 Hi friends, I am Purva and today we will discuss the following question. If vector A, vector B and vector C are mutually perpendicular vectors of equal magnitudes, then show that the vector, vector A plus vector B plus vector C is equally inclined to vector A, vector B and vector C. Let us now begin with the solution. Now since vector A, vector B and vector C are mutually perpendicular, therefore we have vector A dot vector B is equal to vector B dot vector C is equal to vector C dot vector A which is equal to 0. We are also given that vector A, vector B and vector C have equal magnitudes. Therefore we have mod of vector A is equal to mod of vector B which is equal to mod of vector C. Now vector A plus vector B plus vector C will be equally inclined to vector A, vector B and vector C if the angle between them is same. So let alpha be the angle between vector A plus vector B plus vector C and vector A. Then we have cos alpha is equal to vector A plus vector B plus vector C dot vector A upon mod of vector A plus vector B plus vector C into mod of vector A. Now this is equal to vector A dot vector A plus vector B dot vector A plus vector C dot vector A upon mod of vector A plus vector B plus vector C into mod of vector a. This is equal to now vector a dot vector a is mod of vector a square plus now since we know that vector b dot vector a is equal to vector c dot vector a which is equal to 0. So here we get plus 0 upon mod of vector a plus vector b plus vector c into mod of vector a. This is further equal to mod of vector a square upon vector a plus vector b plus vector c into mod of vector a. So we get cos alpha is equal to mod of vector a upon vector a plus vector b plus vector c. Now let beta be the angle between vector a plus vector b plus vector c and vector b therefore we get cos beta is equal to vector a plus vector b plus vector c dot vector b upon mod of vector a plus vector b plus vector c into mod of vector b. Now this is equal to vector a dot vector b plus vector b dot vector b which is equal to mod of vector b square plus vector c dot vector b upon mod of vector a plus vector b plus vector c into mod of vector b. Now since vector a dot vector b is equal to vector c dot vector b which is equal to 0 therefore here we get this is equal to mod of vector b square upon mod of vector a plus vector b plus vector c into mod of vector b and we get cos beta is equal to vector b upon vector a plus vector b plus vector c. Now since mod of vector b is equal to mod of vector a this is given to us so we get here this implies cos beta is equal to mod of vector a upon vector a plus vector b plus vector c. Now finally let gamma be the angle between vector a plus vector b plus vector c and vector c therefore we get cos gamma is equal to vector a plus vector b plus vector c dot vector c upon mod of vector a plus vector b plus vector c into mod of vector c. Now this is equal to vector a dot vector c plus vector b dot vector c plus vector c dot vector c which is equal to mod of vector c square upon mod of vector a plus vector b plus vector c into mod of vector c. Now Now since vector a dot vector c is equal to vector b dot vector c which is equal to 0 therefore we get cos gamma is equal to mod of vector c square upon mod of vector a plus vector b plus vector c into mod of vector c and this is equal to mod of vector c upon mod of vector a plus vector b plus vector c. Now since mod of vector c is equal to mod of vector a therefore we get cos gamma is equal to mod of vector a upon mod of vector a plus vector b plus vector c. Now here let us mark this as equation one this as equation two and this as equation three. So from equation one, two and three we get right hand side are equal so left hand side becomes equal so we get cos alpha is equal to cos beta is equal to cos gamma and this implies alpha is equal to beta is equal to gamma. Hence we get vector a plus vector b plus vector c is equally inclined to vector a vector b and vector c. This is our answer. Hope you have understood the solution. Bye and take care.