Skip to content
Notely Maths 12
Class 12 Maths
🚨 This site is a work in progress. Exciting updates are coming soon!

7.5 Q-5

Question Statement

Part (a) - Find the following values:

  1. 2i^×2j^⋅k^2 \hat{i} \times 2 \hat{j} \cdot \hat{k}
  2. 3j^⋅(k^×i^)3 \hat{j} \cdot (\hat{k} \times \hat{i})
  3. [k^,i^,j^][\hat{k} , \hat{i} , \hat{j}]
  4. [i^,j^,k^][\hat{i} , \hat{j} , \hat{k}]

Part (b) - Prove the identity:

U⋅(V×W)+V⋅(W×U)+W⋅(U×V)=3U⋅(V×W)\mathbf{U} \cdot (\mathbf{V} \times \mathbf{W}) + \mathbf{V} \cdot (\mathbf{W} \times \mathbf{U}) + \mathbf{W} \cdot (\mathbf{U} \times \mathbf{V}) = 3 \mathbf{U} \cdot (\mathbf{V} \times \mathbf{W})

Background and Explanation

In this problem, we deal with vector cross products and dot products. The following properties are important:

  1. Cross product a×b\mathbf{a} \times \mathbf{b} produces a vector perpendicular to both a\mathbf{a} and b\mathbf{b}.
  2. Dot product a⋅b\mathbf{a} \cdot \mathbf{b} produces a scalar representing the magnitude of the projection of a\mathbf{a} onto b\mathbf{b}.
  3. Scalar triple product: The scalar triple product a⋅(b×c)\mathbf{a} \cdot (\mathbf{b} \times \mathbf{c}) gives the volume of the parallelepiped formed by the three vectors a\mathbf{a}, b\mathbf{b}, and c\mathbf{c}, and is invariant under cyclic permutations of the vectors.

Solution

Part (a) - Finding the Values

(i) 2i^×2j^⋅k^2 \hat{i} \times 2 \hat{j} \cdot \hat{k}

We first calculate the cross product 2i^×2j^2 \hat{i} \times 2 \hat{j}:

2i^×2j^=4(i^×j^)2 \hat{i} \times 2 \hat{j} = 4 (\hat{i} \times \hat{j})

We know from the properties of the cross product that:

i^×j^=k^\hat{i} \times \hat{j} = \hat{k}

Thus,

4(i^×j^)=4k^4 (\hat{i} \times \hat{j}) = 4 \hat{k}

Now, we take the dot product with k^\hat{k}:

4k^⋅k^=4(1)=44 \hat{k} \cdot \hat{k} = 4(1) = 4

So, the value is 44.

(ii) 3j^⋅(k^×i^)3 \hat{j} \cdot (\hat{k} \times \hat{i})

First, calculate the cross product k^×i^\hat{k} \times \hat{i}:

k^×i^=−j^\hat{k} \times \hat{i} = -\hat{j}

Now, take the dot product with j^\hat{j}:

3j^⋅(−j^)=3(−1)=−33 \hat{j} \cdot (-\hat{j}) = 3(-1) = -3

Thus, the value is −3-3.

(iii) [k^,i^,j^][\hat{k} , \hat{i} , \hat{j}]

This is the scalar triple product:

[k^,i^,j^]=k^⋅(i^×j^)[\hat{k} , \hat{i} , \hat{j}] = \hat{k} \cdot (\hat{i} \times \hat{j})

We know that i^×j^=k^\hat{i} \times \hat{j} = \hat{k}, so:

k^⋅k^=1\hat{k} \cdot \hat{k} = 1

Thus, the value is 11.

(iv) [i^,j^,k^][\hat{i} , \hat{j} , \hat{k}]

This is another scalar triple product:

[i^,j^,k^]=i^⋅(j^×k^)[\hat{i} , \hat{j} , \hat{k}] = \hat{i} \cdot (\hat{j} \times \hat{k})

We know that j^×k^=i^\hat{j} \times \hat{k} = \hat{i}, so:

i^⋅i^=1\hat{i} \cdot \hat{i} = 1

Thus, the value is 11.

Part (b) - Proving the Identity

We need to prove the following identity:

U⋅(V×W)+V⋅(W×U)+W⋅(U×V)=3U⋅(V×W)\mathbf{U} \cdot (\mathbf{V} \times \mathbf{W}) + \mathbf{V} \cdot (\mathbf{W} \times \mathbf{U}) + \mathbf{W} \cdot (\mathbf{U} \times \mathbf{V}) = 3 \mathbf{U} \cdot (\mathbf{V} \times \mathbf{W})

Consider the left-hand side (LHS):

LHS=U⋅(V×W)+V⋅(W×U)+W⋅(U×V)\text{LHS} = \mathbf{U} \cdot (\mathbf{V} \times \mathbf{W}) + \mathbf{V} \cdot (\mathbf{W} \times \mathbf{U}) + \mathbf{W} \cdot (\mathbf{U} \times \mathbf{V})

By the cyclic property of the scalar triple product, we know:

V⋅(W×U)=W⋅(U×V)=U⋅(V×W)\mathbf{V} \cdot (\mathbf{W} \times \mathbf{U}) = \mathbf{W} \cdot (\mathbf{U} \times \mathbf{V}) = \mathbf{U} \cdot (\mathbf{V} \times \mathbf{W})

So we can simplify the expression as:

LHS=U⋅(V×W)+U⋅(V×W)+U⋅(V×W)\text{LHS} = \mathbf{U} \cdot (\mathbf{V} \times \mathbf{W}) + \mathbf{U} \cdot (\mathbf{V} \times \mathbf{W}) + \mathbf{U} \cdot (\mathbf{V} \times \mathbf{W}) LHS=3(U⋅(V×W))\text{LHS} = 3 (\mathbf{U} \cdot (\mathbf{V} \times \mathbf{W}))

Thus, we have:

LHS=RHS\text{LHS} = \text{RHS}

Therefore, the identity is proven.

Key Formulas or Methods Used

  • Cross Product: i^×j^=k^\hat{i} \times \hat{j} = \hat{k}, and similar results for other unit vectors.
  • Dot Product: The result of a dot product of a unit vector with itself is 1, and with any other unit vector is 0.
  • Scalar Triple Product: The scalar triple product is invariant under cyclic permutations of the vectors:
U⋅(V×W)=V⋅(W×U)=W⋅(U×V) \mathbf{U} \cdot (\mathbf{V} \times \mathbf{W}) = \mathbf{V} \cdot (\mathbf{W} \times \mathbf{U}) = \mathbf{W} \cdot (\mathbf{U} \times \mathbf{V})

Summary of Steps

  1. Calculate each vector product and dot product in Part (a), using properties of cross and dot products.
  2. Simplify the results to find the required values.
  3. For Part (b), use the cyclic property of the scalar triple product to simplify the left-hand side.
  4. Verify that both sides of the equation are equal, thus proving the identity.