 Hello and welcome to the session. In this session we are going to discuss the following question which says that show that the four points A with the coordinates 0, 8, 6, B with the coordinates minus 2, minus 10, minus 6, C with the coordinates minus 4, minus 4, 2, D with the coordinates 2, 2, minus 2 are co-planar. Let us proceed with the solution. Let position vector of A is given by 0 i cap plus 8 j cap plus 6 j cap. Position vector of B is given by minus 2 i cap minus 10 j cap minus 6 j cap. Position vector of C is given by minus 4 i cap minus 4 j cap plus 2 j cap. And position vector of B is given by 2 i cap plus 2 j cap minus 2 j cap. The given points A, B, C, D are co-planar if and only if any one of the following types of vectors are co-planar. That is vector AB, vector AC, vector AB or vector AB, vector BC, vector CD or vector BC, vector BA, vector BD, etc. Let us consider the co-planarity of vectors AB, AC and AB. Then points AB, CD will be co-planar if scalar triple product of vector AB, AC, AD is 0. Vector AB is given by position vector of B minus position vector of A. Position vector of B is given by minus 2 i cap minus 10 j cap minus 6 k cap. Position vector of A is given by 0 i cap plus 8 j cap plus 6 k cap. So we have minus 2 i cap minus 10 j cap minus 6 k cap minus of 0 i cap plus 8 j cap plus 6 k cap. Which is equal to minus 2 i cap minus 0 i cap that is minus 2 i cap minus 10 j cap minus 8 j cap is equal to minus 18 j cap minus 6 k cap minus 6 k cap is minus 12 k cap. Therefore vector AB is given by minus 2 i cap minus 18 j cap minus 12 k cap. Now vector AC is given by position vector of C minus position vector of A. Position vector of C is given by minus 4 i cap minus 4 j cap plus 2 k cap and position vector of A is given by 0 i cap plus 8 j cap plus 6 k cap. So we have minus 4 i cap minus 4 j cap plus 2 k cap minus of 0 i cap plus 8 j cap plus 6 k cap which is equal to minus 4 i cap minus 0 i cap that is minus 4 i cap minus 4 j cap minus 8 j cap that is minus 12 j cap plus 2 k cap minus 6 k cap that is minus 4 k cap. So vector AC is given by minus 4 i cap minus 12 j cap minus 4 k cap. Now vector AB is given by position vector of B minus position vector of A. Position vector of D is given by 2 i cap plus 2 j cap minus 2 k cap and position vector of A is given by 0 i cap plus 8 j cap plus 6 k cap. So we have 2 i cap plus 2 j cap minus 2 k cap minus of 0 i cap plus 8 j cap plus 6 k cap which gives 2 i cap minus 0 i cap that is 2 i cap plus 2 j cap minus 8 j cap that is minus 6 j cap minus 2 k cap minus 6 k cap is minus 8 k cap. Therefore vector AB is equal to 2 i cap minus 6 j cap minus 8 k cap. Now we have vector AB as minus 2 i cap minus 18 j cap minus 12 k cap. Vector AC as minus 4 i cap minus 12 j cap minus 4 k cap and vector AB as 2 i cap minus 6 j cap minus 8 k cap. Therefore scalar triple vector vector AB AC AD is given by the determinant of vector AB AC and AB. Scalar triple vector vector AB AC AD is given by minus 2 into minus 12 into minus 8 that is 96 minus of minus 6 into minus 4 that is 24 minus of minus 18 into minus 4 into minus 8 that is 32 minus of 2 into minus 4 that is minus 8 minus 12 into minus 8. So this is minus 2 minus 4 into minus 6 that is 24 minus of 2 into minus 12 that is minus 24 which is equal to minus 2 into 96 minus 24 that is 72 minus of minus 18 is plus 18 into 32 minus of minus 8 that is 32 plus 8 minus 12 into minus 12 minus of minus 24 that is 24 minus 8. Which is equal to minus 23 which is pulling 24 plus 24 this gives minus 2 into 72 is minus 144 plus 18 into 32 plus 8 is 40 minus 12 into 25 plus 24 is 48 which is equal to minus 144 plus 720 minus 576 to minus 720 plus 720 that is 0. So Taylor triple product of vector AB AC AD is equal to 0 and we know that point ABCD will be co-planar if Taylor triple product of vector AB AC AB is 0. Hence the four points AB, C and D are co-planar. This completes our session. Hope you enjoyed this session.