Find the area of the parallelogram whose diagonals are 2 i ^ - j ^ + k ^ and 3 i ^ + 4 j ^ - k ^ .

Given a → = 2 i ^ - j ^ + k ^ and b → = 3 i ^ + 4 j ^ - k ^ . Area of the parallelogram with diagonals a → and b → is given by 1 2 a → × b → The cross product of these two vectors is a → × b → = i ^ j ^ k ^ 2 - 1 1 3 4 - 1 , ⇒ a → × b → = 1 - 4 i ^ - - 2 - 3 j ^ + 8 + 3 k ^ , ⇒ a → × b → = - 3 i ^ + 5 j ^ + 11 k . ^ And the corresponding magnitude will be ⇒ a → × b → = - 3 2 + 5 2 + 11 2 , ⇒ a → × b → = 9 + 25 + 121 , ⇒ a → × b → = 155 . Then the area of the parallelogram with the given two diagonals will be 1 2 155 sq. units.

Find the area of the parallelogram whose diagonals are 2 i ^ - j ^ + k ^ and 3 i ^ + 4 j ^ - k ^ .

Given a → = 2 i ^ - j ^ + k ^ and b → = 3 i ^ + 4 j ^ - k ^ . Area of the parallelogram with diagonals a → and b → is given by 1 2 a → × b → The cross product of these two vectors is a → × b → = i ^ j ^ k ^ 2 - 1 1 3 4 - 1 , ⇒ a → × b → = 1 - 4 i ^ - - 2 - 3 j ^ + 8 + 3 k ^ , ⇒ a → × b → = - 3 i ^ + 5 j ^ + 11 k . ^ And the corresponding magnitude will be ⇒ a → × b → = - 3 2 + 5 2 + 11 2 , ⇒ a → × b → = 9 + 25 + 121 , ⇒ a → × b → = 155 . Then the area of the parallelogram with the given two diagonals will be 1 2 155 sq. units.

Find the area of the parallelogram whose diagonals are the vectors 2 m → - n → and 4 m → - 5 n → , where m → and n → are unit vectors forming an angle of 45 ° .

Given a → = 2 m → - n → and b → = 4 m → - 5 n → . Area of the parallelogram with diagonals p → and q → is given by 1 2 p → × q → The cross product of these two vectors is a → × b → = 2 m → - n → × 4 m → - 5 n → Applying distributive law, we get a → × b → = 2 m → × 4 m → - 2 m → × 5 n → - n → × 4 m → + n → × 5 n → We know, if r is a scalar quantity, then r a → × b → = a → × r b → = r a → × b → , so above equation becomes, a → × b → = 8 m → × m → - 10 m → × n → - 4 n → × m → + 5 n → × n → But we know a → × a → = 0 , so the above equation becomes, a → × b → = 8 0 - 10 m → × n → - 4 n → × m → + 5 0 ⇒ a → × b → = - 10 m → × n → - 4 n → × m → We also know, a → × b → = - b → × a → , so above equation becomes, a → × b → = 10 n → × m → - 4 n → × m → a → × b → = 6 n → × m → And the corresponding magnitude will be ⇒ a → × b → = 6 n → × m → ⇒ a → × b → = 6 n → × m → ⇒ a → × b → = 6 n → m → sin θ But it is given that the angle between the two vectors is 45 ° , so sin 45 ° = 1 2 , substituting this value in above equation, we get ⇒ a → × b → = 6 n → m → × 1 2 ⇒ a → × b → = 6 2 n → m → It is also given that m → and n → are unit vectors, so above equation becomes, ⇒ a → × b → = 6 2 Then the area of the parallelogram with the given two diagonals will be 1 2 × 6 2 = 3 2 sq. units.

Find the vector area of the triangle, the position vectors of whose vertices are i ^ + j ^ + 2 k ^ , 2 i ^ + 2 j ^ - 3 k ^ and 3 i ^ - j ^ - k ^ . Also, find its scalar area.

Given position vectors are A → = i ^ + j ^ + 2 k ^ , B → = 2 i ^ + 2 j ^ - 3 k , ^ C → = 3 i ^ - j ^ - k ^ . So the sides of the triangle will be A B , B C and A C , as shown in the figure. Area of triangle having adjacent sides a → and b → = 1 2 a → × b → . So to find the area, we just need any two adjacent sides. Now we will find the value of two adjacent sides A B → and A C → vectors. ∴ A B → = B → - A → , ⇒ A B → = 2 i ^ + 2 j ^ - 3 k ^ - i ^ + j ^ + 2 k ^ , ⇒ A B → = i ^ + j ^ - 5 k ^ . ∴ A C → = C → - A → , ⇒ A C → = 3 i ^ - j ^ - k ^ - i ^ + j ^ + 2 k ^ , ⇒ A C → = 2 i ^ - 2 j ^ - 3 k . ^ The cross product of these two vectors is A B → × A C → = i ^ j ^ k ^ 1 1 - 5 2 - 2 - 3 , ⇒ A B → × A C → = - 3 - 10 i ^ - - 3 + 10 j ^ + - 2 - 2 k ^ , ⇒ A B → × A C → = - 13 i ^ - 7 j ^ - 4 k ^ . And the corresponding magnitude will be ⇒ A B → × A C → = - 13 2 + - 7 2 + - 4 2 , ⇒ A B → × A C → = 169 + 49 + 16 , ⇒ A B → × A C → = 234 . So the vector area of the given triangle is 1 2 - 13 i ^ - 7 j ^ - 4 k ^ = 1 2 234 . And the scalar area of the triangle with these three vertices will be 234 2 sq. units.