 Hello and welcome to the session. In this session, we are going to discuss the following question and the question is, if A with coordinates 1, minus 2, 3, B with coordinates 3, 7, 5, C with coordinates minus 4, minus 5, minus 6 and B with coordinates 2, minus 1, 4 are four given points, then find the angle between A, B and C, D. We know that angle between any two lines is given by cos theta equal to L1, L2 plus M1, M2 plus N1, N2, where L1, M1 and L2, M2, N2 are the direction cosines of the two lines. The direction cosines are given by the formula L is equal to plus minus A upon square root of A square plus B square plus C square, M is equal to plus minus B upon square root of A square plus B square plus C square, N is equal to plus minus C upon square root of A square plus B square plus C square. Designs of direction cosines should be taken either all positive or all negative. Also, A, B and C are called direction ratios. Direction ratios of any line say P, Q are given by x2 minus x1, y2 minus y1, z2 minus z1, where x1, y1, z1 and x2, y2, z2 are the coordinates of point P and Q respectively. With this key idea, let us proceed with the solution. Here we are given the points A with coordinates 1, minus 2, 3, B with coordinates 3, 7, 5, C with coordinates minus 4, minus 5, minus 6, D with coordinates 2, minus 1, 4. Now first we shall find out the direction ratios of line A, B and C, D as we need to find the angle between line A, B and line C, D. Using the key idea, we know that the direction ratios of any line say P, Q are given by x2 minus x1, y2 minus y1, z2 minus z1, where x1, y1, z1, z2, minus z1 and x2, y2, z2 are the coordinates of point P and Q. Therefore, direction ratios of line A, B are given by x2 minus x1, that is 3 minus 1, y2 minus y1, that is 7 minus of minus 2, z2 minus z1, that is 5 minus 3, which can be written as 2, 9, 2. Similarly, direction ratios of line C, D are given by x2 minus x1, x2 minus x1, x2 minus x1, x2 minus x1, that is 2 minus of minus 4, y2 minus y1, that is minus 1, minus of minus 5, z2 minus z1, that is 4 minus of minus 6, which can be written as 6, 4, 10. Now we shall find direction cosines of line A, B and C, D using the key idea, which gives the formula for direction cosines as L is equal to plus minus A upon square root of A square plus B square plus C square, M is equal to plus minus B upon square root of A square plus B square plus C square and N is equal to plus minus C upon square root of A square plus B square plus C square, where A, B and C are the direction ratios of that line. Therefore, direction cosines of line A, B having direction ratios as 2, 9, 2, that is 2 minus C upon square root of A is equal to 2, B is equal to 9 and C is equal to 2 are given by L is equal to A, that is 2 upon square root of A square plus B square plus C square, that is 2 square plus 9 square plus 2 square, which is equal to 2 upon square root of 89, M is equal to B that is 9 upon square root of A square plus B square plus C square, that is 2 square plus 9 square plus 2 square, which is equal to 9 upon square root of 89, N is equal to C upon that is 2 upon square root of A square plus B square plus C square, that is 2 square plus 9 square plus 2 square, which is equal to 2 upon square root of 89. Let us denote the direction cosines of line A, B by L1, M1, L2, L3, L4, L4, L5, L5, L5, L5, L5, L5, cosines of line A, B by L1, M1 and N1 and we should also note that the signs of L1, M1 and N1 are taken all positive. Similarly, we shall find direction cosines of line C D having direction ratios 4, 10 that is A is equal to 6, B is equal to 4 and C is equal to 10 are given by L2 is equal to A upon that is 6 upon square root of A square plus B square plus C square that is 6 square plus 4 square plus 10 square, which is equal to 6 upon square root of 152, M2 is equal to B upon that is 4 upon square root of 6 square plus 4 square plus 10 square, which is equal to 4 upon square root of 152, N2 is given by C upon that is 10 upon square root of A square plus B square plus C square that is 6 square plus 4 square plus 10 square which is equal to 10 upon square root of 152. Here also we have taken all positive signs for L2, M2 and N2. Now we shall find the angle between line AB and C D using the key idea which states that angle between any two lines is given by cos theta is equal to L1, L2 plus M1, M2 plus N1, N2 where L1, M1, N1 and L2, M2, N2 are the direction cosines of the two lines. Therefore we can write angle between line AB and line C D is given by cos theta equal to L1, L2 plus M1, M2 where L1, M1, N1 plus N1, N2. Or we can write cos theta equal to L1, L2 that is 2 upon square root of 89 into 6 upon square root of 152 plus M1, M2 that is 9 upon square root of 89 into 4 upon square root of 152 plus N1, N2 that is 2 upon square root of 89 into 10 upon square root of 152 or we can write cos theta is equal to 1 upon square root of 89 into square root of 152 into 2 into 6 plus 9 into 4 plus 2 into 10 in the brackets or we can write cos theta is equal to 1 upon square root of 89 into 2 into square root of 38 into 2 into 2 into 12 plus 36 plus 20 in the brackets. Next we get cos theta is equal to 1 upon 2 into square root of 89 into 38 whole multiplied by 68 which is equal to cos theta is equal to 34 upon square root of 3382 which gives cos theta is equal to 34 upon 58.15 which is equal to cos theta is equal to 0.5847. Now using the cosine table we get the value of theta as 54 degrees 13 minutes approximately. Hence the angle of the cosine table is equal to 0.5847 between line AB and line CD is given by 54 degrees 13 minutes approximately which is our final answer. This completes our session. Hope you enjoyed this session.