If `vec u , vec v and vec w` are three non-cop0lanar vectors, then prove that `( vec u+ vec v- vec w)dot` `[( vec u- vec v)xx( vec v- vec w)]`= `vec u dot` `(vec v xx vec w) `

Options

A. 0
B. `u*(vxxw)`
C. `u*(wxxv)`
D. `3u*(vxxw)`

Answer

To prove the equation

\[ (\vec{u} + \vec{v} - \vec{w}) \cdot [(\vec{u} - \vec{v}) \times (\vec{v} - \vec{w})] = \vec{u} \cdot (\vec{v} \times \vec{w}), \]

we will follow these steps:

### Step 1: Expand the Left-Hand Side

We start with the left-hand side:

\[ \vec{u} + \vec{v} - \vec{w} \text{ and } (\vec{u} - \vec{v}) \times (\vec{v} - \vec{w}). \]

Let's denote:

\[ \vec{A} = \vec{u} + \vec{v} - \vec{w}, \]
\[ \vec{B} = (\vec{u} - \vec{v}) \times (\vec{v} - \vec{w}). \]

We need to compute \(\vec{A} \cdot \vec{B}\).

### Step 2: Compute the Cross Product

Now, we calculate \(\vec{B}\):

\[ \vec{B} = (\vec{u} - \vec{v}) \times (\vec{v} - \vec{w}). \]

Using the distributive property of the cross product, we can expand this:

\[ \vec{B} = \vec{u} \times \vec{v} - \vec{u} \times \vec{w} - \vec{v} \times \vec{v} + \vec{v} \times \vec{w}. \]

Since \(\vec{v} \times \vec{v} = \vec{0}\), we simplify:

\[ \vec{B} = \vec{u} \times \vec{v} - \vec{u} \times \vec{w} + \vec{v} \times \vec{w}. \]

### Step 3: Dot Product Calculation

Now we compute \(\vec{A} \cdot \vec{B}\):

\[ \vec{A} \cdot \vec{B} = (\vec{u} + \vec{v} - \vec{w}) \cdot (\vec{u} \times \vec{v} - \vec{u} \times \vec{w} + \vec{v} \times \vec{w}). \]

Expanding this gives:

\[ \vec{A} \cdot \vec{B} = \vec{u} \cdot (\vec{u} \times \vec{v}) + \vec{v} \cdot (\vec{u} \times \vec{v}) - \vec{w} \cdot (\vec{u} \times \vec{v}) \]
\[ - \vec{u} \cdot (\vec{u} \times \vec{w}) - \vec{v} \cdot (\vec{u} \times \vec{w}) + \vec{w} \cdot (\vec{u} \times \vec{w}) \]
\[ + \vec{u} \cdot (\vec{v} \times \vec{w}) + \vec{v} \cdot (\vec{v} \times \vec{w}) - \vec{w} \cdot (\vec{v} \times \vec{w}). \]

### Step 4: Simplifying Each Term

Using the property that \(\vec{a} \cdot (\vec{b} \times \vec{c}) = 0\) when \(\vec{a}\) is one of the vectors in the cross product:

- \(\vec{u} \cdot (\vec{u} \times \vec{v}) = 0\)
- \(\vec{v} \cdot (\vec{u} \times \vec{v}) = 0\)
- \(\vec{u} \cdot (\vec{u} \times \vec{w}) = 0\)
- \(\vec{w} \cdot (\vec{u} \times \vec{w}) = 0\)
- \(\vec{v} \cdot (\vec{v} \times \vec{w}) = 0\)

Thus, we are left with:

\[ \vec{A} \cdot \vec{B} = -\vec{w} \cdot (\vec{u} \times \vec{v}) + \vec{u} \cdot (\vec{v} \times \vec{w}). \]

### Step 5: Rearranging Terms

Rearranging gives us:

\[ \vec{A} \cdot \vec{B} = \vec{u} \cdot (\vec{v} \times \vec{w}), \]

which is exactly what we wanted to prove.

### Conclusion

Thus, we have shown that:

\[ (\vec{u} + \vec{v} - \vec{w}) \cdot [(\vec{u} - \vec{v}) \times (\vec{v} - \vec{w})] = \vec{u} \cdot (\vec{v} \times \vec{w}). \]