- #1
marutkpadhy
- 9
- 0
Find the radius of the circular section of the sphere of the sphere x^2 + y^2 + z^2 = 49 by the plane 2x+3y-z-5 \sqrt{14}= 0
MHB Calculating Radius of Circular Section from Sphere and Plane Intersection
Click For Summary
To find the radius of the circular section formed by the intersection of the sphere defined by x^2 + y^2 + z^2 = 49 and the plane 2x + 3y - z - 5√14 = 0, one approach is to calculate the distance from the sphere's center to the plane. This distance can be determined using the formula d = |Ax0 + By0 + Cz0 + D| / √(A^2 + B^2 + C^2). Once the distance d is found, the radius r of the circular section can be derived using the right triangle formed by the sphere's radius and the distance to the plane. This method simplifies the problem without needing to substitute z back into the sphere's equation. The final radius of the circular section can then be calculated effectively.
- #2
MarkFL
Gold Member
MHB
- 13,284
- 12
Suppose you solve the equation of the plane for $z$, and then substitute for $z$ into the equation of the sphere...what do you get?
- #3
Evgeny.Makarov
Gold Member
MHB
- 2,434
- 4
View attachment 2708
I think it is easier to find the distance $d$ from the center of the sphere to the plane (recall the the distance from $(x_0,y_0,z_0)$ to $Ax+By+Cz+D=0$ is $\dfrac{Ax_0+By_y+Cz_0+D}{\sqrt{A^2+B^2+C^2}}$) and then find the radius $r$ of the required circular section from the right triangle.
I think it is easier to find the distance $d$ from the center of the sphere to the plane (recall the the distance from $(x_0,y_0,z_0)$ to $Ax+By+Cz+D=0$ is $\dfrac{Ax_0+By_y+Cz_0+D}{\sqrt{A^2+B^2+C^2}}$) and then find the radius $r$ of the required circular section from the right triangle.
Attachments
Similar threads
- · Replies 8 ·
- · Replies 11 ·
- · Replies 2 ·
- · Replies 1 ·
- · Replies 6 ·
- · Replies 1 ·
- · Replies 1 ·
- · Replies 2 ·
- · Replies 3 ·