Q.Find the equation of a plane passing through the points (1, 0, 0) and (0, 2, 0) which is at a distance of 6/7 units from the origin. OR Find the distance of the point (1, −2, 3) from the plane x − y + z = −1 measured along a line parallel to the line x/2 = y/3 = z/−6.
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Start your 14-day free trial to unlock the full solution →Set up the plane's equation using the two given points, then use the distance-from-origin condition to find the unknown coefficient. (Answering the primary version of this question; the OR alternative is not separately solved.)
Let the plane be ax+by+cz=d.
Through (1,0,0): a = d.
Through (0,2,0): 2b = d ⟹ b = d/2.
So the plane is: dx + (d/2)y + cz = d. Dividing by d (d≠0) and letting k=c/d:
x + y/2 + kz = 1, i.e., 2x + y + 2kz = 2 (multiplying by 2).
Distance from the origin (0,0,0) to this plane 2x+y+2kz−2=0:
|−2|/√(2²+1²+(2k)²) = 2/√(5+4k²)
Set this equal to 6/7:
2/√(5+4k²) = 6/7
√(5+4k²) = 14/6 = 7/3
5+4k² = 49/9 …
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