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Question

Find the shortest distance between the following lines :

r=^i+2^j+3^k+λ(2^i+3^j+4^k) and  r=2^i+4^j+5^k+μ(4^i+6^j+8^k)

                                                                                                   OR

Find the equation of the plane passing through the line of intersection of the planes 2x + y - z = 3 and 5x - 3y + 4z + 9 = 0 and is parallel to the line x12=y34=5z5.


Solution

Let L1:r=^i+2^j+3^k+λ(2^i+3^j+4^k)andL2:r=2^i+4^j+5^k+μ(4^i+6^j+8^k)

Here a1=^i+2^j+3^k, b1=2^i+3^j+4^k; a2=2^i+4^j+5^k, b2=4^i+6^j+8^k

Note that, b2=2b1 that implies b1 || b2 so, the lines are parallel.

Therefore, S.D. between two parallel lines, d=(a2a1)×bb

d=(^i+2^j+2^k)×(2^i+3^j+4^k)2^i+3^j+4^k=2^i^k4+9+16=529=529  units.

                                                                      OR

Plane passing through the intersection of the planes 2x + y - z = 3 and 5x - 3y + 4z + 9 = 0 is : π:2x+yz3+λ(5x3y+4z+9)=0         π:x(2+5λ)+y(13λ)+z(1+4λ)3+9λ=0

Since plane π is parallel to the line x12=y34=5z5 with d.r's 2, 4, 5 so, the normal to the plane will be perpendicular to the line.

Therefore,  2(2+5λ)+4(13λ)+5(1+4λ)=0       10λ12λ+20λ+4+45=0

18λ=3       λ=16

Therefore the required plane is π:x(256)+y(1+12)+z(123)332=0

That is 7x + 9y - 10z - 27 = 0.

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