ANGLE BETWEEN TWO VECTORS USING CROSS PRODUCT

Formula to find the angle θ between the two vectors 'a' and 'b' using cross product : 

Example 1 :

Find the angle between the following two vectors using cross product.

2i + j - k 

i + 2j + k

Solution :

  |a x b| =  i[1 + 2] - j[2 + 1] + k[4 - 1]

  |a x b| =  3i - 3j + 3k

|a x b|2 √(32 + (-3)2 + 32= 3√3

|a| = √(22 + 1+ (-1)2) = √6

|b| = √(12 + 2+ 12) = √6

θ = sin-1(3√3/√6√6)

θ = sin-1(3√3/6)

= sin-1(√3/2)

π/3

Example 2 :

Let a vector, b vector, c vector be unit vectors such that a ⋅ b = a ⋅ c = 0 and the angle between b vector and c vector is π/3. Prove that a vector = ±(2/3)(b x c).

Solution :

From given information, we have a ⋅ b = a ⋅ c = 0.

From this we may decide that a vector is perpendicular to b vector and a vector is perpendicular to c vector.

a vector is perpendicular to both b vector and c vector. So, a vector is proportional to (b x c) vector

a vector = ± λ(b x c)

|a| = ±λ|(b x c| ----(1)

|a| = ±λ|b||c|sinθ

1 = ±λsin(π/3)

λ = 2/3

Substituting λ = 2/3 in (1), 

a vector = ±(2/3)(b x c)

Example 3 :

For any vector a vector prove that

|a x i|+ |a x j|2+ |a x k|= 2|a|2

Solution :

Let a vector = xi + yj + zk. 

a x i : 

= i[0 - 0] - j[0 - z] + k[0 - y]

a x i = zj - yk

|a x i| = √(z2 + (-y)2) = √(z2 + y2)

|a x i|2 = z2 + y2 ----(1)

a x j : 

= i[0 - z] - j[0 - 0] + k[x - 0]

a x j = -zi + xk

|a x j| = √((-z)2 + x2) = √(z2 + x2)

|a x j|2 = z2 + x2 ----(2)

Similarly,  

a x k = i[y - 0] - j[x - 0] + k[0 - 0]

a x k = yi - xj

|a x k| = √(y2 + (-x)2) = √(y2 + x2)

|a x k|2 = y2 + x----(3)

(1) + (2) + (3) :  

|a x i|+|a x j|+ |a x k|= (z2 + y2) + (z2 + x2) + (y2 + x2)

= z2 + y2 + z2 + x2 + y2 + x2

= 2x2 + 2y2 + 2z2

= 2(x2 + y2 + z)2

= 2|a|2

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