 Hello and welcome to the session. In this session we discussed the following question which says what is the cosine of the angle which the vector root 2i cap plus j cap plus k cap makes with the y axis. If suppose we have that a, b and c are the direction ratios of a line then the direction cosines of the line are given as plus minus a upon square root of a square plus b square plus c square plus minus b upon square root of a square plus b square plus c square and plus minus c upon square root of a square plus b square plus c square. Then the l1 and l2 and we take l1, l1, l1, l1, dv, direction cosines of the line l1 and l2, l2, l2, dv, direction cosines of the line l2 and this theta is the angle between the lines l1 and l2 then we have cos theta is equal to modulus of l1, l2 plus m1, n2 plus m1, n2. This is the key idea that we use for this question. Let's now move on to the solution. We have given a vector and y axis and we have to find the cosine of the angle which the vector makes with the y axis. For this we suppose let vector a be equal to 2i cap plus j cap plus k cap and let theta be the angle which the vector a makes with the y axis then we are supposed to find cos theta. Now from this vector a we find that the direction ratios of vector a are root 2, 1 and 1 which are the scalar components of vector a. Direction cosines of vector a be given by l1, n1, n1. So l1 would be equal to root 2 square which is 2 plus 1 square that is 1 plus 1 square which is again 1. Now as we know that the direction cosines are given as these. So for all the 3 direction cosines we will either take the plus sign or the minus sign. So here we are considering only the plus sign for all the 3 direction cosines. So n1 would be equal to 1 upon square root of 2 plus 1 plus 1 then n1 would be equal to 1 upon square root of 2 plus 1 plus 1. So we now have n1 is equal to root 2 upon square root of 2 plus 1 plus 1 which is square root of 4 and that is 2. So n1 is equal to root 2 upon 2, n1 would be equal to 1 upon 2 and n1 would be equal to 1 upon 2. So these are the direction cosines of the vector a. Now direction cosines the y axis be equal to l2, n2, n2 where we take l2 as 0, n2 as 1 and n2 as 0. So now that we have the direction cosines of both the vectors a and the y axis. So we can easily find out the cosines of the angle between the vector a and the y axis. We had assumed theta to be the angle which the vector a makes with the y axis and we know the cos theta is equal to modulus of l1, l2 plus n1, n2 plus n1, n2 where this theta is the angle between any two lines l1 and l2 and these are the direction ratios. So using this we can say that cos theta is equal to modulus of l1, l2 plus n1, n2 plus now substituting the values we get this is equal to modulus of root 2 upon 2 into 0 plus 1 upon 2 into 0 and so this is equal to modulus of 1 upon 2 that is 1 upon 2. So we have cos theta equal to 1 upon the cosine of the angle which the vector i cap plus j cap plus k cap with the axis is 1 upon 2 and we will answer. This completes the session but we have understood the solution of this question.