A.) I can first help you out with the second part. [Later: in fact, logically, it must be answered first.]
The distance of a point (x1, y1, z1) from a plane
ax + by + cz + d = 0
is mod (a x1 + b y1 +c z1 + d) / sqrt(a^2 + b^2 + c^2).
So the distance of the point (7, 1, -2) [the sphere's centre] from the plane 2x + 3y - z - 5 = 0 is:
mod(2*7 + 3*1 -1*(-2) - 5) / sqrt(4 + 9 + 1) = mod(14) / sqrt(14) = sqrt(14).
Thus the radius of the sphere is sqrt(14).
But the centre must be the same distance from the other tangent plane, which since it's parallel to the first one must also have an equation of the form:
2x + 3y - z - C = 0
So we must also have
mod(2*7 + 3*1 -1*(-2) - C) / sqrt(14) = sqrt(14), so that
mod(19 - C) must be 14 once again. But C = 5 being already taken, the only other number that will satisfy this is C = 33, which gives mod(-14) i.e. mod(14), or 14, as before.
So the other plane is
2x + 3y - z = 33. ......(1) QED
[As Samuel Pepys would say: "And so to bed."}
B.) Back up with the lark:
The equation of the normal to the tangent planes through the centre, C (7, 1, - 2), is given by:
(x - 7)/2 = t, (y - 1)/3 = t, and (z + 2)/(- 1) = t, ......(2)
where t is a parameter that varies along that normal. Inserting those values for x, y and z along the normal into the equation for the tangent plane 2x + 3y - z = 33 (1), one finds:
2(7 + 2t) + 3 (1 + 3 t) - (- 2 - t) = 33, that is
19 + 4 t = 33, i.e. t = 1.
Putting that value of t back into the parameterized equation for the line (2), one finds that the intersection point is:
(9, 4, - 3). QED
And so we've put the entire problem to bed.
Live long and prosper.
2007-04-29 19:49:33
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answer #1
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answered by Dr Spock 6
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Dr Spock is correct except for one little error on this statement near the end.
19 + 4 t = 33, i.e. t = 1
It should say "14 t".
19 + 14 t = 33, i.e. t = 1
2007-05-01 09:42:59
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answer #2
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answered by Northstar 7
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