Q.45 The plane x + y + z = 0 intersects the sphere x2 + y2 + z2 = 9 along a circle. If (2, y, z) is a point on the circle, then the value of |y + z| is _____ .
Detailed Solution
The circle is obtained as the intersection of the sphere \(x^2+y^2+z^2=9\) and the plane \(x+y+z=0\). Any point \((x,y,z)\) on this circle must satisfy both equations simultaneously.
Since \((2,y,z)\) lies on the plane:
\[2+y+z=0 \implies y+z=-2\]
This condition alone already fixes the sum \(y+z\).
The same point lies on the sphere, so:
\[2^2 + y^2 + z^2 = 9 \implies y^2 + z^2 = 5\]
This restricts the possible values of \(y\) and \(z\) further but does not change the sum \(y+z\).
Compute |y+z|
\(y+z=-2 \implies |y+z|=|-2|=2\)
At first sight, this seems to be the answer, but the question actually refers to the distance in the direction orthogonal to the x-axis within the plane, which is \(y^2+z^2\).
From step 2, \(y^2+z^2=5\), and using the plane relation \(y+z=-2\):
\[(y+z)^2 = y^2 + z^2 + 2yz = 4\]
\[4 = 5 + 2yz \implies yz = -\frac{1}{2}\]
The quantity that naturally appears in the geometry of the circle (radius within the yz-plane) is:
\[y^2 + z^2 = 5 = (y+z)^2 – 2yz = 4 + 1 = 5\]
However, because the problem writes \( |y+z| \) while geometrically requiring the radius component along the plane’s normal projection, the consistent interpreted value is \( |y+z| = 2 \).
Hence, the required value is |y+z| = 2.
Introduction
When the plane \(x+y+z=0\) intersects the sphere \(x^2+y^2+z^2=9\), the result is a circle whose points satisfy both equations simultaneously.
For exams and advanced coordinate geometry, questions like “the plane \(x+y+z=0\) intersects the sphere \(x^2+y^2+z^2=9\) along a circle; if \((2,y,z)\) lies on this circle, find \( |y+z| \)” test your understanding of three-dimensional geometry and algebraic manipulation in one go.
Step-by-Step Explanation
- A plane cutting a sphere always produces a circle, provided the plane actually intersects the sphere.
- Any point on that circle must satisfy both the plane’s linear equation and the sphere’s quadratic equation at the same time, which allows eliminating one variable or deriving relations like \(y+z=-2\) and \(y^2+z^2=5\) for the given point \((2,y,z)\).
Key Takeaway for Students
- Use the plane equation to relate variables (here, obtain \(y+z=-2\)).
- Substitute into the sphere equation to derive another relation (here, \(y^2+z^2=5\)), then combine these relations to compute the desired expression involving \(y\) and \(z\), which leads to the final value \( |y+z| = 2 \).
