2.
(a) Let 3x − 2y + 4z = 11 and 2x − 5y + 3z = 3 be two planes. Then P = (7, 1, −2)
are on both planes. Let L be the line of intersection of these two planes. Find a
parametric equation for L.
(b) With the same two planes as in (a), we know that Q = (21, 0, −13) is on both
planes. Find another parametric equation for ℓ. Find a relation between the two
parametrizations in (a) and (b).
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(a) One plane is
3x - 2y + 4z = 11. (1)
Another plane is
2x − 5y + 3z = 3. (2)
The intersection line satifies both these equations (1) and (2).
By multiplying eq(1) by 2 and by multiplying eq(2) by 3, we find that the intersection line satifies these two equations
6x - 4y + 8z = 22, (1')
6x −15y + 9z = 9. (2')
Hence, the intersection line satisfies the difference of these two equations
11y - z = 13. (3)
Having it, we just can write the intersection line in parametric form
y = t, z = 11t - 13, and then for x, from equation (2)
2x = 3 + 5y - 3z = 3 + 5t - 3(11t-13) = 3 + 5t - 33t + 39 = -28t + 42, which gives
x = -14t + 21.
Thus the parametric equations for L are x= -14t + 21, y= t, z= 11t - 13.
I found them independently of the information about point P.
Part (a) is solved.
In order for do not create a mess, I will stop at this point.
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To be honest, it remains unclear to me, for what reason the point P= (7, 1, −2)
is given in this problem and what is its role.
May be, its role is to confuse a reader ?
With our local problems' creators at this forum, it can easily be so.