  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 x−12=y−34=5−z−5.

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=2→b1 that implies →b1 || →b2 so, the lines are parallel. Therefore, S.D. between two parallel lines, d=∣∣(→a2−→a1)×→b∣∣∣∣→b∣∣ ⇒d=∣∣(^i+2^j+2^k)×(2^i+3^j+4^k)∣∣∣∣2^i+3^j+4^k∣∣=∣∣2^i−^k∣∣√4+9+16=√5√29=√529  units.                                                                       OR Plane passing through the intersection of the planes 2x + y - z = 3 and 5x - 3y + 4z + 9 = 0 is : π:2x+y−z−3+λ(5x−3y+4z+9)=0         ⇒π:x(2+5λ)+y(1−3λ)+z(−1+4λ)−3+9λ=0 Since plane π is parallel to the line x−12=y−34=5−z5 with d.r's 2, 4, 5 so, the normal to the plane will be perpendicular to the line. Therefore,  2(2+5λ)+4(1−3λ)+5(−1+4λ)=0       ⇒10λ−12λ+20λ+4+4−5=0 ⇒18λ=−3       ∴λ=−16 Therefore the required plane is π:x(2−56)+y(1+12)+z(−1−23)−3−32=0 That is 7x + 9y - 10z - 27 = 0.  Suggest corrections   