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Right circular cylinder of radius 2 (axis 2,1,2)

Short answer

Axis through (1,2,3) with direction (2,1,2). Distance condition: |AP|² − (AP·d)²/9 = 4. With AP = (x−1,y−2,z−3), expanding yields 5x²+8y²+5z²−4xy−4yz−8zx+22x−16y−14z−10=0.

Find the equation of the right circular cylinder of radius 2 whose axis is the line (x−1)/2 = (y−2)/1 = (z−3)/2.

Given. Axis: (x − 1)/2 = (y − 2)/1 = (z − 3)/2, radius 2.

The axis passes through A(1, 2, 3) with direction d = (2, 1, 2), |d|² = 9.

Condition

A point P(x, y, z) lies on the cylinder if its perpendicular distance from the axis is 2. With AP = (x − 1, y − 2, z − 3):

distance² = |AP|² − (AP·d)²/|d|² = 4

9[(x − 1)² + (y − 2)² + (z − 3)²] − [2(x − 1) + (y − 2) + 2(z − 3)]² = 36

Expand

Let L = 2x + y + 2z − 10, so L² = 4x² + y² + 4z² + 4xy + 8xz + 4yz − 40x − 20y − 40z + 100.

9[(x−1)² + (y−2)² + (z−3)²] = 9x² + 9y² + 9z² − 18x − 36y − 54z + 126.

Subtract and equate to 36:

5x² + 8y² + 5z² − 4xy − 8xz − 4yz + 22x − 16y − 14z + 126 − 100 − 36 = 0

Answer: 5x² + 8y² + 5z² − 4xy − 4yz − 8zx + 22x − 16y − 14z − 10 = 0.

✓ Answer checked by Claude Opus 5.5 (high effort)

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