 Hello and welcome to the session. In this session we discussed the following question which says find the angle between the vectors vector a equal to i cap minus j cap plus k cap and vector b equal to i cap plus j cap minus k cap. Let's move on to the solution. We are given vector a equal to i cap minus j cap plus k cap and vector b equal to i cap plus j cap minus k cap. Now we take let theta be the angle between vector a and vector b. So we have cos theta is equal to vector a dot vector b upon magnitude of vector a into magnitude of vector b. First let's find out vector a dot vector b. This is equal to i cap minus j cap plus k cap dot i cap plus j cap minus k cap. So this is equal to 1 minus 1 minus 1 which comes out to be equal to minus 1. So we have vector a dot vector b is equal to minus 1. Now magnitude of vector a is equal to square root of 1 square plus minus 1 square plus 1 square. So this is equal to square root of 1 plus 1 plus 1 that is equal to square root 3. So magnitude of vector a is square root 3. The next we have magnitude of vector b is equal to square root of 1 square plus 1 square plus minus 1 square. So this is equal to square root of 1 plus 1 plus 1 that is equal to square root 3. Now we put the values for a dot b magnitude of a and magnitude of vector b in this. So we get cos theta is equal to vector a that is equal to minus 1 upon magnitude of vector a into magnitude of vector b that is root 3 into root 3 and this is equal to minus 1 upon 3. That is we get theta is equal to cos inverse minus 1 upon 3. So our final answer is the angle between the vectors a and b is cos inverse minus 1 upon 3. So this completes the session. Hope you have understood the solution for this question.