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If P(alpha,beta,lambda) is a vertex of an equilateral triangle PQR where vertex Q and R are (-1,0,1) and (1,0,-1) respectively, then P can lie on the plane |
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Answer» `x+y+z+6=0` `therefore QR=2sqrt(2)` Now, the mid point of QR is M(0,0,0). Triangle PQR is equilateral. `therefore PM=MQ XX tan60^(@) = sqrt(2)sqrt(3)=sqrt(6)`. Thus, P can lie on the plane whose DISTANCE from point M is less than or equal to `sqrt(6)`. Hence, P can lie on the plane `x+y+z+3sqrt(2)=0` |
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