^2 + y^2 + z^2}$

$PB = \sqrt{(x - (-4))^2 + (y - 0)^2 + (z - 0)^2} = \sqrt{(x + 4)^2 + y^2 + z^2}$

The given condition is $PA + PB = 10$.

$\sqrt{(x - 4)^2 + y^2 + z^2} + \sqrt{(x + 4)^2 + y^2 + z^2} = 10$

To eliminate the square roots, isolate one square root term:

$\sqrt{(x - 4)^2 + y^2 + z^2} = 10 - \sqrt{(x + 4)^2 + y^2 + z^2}$

Square both sides of the equation:

$(\sqrt{(x - 4)^2 + y^2 + z^2})^2 = (10 - \sqrt{(x + 4)^2 + y^2 + z^2})^2$

$(x - 4)^2 + y^2 + z^2 = 10^2 - 2 \times 10 \times \sqrt{(x + 4)^2 + y^2 + z^2} + (\sqrt{(x + 4)^2 + y^2 + z^2})^2$

$(x^2 - 8x + 16) + y^2 + z^2 = 100 - 20\sqrt{(x + 4)^2 + y^2 + z^2} + (x + 4)^2 + y^2 + z^2$

$x^2 - 8x + 16 + y^2 + z^2 = 100 - 20\sqrt{(x + 4)^2 + y^2 + z^2} + x^2 + 8x + 16 + y^2 + z^2$

Subtract $x^2 + y^2 + z^2 + 16$ from both sides:

$-8x = 100 - 20\sqrt{(x + 4)^2 + y^2 + z^2} + 8x$

Isolate the square root term:

$20\sqrt{(x + 4)^2 + y^2 + z^2} = 100 + 8x + 8x$

$20\sqrt{(x + 4)^2 + y^2 + z^2} = 100 + 16x$

Divide both sides by 4:

$5\sqrt{(x + 4)^2 + y^2 + z^2} = 25 + 4x$

Square both sides again:

$(5\sqrt{(x + 4)^2 + y^2 + z^2})^2 = (25 + 4x)^2$

$25((x + 4)^2 + y^2 + z^2) = 25^2 + 2(25)(4x) + (4x)^2$

$25(x^2 + 8x + 16 + y^2 + z^2) = 625 + 200x + 16x^2$

$25x^2 + 200x + 400 + 25y^2 + 25z^2 = 625 + 200x + 16x^2$

Move all terms to the left side:

$25x^2 - 16x^2 + 25y^2 + 25z^2 + 200x - 200x + 400 - 625 = 0$

$9x^2 + 25y^2 + 25z^2 - 225 = 0$

Rearrange the equation:

$9x^2 + 25y^2 + 25z^2 = 225$

This equation represents an ellipsoid. The given points A and B are the foci of this ellipsoid.

The equation of the set of points P is $\mathbf{9x^2 + 25y^2 + 25z^2 = 225}$.

Alternatively, dividing by 225, we can write the equation in standard form:

$\frac{9x^2}{225} + \frac{25y^2}{225} + \frac{25z^2}{225} = \frac{225}{225}$

$\frac{x^2}{25} + \frac{y^2}{9} + \frac{z^2}{9} = 1$

Given:

Four points A, B, C, and D with coordinates A(1, 2, 3), B(–1, –2, –1), C(2, 3, 2), and D(4, 7, 6).

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To Show:

The points A, B, C, and D are the vertices of a parallelogram ABCD.

The parallelogram ABCD is not a rectangle.

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Solution:

To show that the points form a parallelogram, we need to verify that the lengths of opposite sides are equal.

To show that a parallelogram is not a rectangle, we can verify that the lengths of the diagonals are not equal.

The distance between two points $(x_1, y_1, z_1)$ and $(x_2, y_2, z_2)$ is given by the distance formula:

$\text{Distance} = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2 + (z_2 - z_1)^2}$

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Part 1: Verify that ABCD is a parallelogram.

We calculate the lengths of the sides AB, BC, CD, and DA.

Length of AB (using A(1, 2, 3) and B(–1, –2, –1)):

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

$AB = \sqrt{(-2)^2 + (-4)^2 + (-4)^2}$

$AB = \sqrt{4 + 16 + 16} = \sqrt{36} = 6$

Length of BC (using B(–1, –2, –1) and C(2, 3, 2)):

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

$BC = \sqrt{(2 + 1)^2 + (3 + 2)^2 + (2 + 1)^2}$

$BC = \sqrt{3^2 + 5^2 + 3^2}$

$BC = \sqrt{9 + 25 + 9} = \sqrt{43}$

Length of CD (using C(2, 3, 2) and D(4, 7, 6)):

$CD = \sqrt{(4 - 2)^2 + (7 - 3)^2 + (6 - 2)^2}$

$CD = \sqrt{2^2 + 4^2 + 4^2}$

$CD = \sqrt{4 + 16 + 16} = \sqrt{36} = 6$

Length of DA (using D(4, 7, 6) and A(1, 2, 3)):

$DA = \sqrt{(1 - 4)^2 + (2 - 7)^2 + (3 - 6)^2}$

$DA = \sqrt{(-3)^2 + (-5)^2 + (-3)^2}$

$DA = \sqrt{9 + 25 + 9} = \sqrt{43}$

We compare the lengths of opposite sides:

$AB = 6$ and $CD = 6$. So, $AB = CD$.

$BC = \sqrt{43}$ and $DA = \sqrt{43}$. So, $BC = DA$.

Since the opposite sides are equal in length, the points A, B, C, and D, taken in order, form a parallelogram ABCD.

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Part 2: Verify that ABCD is not a rectangle.

A parallelogram is a rectangle if and only if its diagonals are equal in length.

We calculate the lengths of the diagonals AC and BD.

Length of diagonal AC (using A(1, 2, 3) and C(2, 3, 2)):

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

$AC = \sqrt{1^2 + 1^2 + (-1)^2}$

$AC = \sqrt{1 + 1 + 1} = \sqrt{3}$

Length of diagonal BD (using B(–1, –2, –1) and D(4, 7, 6)):

$BD = \sqrt{(4 - (-1))^2 + (7 - (-2))^2 + (6 - (-1))^2}$

$BD = \sqrt{(4 + 1)^2 + (7 + 2)^2 + (6 + 1)^2}$

$BD = \sqrt{5^2 + 9^2 + 7^2}$

$BD = \sqrt{25 + 81 + 49} = \sqrt{155}$

We compare the lengths of the diagonals:

$AC = \sqrt{3}$ and $BD = \sqrt{155}$.

Since $\sqrt{3} \neq \sqrt{155}$, the lengths of the diagonals AC and BD are not equal.

Therefore, the parallelogram ABCD is not a rectangle.

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We have shown that opposite sides are equal ($AB=CD=6$, $BC=DA=\sqrt{43}$), confirming it is a parallelogram, and that the diagonals are not equal ($AC=\sqrt{3}$, $BD=\sqrt{155}$), confirming it is not a rectangle.

Given:

Point A has coordinates $(3, 4, -5)$.

Point B has coordinates $(-2, 1, 4)$.

We are looking for the set of points P such that the distance from P to A is equal to the distance from P to B.

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To Find:

The equation of the set of points P.

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Solution:

Let the coordinates of the variable point P be $(x, y, z)$.

The condition that P is equidistant from A and B is given by $PA = PB$.

Squaring both sides of the equation, we get $PA^2 = PB^2$.

The square of the distance between two points $(x_1, y_1, z_1)$ and $(x_2, y_2, z_2)$ is given by the formula:

$\text{Distance}^2 = (x_2 - x_1)^2 + (y_2 - y_1)^2 + (z_2 - z_1)^2$

Using this formula for $PA^2$ (with P$(x, y, z)$ and A$(3, 4, -5)$):

$PA^2 = (x - 3)^2 + (y - 4)^2 + (z - (-5))^2$

$PA^2 = (x - 3)^2 + (y - 4)^2 + (z + 5)^2$

Using this formula for $PB^2$ (with P$(x, y, z)$ and B$(-2, 1, 4)$):

$PB^2 = (x - (-2))^2 + (y - 1)^2 + (z - 4)^2$

$PB^2 = (x + 2)^2 + (y - 1)^2 + (z - 4)^2$

Now, set $PA^2 = PB^2$:

$(x - 3)^2 + (y - 4)^2 + (z + 5)^2 = (x + 2)^2 + (y - 1)^2 + (z - 4)^2$

Expand each squared term:

$(x^2 - 6x + 9) + (y^2 - 8y + 16) + (z^2 + 10z + 25) = (x^2 + 4x + 4) + (y^2 - 2y + 1) + (z^2 - 8z + 16)$

Simplify the equation by canceling the $x^2$, $y^2$, and $z^2$ terms from both sides:

$-6x + 9 - 8y + 16 + 10z + 25 = 4x + 4 - 2y + 1 - 8z + 16$

Combine the constant terms on each side:

$-6x - 8y + 10z + 50 = 4x - 2y - 8z + 21$

Move all terms involving x, y, and z to one side (e.g., the left side) and constant terms to the other side:

$-6x - 8y + 10z - 4x + 2y + 8z = 21 - 50$

Combine like terms:

$(-6x - 4x) + (-8y + 2y) + (10z + 8z) = -29$

$-10x - 6y + 18z = -29$

Multiply the entire equation by -1 to make the coefficients positive (optional, but standard form):

$10x + 6y - 18z = 29$

Or, write it with all terms on one side:

$10x + 6y - 18z - 29 = 0$

This is the equation of a plane, which is the perpendicular bisector plane of the line segment joining points A and B.

The equation of the set of points P is $\mathbf{10x + 6y - 18z - 29 = 0}$.

Given:

Coordinates of vertex A are $(3, -5, 7)$.

Coordinates of vertex B are $(-1, 7, -6)$.

Coordinates of the centroid G are $(1, 1, 1)$.

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To Find:

The coordinates of vertex C.

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Solution:

Let the coordinates of the vertices of the triangle ABC be $A(x_1, y_1, z_1)$, $B(x_2, y_2, z_2)$, and $C(x_3, y_3, z_3)$.

The coordinates of the centroid G of the triangle are given by the formula:

$G = \left(\frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3}, \frac{z_1 + z_2 + z_3}{3}\right)$

We are given:

$A = (3, -5, 7) \implies x_1 = 3, y_1 = -5, z_1 = 7$

$B = (-1, 7, -6) \implies x_2 = -1, y_2 = 7, z_2 = -6$

$G = (1, 1, 1)$

Let the coordinates of C be $(x_3, y_3, z_3)$.

Using the centroid formula, we can set up the following equations for each coordinate:

For the x-coordinate:

Multiply both sides by 3:

Solve for $x_3$:

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For the y-coordinate:

Multiply both sides by 3:

Solve for $y_3$:

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For the z-coordinate:

Multiply both sides by 3:

Solve for $z_3$:

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Thus, the coordinates of point C are $(1, 1, 2)$.

The coordinates of point C are $\mathbf{(1, 1, 2)}$.

Given:

The coordinates of three vertices of a parallelogram ABCD are A(3, –1, 2), B(1, 2, –4), and C(–1, 1, 2).

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To Find:

The coordinates of the fourth vertex D.

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Solution:

In a parallelogram, the diagonals bisect each other. This means the midpoint of the diagonal AC is the same as the midpoint of the diagonal BD.

Let the coordinates of the fourth vertex D be $(x, y, z)$.

The formula for the midpoint of a line segment joining points $(x_1, y_1, z_1)$ and $(x_2, y_2, z_2)$ is $\left(\frac{x_1+x_2}{2}, \frac{y_1+y_2}{2}, \frac{z_1+z_2}{2}\right)$.

Let's find the midpoint of diagonal AC, using A(3, –1, 2) and C(–1, 1, 2):

Now, let's find the midpoint of diagonal BD, using B(1, 2, –4) and D(x, y, z):

Since the midpoints are the same, we equate the corresponding coordinates:

Thus, the coordinates of the fourth vertex D are $(1, -2, 8)$.

The coordinates of the fourth vertex are $\mathbf{(1, -2, 8)}$.

Given:

The vertices of the triangle ABC are A(0, 0, 6), B(0, 4, 0), and C(6, 0, 0).

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To Find:

The lengths of the medians of the triangle.

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Solution:

A median of a triangle is a line segment joining a vertex to the midpoint of the opposite side.

Let the vertices be $A(x_1, y_1, z_1)$, $B(x_2, y_2, z_2)$, and $C(x_3, y_3, z_3)$.

The midpoint M of a line segment joining $(x_a, y_a, z_a)$ and $(x_b, y_b, z_b)$ is $\left(\frac{x_a+x_b}{2}, \frac{y_a+y_b}{2}, \frac{z_a+z_b}{2}\right)$.

The distance between two points $(x_a, y_a, z_a)$ and $(x_b, y_b, z_b)$ is $\sqrt{(x_b - x_a)^2 + (y_b - y_a)^2 + (z_b - z_a)^2}$.

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Let D be the midpoint of side BC. The median from vertex A is AD.

Using B(0, 4, 0) and C(6, 0, 0):

Now, calculate the length of median AD using A(0, 0, 6) and D(3, 2, 0):

$AD = \sqrt{(3 - 0)^2 + (2 - 0)^2 + (0 - 6)^2}$

$AD = \sqrt{3^2 + 2^2 + (-6)^2}$

$AD = \sqrt{9 + 4 + 36} = \sqrt{49} = 7$

The length of the median from A is $\mathbf{7}$.

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Let E be the midpoint of side AC. The median from vertex B is BE.

Using A(0, 0, 6) and C(6, 0, 0):

Now, calculate the length of median BE using B(0, 4, 0) and E(3, 0, 3):

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

$BE = \sqrt{3^2 + (-4)^2 + 3^2}$

$BE = \sqrt{9 + 16 + 9} = \sqrt{34}$

The length of the median from B is $\mathbf{\sqrt{34}}$.

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Let F be the midpoint of side AB. The median from vertex C is CF.

Using A(0, 0, 6) and B(0, 4, 0):

Now, calculate the length of median CF using C(6, 0, 0) and F(0, 2, 3):

$CF = \sqrt{(0 - 6)^2 + (2 - 0)^2 + (3 - 0)^2}$

$CF = \sqrt{(-6)^2 + 2^2 + 3^2}$

$CF = \sqrt{36 + 4 + 9} = \sqrt{49} = 7$

The length of the median from C is $\mathbf{7}$.

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The lengths of the medians of the triangle with the given vertices are $\mathbf{7}$, $\mathbf{\sqrt{34}}$, and $\mathbf{7}$.

Given:

The vertices of triangle PQR are P(2a, 2, 6), Q(–4, 3b, –10), and R(8, 14, 2c).

The centroid of the triangle is the origin, O(0, 0, 0).

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To Find:

The values of a, b, and c.

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Solution:

Let the coordinates of the vertices be $P(x_1, y_1, z_1) = (2a, 2, 6)$, $Q(x_2, y_2, z_2) = (-4, 3b, -10)$, and $R(x_3, y_3, z_3) = (8, 14, 2c)$.

Let the coordinates of the centroid G be $(x_G, y_G, z_G) = (0, 0, 0)$.

The coordinates of the centroid of a triangle are given by the formula:

Substitute the given coordinates into the centroid formula:

For the x-coordinate of the centroid:

Multiply both sides by 3:

Solve for a:

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For the y-coordinate of the centroid:

Multiply both sides by 3:

Solve for b:

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For the z-coordinate of the centroid:

Multiply both sides by 3:

Solve for c:

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Thus, the values of a, b, and c are -2, $-\frac{16}{3}$, and 2, respectively.

The values are $\mathbf{a = -2}$, $\mathbf{b = -\frac{16}{3}}$, and $\mathbf{c = 2}$.

Given:

Point A has coordinates $(3, 4, 5)$.

Point B has coordinates $(-1, 3, -7)$.

The condition $PA^2 + PB^2 = k^2$, where P is a variable point and $k$ is a constant.

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To Find:

The equation of the set of points P.

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Solution:

Let the coordinates of the variable point P be $(x, y, z)$.

The square of the distance between two points $(x_1, y_1, z_1)$ and $(x_2, y_2, z_2)$ is given by the formula:

$\text{Distance}^2 = (x_2 - x_1)^2 + (y_2 - y_1)^2 + (z_2 - z_1)^2$

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First, let's calculate $PA^2$ using the coordinates of P$(x, y, z)$ and A$(3, 4, 5)$.

$PA^2 = (x - 3)^2 + (y - 4)^2 + (z - 5)^2$

Expanding the terms:

$PA^2 = (x^2 - 6x + 9) + (y^2 - 8y + 16) + (z^2 - 10z + 25)$

$PA^2 = x^2 + y^2 + z^2 - 6x - 8y - 10z + 9 + 16 + 25$

$PA^2 = x^2 + y^2 + z^2 - 6x - 8y - 10z + 50$

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Next, let's calculate $PB^2$ using the coordinates of P$(x, y, z)$ and B$(-1, 3, -7)$.

$PB^2 = (x - (-1))^2 + (y - 3)^2 + (z - (-7))^2$

$PB^2 = (x + 1)^2 + (y - 3)^2 + (z + 7)^2$

Expanding the terms:

$PB^2 = (x^2 + 2x + 1) + (y^2 - 6y + 9) + (z^2 + 14z + 49)$

$PB^2 = x^2 + y^2 + z^2 + 2x - 6y + 14z + 1 + 9 + 49$

$PB^2 = x^2 + y^2 + z^2 + 2x - 6y + 14z + 59$

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Now, substitute the expressions for $PA^2$ and $PB^2$ into the given condition $PA^2 + PB^2 = k^2$.

$(x^2 + y^2 + z^2 - 6x - 8y - 10z + 50) + (x^2 + y^2 + z^2 + 2x - 6y + 14z + 59) = k^2$

Combine like terms:

$(x^2 + x^2) + (y^2 + y^2) + (z^2 + z^2) + (-6x + 2x) + (-8y - 6y) + (-10z + 14z) + (50 + 59) = k^2$

$2x^2 + 2y^2 + 2z^2 - 4x - 14y + 4z + 109 = k^2$

Rearrange the equation to find the equation of the set of points P:

$2x^2 + 2y^2 + 2z^2 - 4x - 14y + 4z + 109 - k^2 = 0$

This equation represents the set of points P satisfying the given condition. It is the equation of a sphere if the term $109 - k^2$ results in a positive radius squared after completing the square.

The equation of the set of points P is $\mathbf{2x^2 + 2y^2 + 2z^2 - 4x - 14y + 4z + 109 - k^2 = 0}$.