AP EAMCET201922 Apr 2019Morning ShiftMathematicsVector AlgebraActual
For a non-zero real number (x ), if the points with position vectors ((x-u) i +x j +x k , x i +(x-v) j +x k ), (x i +x j +(x-w) k ) and ((x-1) i +(x-1) j +(x-1) k ) are coplanar, then
Options
- A(u+v+w=1 )
- B(u v w=1 )
- C( 1 u + 1 v + 1 w =1 )
- D(u v+v w+u w=1 )
Correct answer
C. ( 1 u + 1 v + 1 w =1 )
Step-by-step solution
Let ( aligned & O A =(x-u) i +x j +x k & O B =x i +(x-v) j +x k & O C =x i +x j +(x-w) k & O D =(x-1) i +(x-1) j +(x-1) k aligned ) and ( O D =(x-1) i +(x-1) j +(x-1) k ) Here, ( D A =( l -u) i + j + k ) ( D B = i +( l -v) j + k ) and ( D C = i + j +( l -w) k ) Since, points are collinear [DA DB DC] (=0 ) ( [ array ccc 1-u & 1 & 1 1 & 1-v & 1 1 & 1 & 1-w array ]=0 ) On solving this, we get ( 1 u + 1 v + 1 w =1 )