 Hello and welcome to the session. In this session we are going to discuss the following question which says that Find the value of lambda from which the four points with position vectors minus 6 i cap minus 2 j cap plus 0 k cap 2 i cap plus 4 j cap plus 4 k cap 2 i cap minus 2 j cap plus 2 k cap lambda i cap minus 2 j cap minus 2 k cap are coplanar Let us proceed with the solution. Let position vector of a be given by minus 6 i cap minus 2 j cap plus 0 k cap position vector of b is given by 2 i cap plus 4 j cap plus 4 k cap position vector of c is given by 2 i cap minus 2 j cap plus 2 k cap and position vector of b is given by lambda i cap minus 2 j cap minus 2 k cap Now vector ab is given by position vector of b minus position vector of a position vector of b is given by 2 i cap plus 4 j cap plus 4 k cap and position vector of a is given by minus 6 i cap minus 2 j cap plus 0 k cap So we have 2 i cap plus 4 j cap plus 4 k cap minus 6 i cap minus 2 j cap plus 0 k cap which is equal to 2 i cap minus of minus 6 i cap that is 8 i cap plus 4 j cap minus of minus 2 j cap that is plus 6 j cap plus 4 k cap minus of 0 k cap that is plus 4 k cap So vector ab is given by 8 i cap plus 6 j cap plus 4 k cap vector ac is given by position vector of c minus position vector of a position vector of c is given by 2 i cap minus 2 j cap plus 2 k cap and position vector of a is given by minus 6 i cap minus 2 j cap plus 0 k cap So we have 2 i cap minus 2 j cap plus 2 k cap minus of minus 6 i cap minus 2 j cap plus 0 k cap which is equal to 2 i cap minus of minus 6 i cap that is 8 i cap minus 2 j cap minus of minus 2 j cap that is 0 j cap plus 2 k cap minus of 0 k cap that is plus 2 k cap So vector ac is given by 8 i cap plus 0 j cap plus 2 k cap vector ab is given by position vector of b minus position vector of a position vector of b is given by lambda i cap minus 2 j cap minus 2 k cap and position vector of a is given by minus 6 i cap minus 2 j cap plus 0 k cap So we have lambda i cap minus 2 j cap minus 2 k cap minus of minus 6 i cap minus 2 j cap plus 0 k cap which is equal to lambda i cap minus of minus 6 i cap that is lambda plus 6 i cap minus 2 j cap minus of plus 2 j cap that is plus 0 j cap minus 2 k cap minus of 0 k cap that is minus 2 k cap Therefore vector ab is given by lambda plus 6 i cap plus 0 j cap minus 2 k cap Now we have vector ab equal to 8 i cap plus 6 j cap plus 4 k cap vector ac is equal to 8 i cap plus 0 j cap plus 2 k cap and vector ad is equal to lambda plus 6 i cap plus 0 j cap minus 2 k cap and it is given that vector abcd are coplanar since vector abcd are coplanar that is vector ab ac ad are coplanar Therefore scalar triple product of vectors ab ac ad must be 0 So scalar triple product of vectors ab ac ad is given by the determinant of vectors ab ac and ad should be equal to 0 which implies that 8 into 0 into minus 2 that is 0 minus of 0 into 2 that is 0 minus 6 into 8 into minus 2 that is minus 16 minus of 2 into lambda plus 6 plus 4 into 8 into 0 that is 0 minus of 0 into lambda plus 6 that is 0 is equal to 0 which implies that 8 into 0 minus 0 that is 0 minus of 6 into minus 16 minus 2 into lambda that is minus 2 lambda minus 2 into 6 that is minus 12 plus 4 into 0 minus 0 that is 0 is equal to 0 which further gives minus 6 into minus 28 minus 2 lambda is equal to 0 which implies that minus 6 cannot be equal to 0 therefore minus 28 minus 2 lambda is equal to 0 that is minus 28 is equal to 2 lambda or we can write 2 lambda is equal to minus 28 which implies that lambda is given by minus 28 by 2 that is minus 14 therefore the value of lambda is equal to minus 14 which is the required answer this completes our session hope you enjoyed this session