**Angle Between Two Vectors Using Cross Product :**

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

**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| = √3^{2} + 3^{2 }+ 3^{2 }= 3√3

|a vector| = √2^{2} + 1^{2 }+ 1^{2 }= √6

|b vector| = √1^{2} + 2^{2 }+ 1^{2 }= √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/√3

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| = √z^{2} + y^{2}

|a vector x i vector|^{2} = (√z^{2} + y^{2})^{2}

|a vector x i vector|^{2} = z^{2} + y^{2 } -----(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| = √z^{2} + x^{2}

|a vector x j vector|^{2} = (√z^{2} + x^{2})^{2}

|a vector x j vector|^{2} = z^{2} + x^{2} -----(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| = √y^{2} + x^{2}

|a vector x k vector|^{2} = (√y^{2} + x^{2})^{2}

|a vector x k vector|^{2} = y^{2} + x^{2} -----(3)

(1) + (2) + (3) ==>

|a vector × i vector |^{2}+|a vector × j vector|^{2}+|a vector × k vector|^{2 }= 2x^{2} + 2y^{2} + 2z^{2}

= 2(x^{2} + y^{2} + z^{2})

= 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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