ANGLE BETWEEN TWO VECTORS USING CROSS PRODUCT

About "Angle Between Two Vectors Using Cross Product"

Angle Between Two Vectors Using Cross Product :

Here we are going to see how to find angle between two vectors using cross product.

Angle between two vectors using cross product - Examples

Question 1 :

Find the angle between the vectors 2i vector + j vector − k vector and i vector+ 2j vector + k vector using vector product.

Solution :

Angle between two vectors using vector product

 θ  =  sin-1 (|a vector x b vector|/|a vector||b vector|)

  =  i[1+2]-j[2+1]+k[4-1]

a vector x b vector= 3i vector + 3j vector + 3k vector

|a vector x b vector| =  √32 + 32 + 32  =  3√3

|a vector|  =   √22 + 1+ 1=  √6

|b vector|  =   √12 + 2+ 12  =  √6

 θ  =  sin-1 (3√3/√6√6)

 θ  =  sin-1 (√3/2)

 θ  =  π/3

Question 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 × 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 vector x c vector)

|a vector|  =  ± λ |(b vector x c vector)|  ----(1)

|a vector|  =  ± λ |b||c| sin θ

1 =  ± λ sin π/3

λ  =  2/

By applying the value λ  =  2/3 in (1), we get

a vector  =  ± (2/3) (b × c)

Hence it is proved.

Question 3 :

For any vector a vector prove that

|a vector × i vector |2+|a vector × j vector|2+|a vector × k vector|2= 2 |a vector|2 .

Solution :

|a vector × i vector |2+|a vector × j vector|2+|a vector × k vector|2= 2 |a vector|2 .

Let a vector  =  xi vector + yj vector + zk vector, then 

a vector x i vector  =  

  =  i[0-0] -j[0-z] + k[0-y]

a vector x i vector  = zj vector - yk vector

|a vector x i vector|  =  √z2 + y2

|a vector x i vector|2  =  (√z2 + y2)2

|a vector x i vector|2  =  z2 + y  -----(1)

||| ly 

a vector x j vector  = 

  =  i[0-z] -j[0-0] + k[x-0]

a vector x j vector  = -zi vector + xk vector

|a vector x j vector|  =  √z2 + x2

|a vector x j vector|2  =  (√z2 + x2)2

|a vector x j vector|2  =  z2 + x2  -----(2)

||| ly 

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

a vector x k vector  = yi vector - xj vector

|a vector x k vector|  =  √y2 + x2

|a vector x k vector|2  =  (√y2 + x2)2

|a vector x k vector|2  =  y2 + x2  -----(3)

(1) + (2) + (3)  ==>

|a vector × i vector |2+|a vector × j vector|2+|a vector × k vector|=  2x2 + 2y2 + 2z2

  =  2(x2 + y2 + z2)

  =  2|z|2

Hence proved.

After having gone through the stuff given above, we hope that the students would have understood, "Angle Between Two Vectors Using Cross Product"

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