Let L₁ be the line of intersection of the planes x+y=2 and 2x-z=1 . If the image of the point P (2,0,0) in the line L₁ is Q , and d is the shortest distance between the line passing through the origin and Q , and the line L₂ passing through

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Let L₁ be the line of intersection of the planes x+y=2 and 2x-z=1 . If the image of the point P (2,0,0) in the line L₁ is Q , and d is the shortest distance between the line passing through the origin and Q , and the line L₂ passing through (3,3,1) and (4,4,2) , then d² is equal to

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Correct answer: C. 2

Let x = . From the given planes, y = 2- and z = 2 -1 . The symmetric equation of L₁ is x 1 = y-2 -1 = z+1 2 = . Let the foot of the perpendicular from P (2,0,0) to L₁ be F ( , 2- , 2 -1) . The direction ratios of PF are ( -2, 2- , 2 -1) . Since PF is perpendicular to L₁ , the dot product of their direction ratios is zero: 1( -2) - 1(2- ) + 2(2 -1) = 0 - 2 - 2 + + 4 - 2 = 0 6 = 6 = 1 Thus, F = (1,1,1) . Since F is the midpoint of P and its image Q : Q = 2 F - P = (2(1)-2, 2(1)-0, 2(1)-0) = (0,2,2) . The line passing

Let L₁ be the line of intersection of the planes x+y=2 and 2x-z=1 . If the image of the point P (2,0,0) in the line L₁ is Q , and d is the shortest distance between the line passing through the origin and Q , and the line L₂ passing through (3,3,1) and (4,4,2) , then d² is equal to

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