If a, b, c are unit vectors such that `a-b+c=0` then `c*a` is equal to

Options

A. `3/2`
B. `-1/2`
C. `1/3`
D. `-1/3`

Answer

To solve the problem, we start with the given equation involving the unit vectors \( \mathbf{a}, \mathbf{b}, \mathbf{c} \):

\[ \mathbf{a} - \mathbf{b} + \mathbf{c} = 0 \]

From this, we can rearrange to express \( \mathbf{c} \):

\[ \mathbf{c} = \mathbf{b} - \mathbf{a} \]

Next, we need to find the dot product \( \mathbf{c} \cdot \mathbf{a} \):

\[ \mathbf{c} \cdot \mathbf{a} = (\mathbf{b} - \mathbf{a}) \cdot \mathbf{a} \]

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

\[ \mathbf{c} \cdot \mathbf{a} = \mathbf{b} \cdot \mathbf{a} - \mathbf{a} \cdot \mathbf{a} \]

Since \( \mathbf{a} \) is a unit vector, we know that:

\[ \mathbf{a} \cdot \mathbf{a} = 1 \]

Thus, we can substitute this into our equation:

\[ \mathbf{c} \cdot \mathbf{a} = \mathbf{b} \cdot \mathbf{a} - 1 \]

Now, we need to find \( \mathbf{b} \cdot \mathbf{a} \). To do this, we can take the dot product of the original equation \( \mathbf{a} - \mathbf{b} + \mathbf{c} = 0 \) with \( \mathbf{b} \):

\[ (\mathbf{a} - \mathbf{b} + \mathbf{c}) \cdot \mathbf{b} = 0 \]

Expanding this gives:

\[ \mathbf{a} \cdot \mathbf{b} - \mathbf{b} \cdot \mathbf{b} + \mathbf{c} \cdot \mathbf{b} = 0 \]

Since \( \mathbf{b} \) is also a unit vector, we have:

\[ \mathbf{b} \cdot \mathbf{b} = 1 \]

So we can rewrite the equation as:

\[ \mathbf{a} \cdot \mathbf{b} - 1 + \mathbf{c} \cdot \mathbf{b} = 0 \]

This simplifies to:

\[ \mathbf{c} \cdot \mathbf{b} = 1 - \mathbf{a} \cdot \mathbf{b} \]

Now, we can also take the dot product of the original equation with \( \mathbf{c} \):

\[ (\mathbf{a} - \mathbf{b} + \mathbf{c}) \cdot \mathbf{c} = 0 \]

Expanding this gives:

\[ \mathbf{a} \cdot \mathbf{c} - \mathbf{b} \cdot \mathbf{c} + \mathbf{c} \cdot \mathbf{c} = 0 \]

Since \( \mathbf{c} \) is a unit vector, we have:

\[ \mathbf{c} \cdot \mathbf{c} = 1 \]

Thus, we can rewrite the equation as:

\[ \mathbf{a} \cdot \mathbf{c} - \mathbf{b} \cdot \mathbf{c} + 1 = 0 \]

Rearranging gives:

\[ \mathbf{a} \cdot \mathbf{c} = \mathbf{b} \cdot \mathbf{c} - 1 \]

Now we have two equations:

1. \( \mathbf{c} \cdot \mathbf{a} = \mathbf{b} \cdot \mathbf{a} - 1 \)
2. \( \mathbf{a} \cdot \mathbf{c} = \mathbf{b} \cdot \mathbf{c} - 1 \)

From equation 1, substituting \( \mathbf{b} \cdot \mathbf{c} = 1 - \mathbf{a} \cdot \mathbf{b} \) into equation 2, we can find \( \mathbf{c} \cdot \mathbf{a} \):

Substituting \( \mathbf{b} \cdot \mathbf{c} \):

\[ \mathbf{c} \cdot \mathbf{a} = (1 - \mathbf{a} \cdot \mathbf{b}) - 1 \]

This simplifies to:

\[ \mathbf{c} \cdot \mathbf{a} = -\mathbf{a} \cdot \mathbf{b} \]

Now, substituting back into our earlier equation gives:

\[ 2(\mathbf{c} \cdot \mathbf{a}) = -2 + \mathbf{a} \cdot \mathbf{b} + (1 - \mathbf{a} \cdot \mathbf{b}) \]

This leads us to:

\[ 2(\mathbf{c} \cdot \mathbf{a}) = -1 \]

Thus:

\[ \mathbf{c} \cdot \mathbf{a} = -\frac{1}{2} \]

Therefore, the final answer is:

\[ \mathbf{c} \cdot \mathbf{a} = -\frac{1}{2} \]