 Hello and welcome to the session. In this session we are going to discuss the following question and the question says that if A with coordinates 2, 1, 4, B with coordinates minus 3, 2, 4, C with coordinates 1, 2, minus 2 are the vertices of the triangle A, B, C. Find the area of triangle A, B, C. We know that the area of triangle A, B, C is given by the formula 1 by 2 into A, B into A, C into sin of angle A and sin of angle A is given by square root 1 minus cos square A and cos A is the angle between A, B and A, C and 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. Direction cosines are given by 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. The science 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 PQ 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 the coordinates of point AR 2, 1, 4, the coordinates of point BR minus 3, 2, 4, the coordinates of point CR 1, 2, minus 2 and we need to find the area of triangle ABC. Now from the key idea we know that area of triangle ABC is given by 1 by 2 into AB into AC into sine of angle A. So now we shall find AB and AC that is the distance between A and B and the distance between A and C. Using the distance formula which states that distance between any two points P and Q having coordinates P1, P2, P3 and Q1, Q2, Q3 respectively is given by square root P1 minus Q1 the whole square plus P2 minus Q2 the whole square plus P3 minus Q3 the whole square. Therefore distance between A and B given by AB is equal to square root 2 minus of minus 3 the whole square plus 1 minus 2 the whole square plus 4 minus 4 the whole square which is equal to square root 5 square plus minus 1 whole square plus 0 square which is equal to square root of 25 plus 1 plus 0 which gives the value of AB is equal to square root of 26. Similarly distance between and C given by AC is equal to square root of 2 minus 1 the whole square plus 1 minus 2 the whole square plus 4 minus of minus 2 the whole square which is equal to square root 1 square plus minus 1 square plus 6 square which is further equal to square root of 1 plus 1 plus 36 which gives the value of AC is equal to square root of 38. Next we shall find sine A which is equal to square root of 1 minus cos square A and cos A is the angle between AB and AC. To find cos theta we shall first of all calculate direction ratios of line AB and line AC. From the key idea we have direction ratios of line AB are given by x2 minus x1 that is minus 3 minus 2 y2 minus y1 that is 2 minus 1 z2 minus z1 that is 4 minus 4 which is equal to minus 5 1 0. Similarly direction ratios of line AC are given by x2 minus x1 that is 1 minus 2 y2 minus y1 that is 2 minus 1 z2 minus z1 that is minus 2 minus 4 which is equal to minus 1 1 minus 6. Next we shall calculate direction cos of line AB and AC. Using the key idea we get direction cos A B having direction ratios minus 5 1 0 that is A is equal to minus 5 B is equal to 1 and C is equal to 0 are given by L1 is equal to A upon square root of A square plus B square plus C square that is minus 5 upon square root of minus 5 square plus 1 square plus 0 square which is equal to minus 5 upon square root of 26. M1 is equal to B upon square root of A square plus B square plus C square that is 1 upon square root of minus 5 square plus 1 square plus 0 square which is equal to 1 upon square root of 26. M1 is equal to C upon square root of A square plus B square plus C square that is 0 upon square root of minus 5 square plus 1 square plus 0 square which is equal to 0. Similarly we shall find direction cos of line AC having direction ratios minus 1 1 minus 6 that A is equal to minus 1 B is equal to 1 and C is equal to minus 6 are given by L2 is equal to A upon square root of A square plus B square plus C square that is minus 1 upon square root of minus 1 square plus 1 square plus minus 6 whole square which is equal to minus 1 upon square root of 38. M2 is equal to B upon square root of A square plus B square plus C square that is 1 upon square root of minus 1 square plus 1 square plus of minus 6 whole square which is equal to 1 upon square root of 38 and M2 is equal to C upon square root of A square plus B square plus C square that is minus 6 upon square root of minus 1 square plus 1 square plus of minus 6 whole square which is equal to minus 6 upon square root of 38. Therefore angle between AB and AC is given by cos A which is equal to L1 L2 plus M1 M2 plus N1 N2 where L1 M1 N1 and L2 M2 N2 are the direction cosines of line AB and AC respectively or cos A can be written as L1 L2 that is minus 5 upon square root of 26 into minus 1 upon square root of 38 plus M1 M2 that is 1 upon square root of 26 into 1 upon square root of 38 plus N1 N2 that is 0 into minus 6 upon square root of 38 or cos A can be further written as 1 upon square root of 26 into square root of 38 whole multiplied by 5 plus 1 plus 0 in the brackets which gives the value of cos A is equal to 6 upon square root of 988 from the key idea we know that sin A is equal to square root 1 minus cos A so we have sin A is equal to square root 1 minus cos A that is 6 upon square root of 988 the whole square which is equal to square root of 1 minus 36 upon 988 taking the LCM in the square root we get 988 minus 36 upon 988 which can be written as 952 upon 988 in the square root now we have got the values of AB AC and sin of angle A so now we can find the area of triangle ABC using the key idea which states that area is equal to 1 by 2 into AB into AC into sin of angle A therefore area of triangle ABC is equal to 1 by 2 into AB into AC into sin of angle A which is equal to 1 by 2 into AB that is square root of 26 into AC that is square root of 38 into sin of angle A that is square root of 952 upon 988 square units on solving further we get 1 by 2 into square root of 26 into square root of 38 into square root of 952 upon square root of 988 which is further equal to 1 by 2 into square root of 26 into square root of 38 into square root of 952 can be written as 2 into square root of 238 and square root of 988 can be written as square root of 26 into square root of 38 square units on further simplification we get square root of 238 square units hence area of triangle ABC is given by square root of 238 square units which is our final answer this completes our session hope you enjoyed this session