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

Created on: April 11, 2025

Answered using O1 by Chat01

Question

A vector -i+j-k is angle bisector of c vector and 3i+4j. Find unit vector in direction of c vector. I have tried the usual solution and that included a lot of calculation. I want you to show me more smarter solutions of this question.

Thought

Drawing connections

The vector -i + j - k is the angle bisector of vectors c and 3i+4j. Using vector geometry, I determine the unit vector in the direction of vector c by combining the unit vectors of c and 3i+4j.

Analyzing angle bisector

I explore that the angle bisector, -i + j - k, aligns with the sum of the unit vectors of c and 3i+4j, achieving a proportional relationship.

Determining unit vector

I conclude that the unit vector for c must be c^=i+jkd^\hat c = -i + j - k - \hat d scaled by a factor of 1, ensuring it has a magnitude of 1.

Identifying vector relationships

I determine that the angle bisector, d^\hat d, is 15(3,4,0)\frac{1}{5} (3, 4, 0) and seek c^=(x,y,z)\hat c = ( x, y, z ) with a magnitude of 1.

Determining unit vector

I conclude that the unit vector c^\hat c should satisfy (x,y,z)+15(3,4,0)(x, y, z) + \frac{1}{5}(3, 4, 0) being parallel to (1,1,1)(-1, 1, -1), with c^\hat c having a magnitude of 1.

Breaking down the problem

I set up equations for xx, yy, and zz using scalar kk, ensuring the unit vector condition is met. This involves solving (x,y,z)+15(3,4,0)(x, y, z) + \frac{1}{5}(3, 4, 0) parallel to (1,1,1)(-1, 1, -1).

Isolating variables

I express xx, yy, and zz in terms of kk and derive the function f(k)=(k3/5)2+(k4/5)2+(k)21f(k) = (-k - 3/5)^2 + (k - 4/5)^2 + (-k)^2 -1 to solve for kk.

Reasoned about vector and angle bisector for 11 seconds

Answer

First, let’s restate the problem clearly:

We have two vectors:

  1. c\mathbf{c} (unknown), in (presumably) 3-dimensional space.
  2. 3i+4j3\mathbf{i} + 4\mathbf{j} (which lies purely in the xyxy-plane).

We are told that i+jk\,-\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 c\mathbf{c}.


Key angle-bisector fact

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

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

Equivalently, there is some scalar λ0\lambda\neq0 such that

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

In our case:

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

Hence

i+jk    cc  +  3i+4j3i+4j. -\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 3i+4j3\mathbf{i} + 4\mathbf{j} has length 55, its unit vector is

d^  =  3i+4j5=(35,45,0). \hat{\mathbf{d}} \;=\; \frac{3\mathbf{i} + 4\mathbf{j}}{5} = \Bigl(\tfrac{3}{5}, \tfrac{4}{5}, 0\Bigr).

And let

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

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

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

is written as

c^+d^  =  (x,y,z)+(35,45,0)    (1,1,1). \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 kk so that

(x+35,y+45,z)  =  k(1,1,1). (\,x + \tfrac{3}{5},\,y + \tfrac{4}{5},\,z\,) \;=\; k\,(-1,\,1,\,-1).

And we must also enforce that c^=(x,y,z)\hat{\mathbf{c}} = (x,\,y,\,z) is a unit vector, i.e. x2+y2+z2=1x^2 + y^2 + z^2 = 1.


Step 1. Write the system in terms of kk

From

(x+35,  y+45,  z)  =  k(1,1,1), (x + \tfrac{3}{5},\; y + \tfrac{4}{5},\; z) \;=\; k(-1,\,1,\,-1),

we get the three equalities:

  1. x+35=kx + \tfrac{3}{5} = -\,k, i.e. x=k35x = -k - \tfrac{3}{5}.
  2. y+45=  ky + \tfrac{4}{5} = \;k, i.e. y=k45y = k - \tfrac{4}{5}.
  3. z  =  kz \;=\; -\,k.

Thus

x=k35,y=k45,z=k. x = -k - \tfrac{3}{5}, \quad y = k - \tfrac{4}{5}, \quad z = -k.

Step 2. Impose the unit-length condition

x2+y2+z2  =  1. x^2 + y^2 + z^2 \;=\; 1.

Substitute the expressions for x,y,zx,y,z in terms of kk:

(k35)2  +  (k45)2  +  (k)2  =  1. \bigl(-k - \tfrac{3}{5}\bigr)^{2} \;+\; \bigl(k - \tfrac{4}{5}\bigr)^{2} \;+\; (-k)^{2} \;=\; 1.

So we want to solve

(k35)2+(k45)2+(k)2  =  1. (-k - \tfrac{3}{5})^{2} + (k - \tfrac{4}{5})^{2} + (-k)^{2} \;=\; 1.

One can expand and simplify to find kk. 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=k2.(-k)^2 = k^2.
  • We can group like terms: each bracket is (±k+constant)2(\pm k + \text{constant})^2.

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

(k35)2=k2+925+65k, ( -k - \tfrac{3}{5} )^2 = k^2 + \tfrac{9}{25} + \tfrac{6}{5}k,

but watch the sign carefully: (k35)2=k2+2k35+(35)2( -k - \frac{3}{5})^2 = k^2 + 2\cdot k\cdot\frac{3}{5} + \bigl(\tfrac{3}{5}\bigr)^2, that middle term is +65k \color{red}{+}\frac{6}{5}k but with negative kk, so be sure of the sign. Actually:

(k35)2=k2+925+2(k) ⁣(35)=k2+925+65k. ( -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. (k45)2=k2+16252k45=k2+162585k. ( 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=k2. (-k)^2 = k^2.

So summing:

k2+925+65k  +  k2+162585k  +  k2  =  1. k^2 + \frac{9}{25} + \frac{6}{5}k \;+\; k^2 + \frac{16}{25} - \frac{8}{5}k \;+\; k^2 \;=\; 1.

Combine like terms:

  • k2+k2+k2=3k2.k^2 + k^2 + k^2 = 3\,k^2.
  • 65k    85k=25k.\frac{6}{5}k \;-\;\frac{8}{5}k = -\frac{2}{5}k.
  • Constants: 925+1625=2525=1.\frac{9}{25} + \frac{16}{25} = \frac{25}{25} = 1.

Hence

3k2    25k  +  1  =  1. 3\,k^2 \;-\;\tfrac{2}{5}\,k \;+\; 1 \;=\; 1.

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

3k2    25k+1  =  13k2    25k  =  0. 3\,k^2 \;-\;\tfrac{2}{5}\,k + 1 \;=\; 1 \quad\Longrightarrow\quad 3\,k^2 \;-\;\tfrac{2}{5}\,k \;=\; 0.

Factor out kk:

k(3k25)=0. k\,\Bigl(3k - \tfrac{2}{5}\Bigr) = 0.

So either

k=0or3k25=0. k = 0 \quad\text{or}\quad 3k - \tfrac{2}{5} = 0.
  1. Case k=0k=0. Then (x,y,z)=(35,  45,  0)(x,y,z) = \bigl(-\tfrac{3}{5},\;-\tfrac{4}{5},\;0\bigr). That’s a unit vector indeed (since (35)2+(45)2=1\left(-\frac{3}{5}\right)^2 + \left(-\frac{4}{5}\right)^2 = 1).

    • But check whether this actually makes i+jk-\mathbf{i}+\mathbf{j}-\mathbf{k} proportional to c^+d^\hat{\mathbf{c}} + \hat{\mathbf{d}}. If k=0k=0, then c^+d^=(x,y,z)+(3/5,4/5,0)=(35+35,  45+45,  0)=(0,0,0)\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=0k=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=0k=0 is an extraneous solution from forcing magnitude=1.
  2. Case 3k25=03k - \tfrac{2}{5} = 0. Then

    k=215. k = \frac{2}{15}.

    Let’s take that.

Step 3. Plug in k=215k = \frac{2}{15}

We then get

x=k35=215915=1115, x = -k - \tfrac{3}{5} = -\tfrac{2}{15} - \tfrac{9}{15} = -\tfrac{11}{15}, y=k45=2151215=1015=23, y = k - \tfrac{4}{5} = \tfrac{2}{15} - \tfrac{12}{15} = -\tfrac{10}{15} = -\tfrac{2}{3}, z=k=215. z = -k = -\tfrac{2}{15}.

Hence

c^  =  (1115,  23,  215). \hat{\mathbf{c}} \;=\; \bigl(-\tfrac{11}{15},\; -\tfrac{2}{3},\; -\tfrac{2}{15}\bigr).

You can check quickly that x2+y2+z2=1x^2 + y^2 + z^2 = 1. Indeed,

  • (1115)2=121225,\left(-\frac{11}{15}\right)^2 = \frac{121}{225},
  • (23)2=49=100225,\left(-\frac{2}{3}\right)^2 = \frac{4}{9} = \frac{100}{225},
  • (215)2=4225.\left(-\frac{2}{15}\right)^2 = \frac{4}{225}.

Summing yields 121225+100225+4225=225225=1.\frac{121}{225} + \frac{100}{225} + \frac{4}{225} = \frac{225}{225} = 1. Perfect.

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

c^=(1115,  23,  215). \boxed{ \hat{\mathbf{c}} = \left(-\frac{11}{15},\; -\frac{2}{3},\; -\frac{2}{15}\right). }

That solves the problem.


Smarter (and less) computation – conceptual shortcuts

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

    c^+3i+4j5    i+jk. \hat{\mathbf{c}} + \frac{3\mathbf{i}+4\mathbf{j}}{5} \;\propto\; -\mathbf{i}+\mathbf{j}-\mathbf{k}.

    Then set

    c^+(35,45,0)=k(1,1,1). \hat{\mathbf{c}} + \Bigl(\frac{3}{5}, \,\frac{4}{5},\,0\Bigr) = k\,(-1,\,1,\,-1).

    From there, simply impose c^=1\|\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), (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=0k=0. Then c^\hat{\mathbf{c}} would be (35,45,0)\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 c^\hat{\mathbf{c}} is exactly opposite to d^\hat{\mathbf{d}}. So that cannot be an angle bisector in the usual sense. Hence you ignore that and take the nonzero solution for kk.

  3. Magnitude of 3i+4j3\mathbf{i} + 4\mathbf{j} is obviously 5. So the presence of 32+42\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:

Angle bisector    cc  +  bb, \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.


Final answer

A unit vector in the direction of c\mathbf{c} (the one consistent with i+jk-\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 1-1 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.

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