A vector -i+j-k is angle bisector of c vector and ...

We can “shortcut” the heavy algebra by recalling a key fact from vector geometry: if a vector bisects the angle between two nonzero vectors, then its direction is proportional to the sum of the unit vectors in the directions of those two vectors.

In our problem, we are given that the vector

$$\mathbf{u} = -\mathbf{i} + \mathbf{j} - \mathbf{k}$$

bisects the angle between an unknown vector $\mathbf{c}$ (whose unit direction we need) and

$$\mathbf{b} = 3\mathbf{i} + 4\mathbf{j}.$$

Because $\mathbf{u}$ is the internal bisector, there is a scalar $\lambda$ such that

$$\frac{\mathbf{c}}{\|\mathbf{c}\|} + \frac{\mathbf{b}}{\|\mathbf{b}\|} = \lambda\,\mathbf{u}.$$

Since we want the unit vector in the direction of $\mathbf{c}$, we can take

$$\frac{\mathbf{c}}{\|\mathbf{c}\|} = \mathbf{c}$$

(with the understanding that $\|\mathbf{c}\|=1$) and note that

$$\|\mathbf{b}\| = \sqrt{3^2 + 4^2} = 5.$$

Thus, our “smart” step is to write

$$\mathbf{c} + \frac{1}{5}(3\mathbf{i} + 4\mathbf{j}) = \lambda\,( -\mathbf{i} + \mathbf{j} - \mathbf{k}).$$

Step 1. Express $\mathbf{c}$ in terms of $\lambda$:

Rearrange the equation:

$$\mathbf{c} = \lambda\,( -\mathbf{i} + \mathbf{j} - \mathbf{k}) - \frac{1}{5}(3\mathbf{i} + 4\mathbf{j}).$$

Writing this componentwise gives

$$\mathbf{c} = \left(-\lambda - \frac{3}{5}\right)\mathbf{i} + \left(\lambda - \frac{4}{5}\right)\mathbf{j} - \lambda\,\mathbf{k}.$$

Step 2. Use the unit vector condition for $\mathbf{c}$:

Since $\mathbf{c}$ is a unit vector, we require

$$\left(-\lambda - \frac{3}{5}\right)^2 + \left(\lambda - \frac{4}{5}\right)^2 + (-\lambda)^2 = 1.$$

Let’s compute each term:

• $\left(-\lambda - \frac{3}{5}\right)^2 = \lambda^2 + \frac{6}{5}\lambda + \frac{9}{25}$,
• $\left(\lambda - \frac{4}{5}\right)^2 = \lambda^2 - \frac{8}{5}\lambda + \frac{16}{25}$,
• $(-\lambda)^2 = \lambda^2$.

Summing them up we get:

$$\lambda^2 + \frac{6}{5}\lambda + \frac{9}{25} + \lambda^2 - \frac{8}{5}\lambda + \frac{16}{25} + \lambda^2 = 3\lambda^2 - \frac{2}{5}\lambda + \frac{25}{25}.$$

Since $\frac{25}{25} = 1$, the norm condition becomes:

$$3\lambda^2 - \frac{2}{5}\lambda + 1 = 1.$$

Subtract 1 from both sides:

$$3\lambda^2 - \frac{2}{5}\lambda = 0,$$

or

$$\lambda\left(3\lambda - \frac{2}{5}\right) = 0.$$

We discard the trivial solution $\lambda=0$ (which would force $\mathbf{c}$ to lie directly opposite to $\mathbf{b}$), so we have

$$3\lambda - \frac{2}{5} = 0 \quad\Longrightarrow\quad \lambda = \frac{2}{15}.$$

Step 3. Substitute $\lambda$ back to get $\mathbf{c}$:

Substitute $\lambda = \frac{2}{15}$ into the expression for $\mathbf{c}$:

$$\mathbf{c} = \frac{2}{15}\,(-\mathbf{i} + \mathbf{j} - \mathbf{k}) - \frac{1}{5}(3\mathbf{i} + 4\mathbf{j}).$$

It is often easier to use a common denominator (15) for the calculation:

$$\mathbf{c} = \left(-\frac{2}{15} - \frac{9}{15}\right)\mathbf{i} + \left(\frac{2}{15} - \frac{12}{15}\right)\mathbf{j} - \frac{2}{15}\mathbf{k}.$$

That is,

$$\mathbf{c} = -\frac{11}{15}\,\mathbf{i} -\frac{10}{15}\,\mathbf{j} - \frac{2}{15}\,\mathbf{k}.$$

Thus, the unit vector in the direction of $\mathbf{c}$ is

$$\boxed{\mathbf{c} = \frac{-11\,\mathbf{i} - 10\,\mathbf{j} - 2\,\mathbf{k}}{15}}.$$

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Summary of the “Smart” Approach

1. Use the angle bisector property: Recognize that the unit bisector is proportional to the sum of the unit vectors of $\mathbf{c}$ and $\mathbf{b}$.
2. Set up the equation: Write $$\mathbf{c} + \frac{1}{5}(3\mathbf{i}+4\mathbf{j}) = \lambda(-\mathbf{i}+\mathbf{j}-\mathbf{k}),$$ which directly relates the unknown unit vector $\mathbf{c}$ to the known bisector.
3. Solve with the unit vector condition: Express $\mathbf{c}$ in terms of $\lambda$ and impose $\|\mathbf{c}\|=1$ to solve for $\lambda$.
4. Substitute back: Finally, substitute back to obtain the components of $\mathbf{c}$.

This method avoids the more complicated vector projection or reflection methods and uses symmetry from the angle bisector concept to reach a quick and elegant solution.

The final answer is:

$$\boxed{\frac{-11\,\mathbf{i} - 10\,\mathbf{j} - 2\,\mathbf{k}}{15}}.$$