 Hello and welcome to the session. Let us discuss the following question. Question says, find a unit vector perpendicular to each of the vector A vector plus B vector and A vector minus B vector where A vector is equal to 3i plus 2j plus 2k and B vector is equal to i plus 2j minus 2k. Let us now start with the solution. Now we know A vector is equal to 3i plus 2j plus 2k and B vector is equal to i plus 2j minus 2k. Now first of all let us find out A vector plus B vector. Now A vector plus B vector is equal to 4i plus 4j adding corresponding elements of A vector and B vector. We get 4i plus 4j. Now we will find out difference of A vector and B vector that is A vector minus B vector it is equal to 2i plus 4k subtracting the corresponding elements of vector A and vector B we get 2i plus 4k is equal to vector A minus vector B. Now we have to find the unit vector perpendicular to both of these vectors. Unit vector perpendicular to A vector plus B vector and A vector minus B vector is equal to A vector plus B vector cross A vector minus B vector upon magnitude of A vector plus B vector cross A vector minus B vector first of all we will find out cross product of A vector plus B vector and A vector minus B vector it is equal to determinant of unit vectors i j and k. Here we will write 4 here we will write 4 and here we will write 0. Now here we will write 2 here we will write 0 and here we will write 4. Let us understand how we write this determinant to find the cross product first of all we will write unit vectors i j k in a single row. Now we will write coefficient of vector i in a single column or we can say we will write coefficient of vector i below vector i only then we will write coefficient of vector j below vector j only and coefficient of vector k below vector k only. Now we will find out this determinant this determinant is equal to i vector multiplied by 16 minus 0 minus j vector multiplied by 16 minus 0 plus k vector multiplied by 0 minus 8 expanding this determinant with respect to first row we get this expansion. Now this is further equal to 16i minus 16j minus 8k. Now we will find out magnitude of this vector so we can write magnitude of A vector plus B vector cross A vector minus B vector is equal to square root of 16 square plus minus 16 square plus minus 8 square. Now this is further equal to square root of 256 plus 256 plus 64 below square of 16 is 256 and square of 8 is 64 and negative sign raised to even power is equal to positive number only so here we get 256 and here also we get 64. Now adding these three terms we get square root of 576. Now this is further equal to plus minus 24. Now using this expression we get the required unit vector is equal to plus minus 16i minus 16j minus 8k upon 24 we know this is the cross product of the two vectors and this is the magnitude of the cross product substituting the value of cross product and magnitude of the cross product in this expression we get the required unit vector. Now this unit vector is further equal to plus minus 16 upon 24 i minus plus 16 upon 24 j minus plus 8 upon 24k. Now we will cancel common factor 8 in each of the term and we get required unit vector is equal to plus minus 2 upon 3 i minus plus 2 upon 3 j minus plus 1 upon 3 k. So this is our required unit vector this completes the session hope you understood the solution take care and have a nice day.