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

First, let’s restate the problem clearly:

We have two vectors:

1. $\mathbf{c}$ (unknown), in (presumably) 3-dimensional space.
2. $3\mathbf{i} + 4\mathbf{j}$ (which lies purely in the $xy$-plane).

We are told that $\,-\mathbf{i} + \mathbf{j} - \mathbf{k}$ is the angle bisector of these two vectors. We want to find the unit vector in the direction of $\mathbf{c}$.

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Key angle-bisector fact

If $\mathbf{v}$ is the angle bisector of two nonzero vectors $\mathbf{a}$ and $\mathbf{b}$, that means

$$\mathbf{v} \;\text{is parallel to}\; \frac{\mathbf{a}}{\|\mathbf{a}\|} \;+\; \frac{\mathbf{b}}{\|\mathbf{b}\|}.$$

Equivalently, there is some scalar $\lambda\neq0$ such that

$$\mathbf{v} \;=\;\lambda \Bigl(\tfrac{\mathbf{a}}{\|\mathbf{a}\|} \;+\; \tfrac{\mathbf{b}}{\|\mathbf{b}\|}\Bigr).$$

In our case:

• $\mathbf{a} = \mathbf{c}$ (unknown but we want a unit vector in its direction).
• $\mathbf{b} = 3\mathbf{i} + 4\mathbf{j}$.

Hence

$$-\mathbf{i} + \mathbf{j} - \mathbf{k} \;\propto\; \frac{\mathbf{c}}{\|\mathbf{c}\|} \;+\; \frac{3\mathbf{i} + 4\mathbf{j}}{\|3\mathbf{i} + 4\mathbf{j}\|}.$$

Since $3\mathbf{i} + 4\mathbf{j}$ has length $5$, its unit vector is

$$\hat{\mathbf{d}} \;=\; \frac{3\mathbf{i} + 4\mathbf{j}}{5} = \Bigl(\tfrac{3}{5}, \tfrac{4}{5}, 0\Bigr).$$

And let

$$\hat{\mathbf{c}} \;=\; \frac{\mathbf{c}}{\|\mathbf{c}\|} = (x,\,y,\,z)$$

denote the unit vector in the direction of $\mathbf{c}$. Then the condition that

$$-\mathbf{i} + \mathbf{j} - \mathbf{k} \;\propto\; \hat{\mathbf{c}} \;+\; \hat{\mathbf{d}}$$

is written as

$$\hat{\mathbf{c}} + \hat{\mathbf{d}} \;=\; (x,\,y,\,z) + \Bigl(\tfrac{3}{5}, \tfrac{4}{5}, 0\Bigr) \;\propto\; (-1,\,1,\,-1).$$

In other words, there is some scalar $k$ so that

$$(\,x + \tfrac{3}{5},\,y + \tfrac{4}{5},\,z\,) \;=\; k\,(-1,\,1,\,-1).$$

And we must also enforce that $\hat{\mathbf{c}} = (x,\,y,\,z)$ is a unit vector, i.e. $x^2 + y^2 + z^2 = 1$.

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Step 1. Write the system in terms of $k$

From

$$(x + \tfrac{3}{5},\; y + \tfrac{4}{5},\; z) \;=\; k(-1,\,1,\,-1),$$

we get the three equalities:

1. $x + \tfrac{3}{5} = -\,k$, i.e. $x = -k - \tfrac{3}{5}$.
2. $y + \tfrac{4}{5} = \;k$, i.e. $y = k - \tfrac{4}{5}$.
3. $z \;=\; -\,k$.

Thus

$$x = -k - \tfrac{3}{5}, \quad y = k - \tfrac{4}{5}, \quad z = -k.$$

Step 2. Impose the unit-length condition

$$x^2 + y^2 + z^2 \;=\; 1.$$

Substitute the expressions for $x,y,z$ in terms of $k$:

$$\bigl(-k - \tfrac{3}{5}\bigr)^{2} \;+\; \bigl(k - \tfrac{4}{5}\bigr)^{2} \;+\; (-k)^{2} \;=\; 1.$$

So we want to solve

$$(-k - \tfrac{3}{5})^{2} + (k - \tfrac{4}{5})^{2} + (-k)^{2} \;=\; 1.$$

One can expand and simplify to find $k$. That expansion is somewhat tedious by hand—but it is straightforward to do quickly if you are systematic or use a bit of algebraic help. Let’s outline a “smart” approach before doing expansions:

• Notice that $(-k)^2 = k^2.$
• We can group like terms: each bracket is $(\pm k + \text{constant})^2$.

You could do it directly, or do a quick calculation. Let’s illustrate a short, neat approach by grouping:

$$( -k - \tfrac{3}{5} )^2 = k^2 + \tfrac{9}{25} + \tfrac{6}{5}k,$$

but watch the sign carefully: $( -k - \frac{3}{5})^2 = k^2 + 2\cdot k\cdot\frac{3}{5} + \bigl(\tfrac{3}{5}\bigr)^2$, that middle term is $\color{red}{+}\frac{6}{5}k$ but with negative $k$, so be sure of the sign. Actually:

$$( -k - \tfrac{3}{5} )^2 = k^2 + \tfrac{9}{25} + 2\bigl(-k \bigr)\!\bigl(-\tfrac{3}{5}\bigr) = k^2 + \tfrac{9}{25} + \tfrac{6}{5}k.$$ $$( k - \tfrac{4}{5} )^2 = k^2 + \tfrac{16}{25} - 2k\,\tfrac{4}{5} = k^2 + \tfrac{16}{25} - \tfrac{8}{5}k.$$ $$(-k)^2 = k^2.$$

So summing:

$$k^2 + \frac{9}{25} + \frac{6}{5}k \;+\; k^2 + \frac{16}{25} - \frac{8}{5}k \;+\; k^2 \;=\; 1.$$

Combine like terms:

• $k^2 + k^2 + k^2 = 3\,k^2.$
• $\frac{6}{5}k \;-\;\frac{8}{5}k = -\frac{2}{5}k.$
• Constants: $\frac{9}{25} + \frac{16}{25} = \frac{25}{25} = 1.$

Hence

$$3\,k^2 \;-\;\tfrac{2}{5}\,k \;+\; 1 \;=\; 1.$$

We see the “$+1$” on the LHS and “$=1$” on the RHS conveniently cancel:

$$3\,k^2 \;-\;\tfrac{2}{5}\,k + 1 \;=\; 1 \quad\Longrightarrow\quad 3\,k^2 \;-\;\tfrac{2}{5}\,k \;=\; 0.$$

Factor out $k$:

$$k\,\Bigl(3k - \tfrac{2}{5}\Bigr) = 0.$$

So either

$$k = 0 \quad\text{or}\quad 3k - \tfrac{2}{5} = 0.$$

1. Case $k=0$. Then $(x,y,z) = \bigl(-\tfrac{3}{5},\;-\tfrac{4}{5},\;0\bigr)$. That’s a unit vector indeed (since $\left(-\frac{3}{5}\right)^2 + \left(-\frac{4}{5}\right)^2 = 1$).

   • But check whether this actually makes $-\mathbf{i}+\mathbf{j}-\mathbf{k}$ proportional to $\hat{\mathbf{c}} + \hat{\mathbf{d}}$. If $k=0$, then $\hat{\mathbf{c}} + \hat{\mathbf{d}} = (x,y,z) + (3/5,4/5,0) = \bigl(-\tfrac{3}{5}+ \tfrac{3}{5},\; -\tfrac{4}{5} + \tfrac{4}{5},\; 0\bigr) = (0,0,0)$. That would be the zero vector, which makes no sense for a direction.
   • So $k=0$ would give the sum of the two unit vectors = zero – i.e., the two vectors would be exact opposites. That can’t be the angle-bisector scenario we want.
   • Hence $k=0$ is an extraneous solution from forcing magnitude=1.

2. Case $3k - \tfrac{2}{5} = 0$. Then

   $$k = \frac{2}{15}.$$

   Let’s take that.

Step 3. Plug in $k = \frac{2}{15}$

We then get

$$x = -k - \tfrac{3}{5} = -\tfrac{2}{15} - \tfrac{9}{15} = -\tfrac{11}{15},$$ $$y = k - \tfrac{4}{5} = \tfrac{2}{15} - \tfrac{12}{15} = -\tfrac{10}{15} = -\tfrac{2}{3},$$ $$z = -k = -\tfrac{2}{15}.$$

Hence

$$\hat{\mathbf{c}} \;=\; \bigl(-\tfrac{11}{15},\; -\tfrac{2}{3},\; -\tfrac{2}{15}\bigr).$$

You can check quickly that $x^2 + y^2 + z^2 = 1$. Indeed,

• $\left(-\frac{11}{15}\right)^2 = \frac{121}{225},$
• $\left(-\frac{2}{3}\right)^2 = \frac{4}{9} = \frac{100}{225},$
• $\left(-\frac{2}{15}\right)^2 = \frac{4}{225}.$

Summing yields $\frac{121}{225} + \frac{100}{225} + \frac{4}{225} = \frac{225}{225} = 1.$ Perfect.

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

$$\boxed{ \hat{\mathbf{c}} = \left(-\frac{11}{15},\; -\frac{2}{3},\; -\frac{2}{15}\right). }$$

That solves the problem.

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Smarter (and less) computation – conceptual shortcuts

1. Angle-Bisector Summation Shortcut. You do not need to find $\mathbf{c}$ itself—only its unit direction, $\hat{\mathbf{c}}$. Directly start from

   $$\hat{\mathbf{c}} + \frac{3\mathbf{i}+4\mathbf{j}}{5} \;\propto\; -\mathbf{i}+\mathbf{j}-\mathbf{k}.$$

   Then set

   $$\hat{\mathbf{c}} + \Bigl(\frac{3}{5}, \,\frac{4}{5},\,0\Bigr) = k\,(-1,\,1,\,-1).$$

   From there, simply impose $\|\hat{\mathbf{c}}\|=1$. That’s exactly what we did, but notice that you can jump straight to that relationship without using separate dot-product definitions or trigonometric definitions of “angle bisector.”

2. Check for a “zero sum.” Once you see that

   $$(x + 3/5,\; y + 4/5,\; z) = k(-1,\,1,\,-1),$$

   you realize that if that sum is the zero vector, that would require $k=0$. Then $\hat{\mathbf{c}}$ would be $\bigl(-\frac{3}{5},-\frac{4}{5},0\bigr)$. That is indeed a unit vector, but it would make the “sum of unit vectors” be zero. That means $\hat{\mathbf{c}}$ is exactly opposite to $\hat{\mathbf{d}}$. So that cannot be an angle bisector in the usual sense. Hence you ignore that and take the nonzero solution for $k$.

3. Magnitude of $3\mathbf{i} + 4\mathbf{j}$ is obviously 5. So the presence of $\sqrt{3^2 +4^2}$ is quick and easy.

4. Sometimes checking “dot-product = 0” conditions can help, but in angle-bisector problems the “sum of unit directions” method is typically the fastest.

Thus, the big conceptual “smartness” in many angle bisector problems is to jump directly to:

$$\text{Angle bisector} \;\propto\; \frac{\mathbf{c}}{\|\mathbf{c}\|} \;+\; \frac{\mathbf{b}}{\|\mathbf{b}\|},$$

and then use minimal linear algebra to solve for the unknown direction, enforcing any unit-vector constraints at the end.

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Final answer

A unit vector in the direction of $\mathbf{c}$ (the one consistent with $-\mathbf{i}+\mathbf{j}-\mathbf{k}$ as angle bisector) is

\boxed{\, \left(-\frac{11}{15},\; -\frac{2}{3},\; -\frac{2}{15}\right).

(You could factor out $-1$ if you wanted a positive first component, but typically any one of the two opposite directions is “the” direction—this is the standard solution matching the given bisector.)

That concludes the “less messy” and more “direct geometric” approach.