If (2, 3, 5) is one end of a diameter of the sphere x² + y² + z² - 6x - 12y - 2z + 20 = 0, then the

Q: Wait, why is the center $(-u, -v, -w)$ and not $(u, v, w)$? I keep getting the signs wrong!

It is a classic trap! The general equation is $x^2 + y^2 + z^2 + 2ux + 2vy + 2wz + d = 0$. If you complete the square for $(x^2 + 2ux)$, you get $(x+u)^2 - u^2$. So the center is at $x = -u$. Always remember to divide the coefficient of $x, y,$ and $z$ by $2$ and then negate it.

Q: Can I use the distance formula to solve this?

You could, but it would be much more complicated. The distance formula would give you a sphere of possible points, not a single point. The midpoint property is the most direct and elegant path.

Q: What if the equation was $x^2 + y^2 + z^2 - 6x - 12y - 2z + 20 = 0$ but the coefficients of $x^2, y^2, z^2$ weren't $1$?

Excellent question! If the coefficients were not $1$, you would first need to divide the entire equation by that coefficient to normalize it to the standard form. Always ensure the $x^2, y^2,$ and $z^2$ terms have a coefficient of $1$ before comparing.

Question

If $(2, 3, 5)$ is one end of a diameter of the sphere $x^2 + y^2 + z^2 - 6x - 12y - 2z + 20 = 0$, then the coordinates of the other end of the diameter are

Options:

(A) $(4, 3, 5)$

(B) $(4, 3, -3)$

(C) $(4, 9, -3)$

(D) $(4, -3, 3)$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine a perfect three-dimensional sphere floating in the vast, silent void of coordinate space. It is a shape of absolute symmetry, defined by the equation:

$$x^2 + y^2 + z^2 - 6x - 12y - 2z + 20 = 0$$

When we look at this equation, we are not just looking at numbers; we are looking at the blueprint of a sphere. Our goal is to find the other end of a diameter, given that one end is at $A(2, 3, 5)$.

### Decoding the General Equation

The general equation of a sphere is given by:

$$x^2 + y^2 + z^2 + 2ux + 2vy + 2wz + d = 0$$

This is our master key. By comparing our specific equation to this general form, we can extract the values of $u$, $v$, and $w$.

For the $x$-term, we have $2u = -6$, which tells us $u = -3$. For the $y$-term, $2v = -12$, so $v = -6$. For the $z$-term, $2w = -2$, meaning $w = -1$.

The center of the sphere is located at $C(-u, -v, -w)$. Substituting our values, we get $C(3, 6, 1)$. This point is the absolute center of our sphere.

### The Geometric Bridge

Now, let us visualize the diameter. A diameter is a line segment that passes through the center of the sphere and connects two points on its surface.

Because the center is equidistant from every point on the surface, it must be the midpoint of any diameter. This is the geometric bridge that connects our known point $A(2, 3, 5)$ to our unknown point $B(x, y, z)$.

The midpoint formula is our tool:

$$C = \left( \frac{x_A + x_B}{2}, \frac{y_A + y_B}{2}, \frac{z_A + z_B}{2} \right)$$

We know $C$ is $(3, 6, 1)$ and $A$ is $(2, 3, 5)$.

### The Final Calculation

We can now set up three simple linear equations to solve for the coordinates of $B(x, y, z)$.

For the $x$-coordinate:
$$\frac{2 + x}{2} = 3 \Rightarrow 2 + x = 6 \Rightarrow x = 4$$

For the $y$-coordinate:
$$\frac{3 + y}{2} = 6 \Rightarrow 3 + y = 12 \Rightarrow y = 9$$

For the $z$-coordinate:
$$\frac{5 + z}{2} = 1 \Rightarrow 5 + z = 2 \Rightarrow z = -3$$

The coordinates of the other end of the diameter are **$(4, 9, -3)$**. We have successfully navigated the geometry of the sphere and found our destination.

Animated Solution for Mathematics - Three Dimensional Geometry: If $(2, 3, 5)$ is one end of a diameter of the sphere $x^{2} + y^{2} + z^{2} - 6 x - 12 y - 2 z + 20 = 0$ , then the coordinates of the other end of the diameter are

Select Answer:

Visualized Solution

Visualizing the Sphere

The General Equation of a Sphere

Comparing Coefficients

Calculating $u, v, w$

Finding the Center $C$

The Diameter and Point $B$

The Midpoint Formula

Setting up the Equation for $x$

Solving for $x$

Setting up the Equation for $y$

Solving for $y$

Setting up the Equation for $z$

Solving for $z$

The Final Coordinates

The Sigma Insight: Equation of a Line in Space

Analyzing the Setup

Decoding the General Equation

The Geometric Bridge

The Final Calculation

Similar Questions

Consider the line $L$ passing through the points $(1, 2, 3)$ and $(2, 3, 5)$ . The distance of the point $(\frac{11}{3}, \frac{11}{3}, \frac{19}{3})$ from the line $L$ along the line $\frac{3 x - 11}{2} = \frac{3 y - 11}{1} = \frac{3 z - 19}{2}$ is equal to

If the length of the perpendicular from the point $(β, 0, β) (β \neq = 0)$ to the line $\frac{x - 1}{1} = \frac{y - 0}{0} = \frac{z + 1}{- 1}$ is $3/2$ , then $β$ is equal to :

Let the line $L$ be the projection of the line $\frac{x - 1}{2} = \frac{y - 3}{1} = \frac{z - 4}{2}$ in the plane $x - 2 y - z = 3$ . If $d$ is the distance of the point $(0, 0, 6)$ from $L$ , then $d^{2}$ is equal to .

The square of the distance of the image of the point $(6, 1, 5)$ in the line $\frac{x - 1}{3} = \frac{y}{2} = \frac{z - 2}{4}$ , from the origin is

The perpendicular distance, of the line $\frac{x - 1}{2} = \frac{y + 2}{- 1} = \frac{z + 3}{2}$ from the point $P (2, - 10, 1)$ , is :

The distance of line $3 y - 2 z - 1 = 0 = 3 x - z + 4$ from the point $(2, - 1, 6)$ is :

The distance of the line $\frac{x - 2}{2} = \frac{y - 6}{3} = \frac{z - 3}{4}$ from the point (1,4,0) along the line $\frac{x}{1} = \frac{y - 2}{2} = \frac{z + 3}{3}$ is:

If a point R(4,y,z) lies on the line segment joining the points P(2,-3,4) and Q(8,0,10), then the distance of R from the origin is :

The length of the perpendicular from the point $(2, - 1, 4)$ on the straight line, $\frac{x + 3}{10} = \frac{y - 2}{- 7} = \frac{z}{1}$ is :

The distance, of the point $(7, - 2, 11)$ from the line $\frac{x - 6}{1} = \frac{y - 4}{0} = \frac{z - 8}{3}$ along the line $\frac{x - 5}{2} = \frac{y - 1}{- 3} = \frac{z - 5}{6}$ , is :

Select Answer:

Visualized Solution

Visualizing the Sphere

The General Equation of a Sphere

Comparing Coefficients

Calculating .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } u,v,w

Finding the Center .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } C

The Diameter and Point .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } B

The Midpoint Formula

Setting up the Equation for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } x

Solving for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } x

Setting up the Equation for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } y

Solving for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } y

Setting up the Equation for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } z

Solving for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } z