 Hello and welcome to the session. In this session we are going to discuss the following question which says that find the area of the triangle whose vertices are a with the coordinates 0, 7, 10, b with the coordinates minus 1, 6, 6 and c with the coordinates minus 4, 9, 6. Area of any triangle, say a, b, c is given by 1 by 2 into modulus of cross product of vector a, b and a, c. With this key idea let us proceed with the solution. We are given point a with the coordinates 0, 7, 10, b with the coordinates minus 1, 6, 6, c with the coordinates minus 4, 9, 6. Then position vector of a is given by 0 i cap plus 7 j cap plus 10 k cap. Position vector of b is given by minus 1 i cap plus 6 j cap plus 6 k cap and position vector of c is given by minus 4 i cap plus 9 j cap plus 6 k cap. Then vector a, b is equal to position vector of b minus position vector of a. Position vector of b is given by minus 1 i cap plus 6 j cap plus 6 k cap and position vector of a is given by 0 i cap plus 7 j cap plus 10 k cap. Vector a, b is equal to minus 1 i cap plus 6 j cap plus 6 k cap minus 6 0 i cap plus 7 j cap plus 10 k cap which is equal to minus i cap minus j cap minus 4 k cap. So vector a, b is given by minus i cap minus j cap minus 4 k cap. Similarly vector a, c is given by position vector of c minus position vector of a. Position vector of c is given by minus 4 i cap plus 9 j cap plus 6 k cap and position vector of a is given by 0 i cap plus 7 j cap plus 10 k cap. Vector a, c is given by minus 4 i cap plus 9 j cap plus 6 k cap minus 0 i cap plus 7 j cap plus 10 k cap which is equal to minus 4 i cap plus 2 j cap minus 4 k cap. So vector a, c is given by minus 4 i cap plus 2 j cap minus 4 k cap. Now vector a, b is given by minus i cap minus j cap minus 4 k cap and vector a, c is given by minus 4 i cap plus 2 j cap minus 4 k cap. Now we shall find cross to the top vector a, b and vector a, c which is equal to i cap into minus 1 into minus 4 that is 4 minus f 2 into minus 4 that is minus 8 minus j cap into minus 1 into minus 4 that is 4 minus f minus 4 into minus 4 that is 16 plus k cap into 2 into to minus 1 that is minus 2 minus f minus 4 into minus 1 that is 4 which is equal to i cap into 4 plus 8 minus j cap into minus 12 plus k cap into minus 6 which is equal to 12 i cap plus 12 j cap minus 6 k cap and we know that area of any triangle say a b c is given by 1 by 2 into modulus of cross product of vector a b and vector a c therefore area of triangle a b c is equal to 1 by 2 into modulus of cross product of vector a b and a c which is equal to 1 by 2 into modulus of cross product of vector a b and a c is given by 12 i cap plus 12 j cap minus 6 k cap so we have 12 i cap plus 12 j cap minus 6 k cap which is equal to 1 by 2 into square root of 12 square plus 12 square plus of minus 6 b whole square that is equal to 1 by 2 into square root of 144 plus 144 plus 36 which is equal to 1 by 2 into square root of 324 which is equal to 1 by 2 into 18 that is equal to 9 therefore area of triangle a b c is equal to 9 square units which is the required answer this completes our session hope you enjoyed this session.