If A → + B → + C → = 0 , show that A × B → = B → × C →

Given, A → + B → + C → = 0 Pre - Multiplying B → on both sides. B → × A → + B → + C → = B → × 0 ⇒ B → × A → + B → × B → + B → × C → = 0 ⇒ - A → × B → + 0 + B → × C → = 0 ⇒ B → × C → = A → × B → Hence, proved.

The magnitude of the resultant of a → and b → is equal to that of a → and is perpendicular to a → . Show that | b → | = 2 · a .

Let R be the resultant of two vectors a → and b → R = a 2 + b 2 + 2 a b cos θ According to the question, R = a Also if R is perpendicular to the vector a → a . ( a + b ) = 0 a . a + a b cos θ = 0 a 2 + a b cos θ = 0 cos θ = - a / b Therefore, resultant equation becomes a = a 2 + b 2 + 2 a b ( - a / b ) Squaring both sides a 2 = a 2 + b 2 - 2 a 2 b 2 = 2 a 2 b = 2 a 2 Hence proved.

If P → + Q → = R → and P + Q = R , what is the angle P → and Q → ?

As Given |P| = |Q| = |R| From vector's Formula, Now, Substituting the values, Let |P| = |Q| = |R| = x [ Only in Magnitude ] &there4; The Angle Between P & Q is 120°

Find the angle between A → = 2 i ^ + 4 j ^ + 4 k ^ and B → = 4 i ^ + 2 j ^ - 4 k ^

Given, A → = 2 i ^ + 4 j ^ + 4 k ^ B → = 4 i ^ + 2 j ^ - 4 k ^ Dot product of two vectors is given by A → . B → = ( 2 i ^ + 4 j ^ + 4 k ^ ) . ( 4 i ^ + 2 j ^ - 4 k ^ ) = 8 + 8 + 16 = 32 Magnitude of A → and B → is given by A → = 2 2 + 4 2 + 4 2 = 4 + 16 + 16 = 36 = 6 B → = 4 2 + 2 2 + ( - 4 ) 2 = 16 + 4 + 16 = 36 = 6 Therefore, angle between the two vectors is given by cos θ = A → . B → A → B → = 32 6 × 6 = 32 36 ⇒ cos θ = 0 . 88 ⇒ θ = 28 . 4 ∘

A triangle is formed by vectors a → , b → and c → = a → + b → . Prove a sin A = b sin B = c sin C

In ΔABC, Let BC= a,CA= b and AB= c, then a = B C , b = C A a n d c = A B From ΔABC, we get ⇒ B C → + C A → = B A → ⇒ B C → + C A → + A B → = 0 ⇒ a → + b → + c → = 0 ⇒ a → × ( a → + b → + c → ) = a → × 0 → ⇒ a → × a → + a → × b → + a → × c → = 0 ⇒ a → × b → = − a → × c → ⇒ a → × b → = c → × a → . . . . . . ( i ) Similarly, a → × b → = b → × c → . . . . . . ( i i ) From (i) and (ii), we get a → × b → = b → × c → = c → × a → a → × b → = b → × c → = c → × a → a b sin ( π − C ) = b c sin ( π − A ) = c a sin ( π − B ) a b sin C = b c sin A = c a sin B Dividing by abc throughout, we get sin C c = sin A a = sin B b ⇒ a sin A = b sin B = c sin C