Find a unit vector in the direction of the resultant of the vectors i ^ - j ^ + 3 k ^ , 2 i ^ + j ^ - 2 k ^ and i ^ + 2 j ^ - 2 k ^ .

Given: a → = i ^ - j ^ + 3 k ^ , b → = 2 i ^ + j ^ - 2 k ^ and c → = i ^ + 2 j ^ - 2 k ^ are the position vectors. Then resultant of the vectors = a → + b → + c → = i ^ - j ^ + 3 k ^ + 2 i ^ + j ^ - 2 k ^ + i ^ + 2 j ^ - 2 k ^ = 4 i ^ + 2 j ^ - k ^ So, a → + b → + c → = ( 4 ) 2 + ( 2 ) 2 + ( - 1 ) 2 = 21 Hence, Unit vector in the direction of the resultant vector = a → + b → + c → a → + b → + c → = 1 21 4 i ^ + 2 j ^ - k ^ .

The adjacent sides of a parallelogram are represented by the vectors a → = i ^ + j ^ - k ^ and b → = - 2 i ^ + j ^ + 2 k ^ . Find unit vectors parallel to the diagonals of the parallelogram.

Given a → = i ^ + j ^ - k ^ and b → = - 2 i ^ + j ^ + 2 k ^ . We have to find unit vectors parallel to the diagonals of the parallelogram. A C → = a → + b → = - i ^ + 2 j ^ + k ^ Let the unit vector along the diagonals A C and B D of the parallelogram be A C → and B D → ⇒ A C → = - i ^ + 2 j ^ + k ^ 6 Similarly, B D → = b → - a → = - 3 i ^ + 3 k ^ ⇒ B D → = - 3 i ^ + 3 k ^ 3 2 = - i ^ + k 2 Hence, the unit vectors parallel to the diagonals of the parallelogram are - i ^ + 2 j ^ + k ^ 6 and - i ^ + k 2 .

Prove that the points i ^ - j ^ , 4 i ^ + 3 j ^ + k ^ and 2 i ^ - 4 j ^ + 5 k ^ are the vertices of a right-angled triangle.

Given the points i ^ - j ^ , 4 i ^ + 3 j ^ + k ^ and 2 i ^ - 4 j ^ + 5 k ^ are A , B and C respectively. Then, A B → = 4 i ^ + 3 j ^ + k ^ - i ^ + j ^ = 3 i ^ + 4 j ^ + k ^ B C → = 2 i ^ - 4 j ^ + 5 k ^ - 4 i ^ - 3 j ^ - k ^ = - 2 i ^ - 7 j ^ + 4 k ^ C A → = i ^ - j ^ - 2 i ^ + 4 j ^ - 5 k ^ = - i ^ + 3 j ^ - 5 k ^ A B → + B C → + C A → = 3 i ^ + 4 j ^ + k ^ - 2 i ^ - 7 j ^ + 4 k ^ - i ^ + 3 j ^ - 5 k ^ = 0 → The given points forms a vertices of a triangle. Now, A B → = 9 + 16 + 1 = 26 B C → = 4 + 49 + 16 = 69 C A → = 1 + 9 + 25 = 35 A B → 2 + C A → 2 = 26 + 35 = 61 ≠ B C → 2 The given triangle is not right-angled.

If the vertices A , B , C of a triangle A B C are the points with position vectors a 1 i ^ + a 2 j ^ + a 3 k ^ , b 1 i ^ + b 2 j ^ + b 3 k ^ , c 1 i ^ + c 2 j ^ + c 3 k ^ respectively, what are the vectors determined by its sides? Find the length of these vectors.

Given the vertices of a triangle A , B and C with position vectors a 1 i ^ + a 2 j ^ + a 3 k ^ , b 1 i ^ + b 2 j ^ + b 3 k ^ and c 1 i ^ + c 2 j ^ + c 3 k ^ , respectively. Then, A B → = ( b 1 - a 1 ) i ^ + ( b 2 - a 2 ) j ^ + ( b 3 - a 3 ) k ^ B C → = ( c 1 - b 1 ) i ^ + ( c 2 - b 2 ) j ^ + ( b 3 - a 3 ) k ^ C A → = ( a 1 - c 1 ) i ^ + ( a 2 - c 2 ) j ^ + ( a 3 - c 3 ) k ^ Therefore, the length of these vectors are: A B → = ( b 1 - a 1 ) 2 + ( b 2 - a 2 ) 2 + ( b 3 - a 3 ) 2 B C → = ( c 1 - b 1 ) 2 + ( c 2 - b 2 ) 2 + ( c 3 - b 3 ) 2 C A → = ( a 1 - c 1 ) 2 + ( a 2 - c 2 ) 2 + ( a 3 - c 3 ) 2 .