Find the vector area and area of the parallelogram having a → = i ^ + 2 j ^ − k ^ and b → = 2 i ^ - j ^ + 2 k ^ .

Given a → = i ^ + 2 j ^ − k ^ and b → = 2 i ^ - j ^ + 2 k ^ Now, a → × b → = i ^ j ^ k ^ 1 2 − 1 2 − 1 2 = i ^ 4 - 1 - j ^ 2 + 2 + k ^ - 1 - 4 Vector area = 3 i ^ - 4 j ^ - 5 k ^ Area of the parallelogram = a → × b → = 9 + 16 + 25 = 50 = 5 2 Sq units.

Find the vector area and area of the parallelogram having a → = i ^ + 2 j ^ − k ^ and b → = 2 i ^ - j ^ + 2 k ^ .

Given a → = i ^ + 2 j ^ − k ^ and b → = 2 i ^ - j ^ + 2 k ^ Now, a → × b → = i ^ j ^ k ^ 1 2 − 1 2 − 1 2 = i ^ 4 - 1 - j ^ 2 + 2 + k ^ - 1 - 4 Vector area = 3 i ^ - 4 j ^ - 5 k ^ Area of the parallelogram = a → × b → = 9 + 16 + 25 = 50 = 5 2 Sq units.

If a → × b → = b → × c → ≠ 0 → , then show that a → + c → = p b → , where p is some scalar.

Given a → × b → = b → × c → ⇒ a → × b → = - c → × b → ⇒ a → × b → + c → × b → = 0 ⇒ a → + c → × b → = 0 It means a → + c → and b → are parallel vectors. So they lie on the same plane. So a vector can be represented in a scalar multiple of other. a → + c → = p b → , where p is some scalar Hence, proved.

Find the vector having magnitude 6 units and perpendicular to both 2 i ^ − k ^ and 3 j ^ − i ^ − k ^ .

Let a → = 2 i ^ − k ^ and b → = 3 j ^ − i ^ − k ^ Now, a → × b → = i ^ j ^ k ^ 2 0 − 1 - 1 3 - 1 = i ^ 0 + 3 - j ^ - 2 - 1 + k ^ 6 - 0 = 3 i ^ + 3 j ^ + 6 k ^ = 3 i ^ + j ^ + 2 k ^ a → × b → = 3 1 + 1 + 4 = 3 6 A vector of magnitude 6 and perpendicular to both a → and b → = ± 6 a → × b → a → × b → = ± 3 6 i ^ + j ^ + 2 k ^ 3 6 = ± i ^ + j ^ + 2 k ^ .

If a → · b → = a → ⋅ c → and a → × b → = a → × c → , a → ≠ 0 , then show that b → = c → .

Given a → · b → = a → ⋅ c → and a → × b → = a → × c → , a → ≠ 0 a → · b → = a → ⋅ c → ⇒ a → · b → - a → ⋅ c → = 0 ⇒ a → · b → - c → = 0 ⇒ a → b → - c → cos θ = 0 , θ is the angle between a → , b → - c → a → × b → = a → × c → ⇒ a → × b → - a → × c → = 0 ⇒ a → × b → - c → = 0 . . . . . . . . . 2 ⇒ a → b → - c → sin θ = 0 , θ is the angle between a → , b → - c → From 1 and 2 it is clear that, the vector b → - c → cannot be both perpendicular to a → and parallel to a → ⇒ b → - c → = 0 ⇒ b → = c → Hence, proved.

Find a vector of magnitude 3 units and perpendicular to both b → = 2 i ^ − 2 j ^ + k ^ and c → = 2 i ^ + 2 j ^ + 3 k ^ .

Given b → = 2 i ^ − 2 j ^ + k ^ and c → = 2 i ^ + 2 j ^ + 3 k ^ Now, b → × c → = i ^ j ^ k ^ 2 - 2 1 2 2 3 = i ^ - 6 - 2 - j ^ 6 - 2 + k ^ 4 + 4 = - 8 i ^ - 4 j ^ + 8 k ^ = 4 - 2 i ^ - j ^ + 2 k ^ b → × c → = 4 4 + 1 + 4 = 4 × 3 = 12 A vector of magnitude 3 and perpendicular to both b → and c → = ± 3 b → × c → b → × c → = ± 3 × 4 - 2 i ^ - j ^ + 2 k ^ 12 = ± - 2 i ^ - j ^ + 2 k ^ .