 Hi and welcome to the session, let's work out the following question. The question says if vector A, vector B and vector C are three mutually perpendicular vectors of equal magnitude, find the angle between vector A and vector A plus vector B plus vector C. So let us see the solution to this question. Let mod of vector A be equal to mod of vector B be equal to mod of vector C be equal to some quantity say lambda, since vector A, vector B and vector C are mutually perpendicular vectors. A dot vector B is equal to vector B dot vector C is equal to vector C dot vector A is equal to zero and this we call equation one. Now mod of vector A plus vector B plus vector C the whole square is equal to vector A plus vector B plus vector C into vector A plus vector B plus vector C which is equal to vector A dot vector A plus vector B dot vector B plus vector C dot vector C plus twice of vector A dot vector B plus twice of vector B dot vector C plus twice of vector C dot vector A and this we call two. Now from one and two we can say that mod of vector A plus vector B plus vector C the whole square is equal to mod of vector A square plus mod of vector B square plus mod of vector C square because this, this and this they become zero and this is equal to three times lambda square and this further implies that mod of vector A plus vector B plus vector C is equal to square root three lambda. Now suppose vector A plus vector B plus vector C they make angle theta with vector A so we will have cos theta will be equal to vector A dot vector A plus vector B plus vector C divided by mod of vector A into mod of vector A plus vector B plus vector C which is further equal to vector A dot vector A plus vector A dot vector B plus vector A dot vector C divided by mod of vector A into square root three lambda because this is equal to square root three lambda this we have just found out this is equal to mod of vector A square plus this is zero this is zero divided by mod of vector A plus square root three lambda sorry this is mod of vector A into square root three lambda this becomes further equal to mod of vector A divided by square root three lambda now mod of vector A is lambda so we have lambda divided by square root three lambda which is further equal to one upon root three so we have cos theta is equal to one upon square root three and this implies that theta is equal to cos inverse one upon root three which is also our answer to this question I hope that you understood the solution and enjoyed the session have a good day