Let lambda1 lambda2 be the values of lambda for which the po

Question

Let lambda_1, lambda_2 be the values of lambda for which the points ((5 / 2), 1, lambda) and (-2,0,1) are at equal distance from the plane 2 x+3 y-6 z+7=0. If lambda_1 > lambda_2, then the distance of the point (lambda_1-lambda_2, lambda_2, lambda_1) from the line (x-5 / 1)=(y-1 / 2)=(z+7 / 2) is ____________.

DivineJEE · Accepted Answer

Since ((5 / 2), 1, lambda) and (-2,0,1) are equidistant from plane 2 x+3 y-6 z+7=0 Therefore |frac2((5 / 2))+3(1)-6(lambda)+7sqrt(2^2+3^2+6^2)|=|frac2(-2)+3(0)-6(1)+7sqrt(2^2+3^2+6^2)| \ => |5+3-6 lambda+7|=|-4-6+7| \ => |15-6 lambda|=|-3| \ => 15-6 lambda= ± 3 \ => 15-6 lambda=3 or 15-6 lambda=-3 \ => 6 lambda=12 or 6 lambda=18 \ => lambda=2 or lambda=3 Because lambda_1>lambda_2 \ Therefore lambda_1=3 and lambda_2=2 So, point will be (1,2,3) Let M_0=(1,2,3) M_1 is the point through which line passes i.e, (5,1,-7) and vec s=hat i+2 hat j+2 hat k Therefore -> M_0 M_1=4 hat i-hat j-10 hat k Now, required distance =|frac-> M_0 M_1 x vec s|vec s|| =frac|(4 hat i-hat j-10 hat k) x (hat i+2 hat j+2 hat k)|sqrt(1+4+4) \ =frac|18 hat i-18 hat j+9 hat k|3=9

Explanation

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