 Hi and welcome to the session. Let's work out the following question. The question says, find the vector and Cartesian equations of the sphere described on the joint of the points 2 minus 3 4 and minus 5 6 minus 7 as the extremity is of the diameter. Let us start with a solution to this question. We know that the vector equation of a sphere with vector A and vector B as the position vectors of the extremities of a diameter is vector R minus vector A multiplied by vector R minus vector B is equal to 0. Here vector A will be 2i cap minus 3j cap plus 4k cap. This we get from this point and vector B is equal to minus 5i cap plus 6j cap minus 7k cap and this we get from this one. So the vector equation of the required sphere is vector R minus 2i cap minus 3j cap plus 4k cap multiplied by vector R minus minus 5i cap minus 6j cap plus 7k cap is equal to 0. Here we have minus 5i cap plus 6j cap minus 7k cap. Now let us see the reduction to Cartesian form taking vector R to be equal to xi cap plus yj cap plus zk cap. The above equation becomes xi cap plus yj cap plus zk cap minus 2i cap minus 3j cap plus 4k cap multiplied by xi cap plus yj cap plus zk cap minus 5i cap plus 6j cap minus 7k cap is equal to 0. Now this implies x minus 2 into i cap plus y plus 3j cap plus z minus 4k cap multiplied by x plus 5i cap plus y minus 6j cap plus z plus 7 to k cap is equal to 0. This implies x minus 2 into x plus 5 that is this into this plus this into this y plus 3 into y minus 6 plus z minus 4 into z plus 7 is equal to 0. This implies x square plus 5x minus 2x minus 10 plus y square minus 6y plus 3y minus 18 plus z square plus 7z minus 4z minus 28 is equal to 0. This implies x square plus y square plus z square. Now this is 3x so plus 3x this is minus 3y this is 3z minus now this gives us 56 is equal to 0 So this is our answer to this question. I hope that you understood the solution and enjoyed the session. Have a good day.