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

Created on: April 11, 2025

Answered using O3 Mini High 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. Temperature=1.0.

Thought

Exploring smarter solutions
Okay, we’re given that the vector -i+j-k is the angle bisector of vector c and 3i+4j. We need the unit vector in the direction of c. So, vector c is the one whose angle with 3i+4j is bisected by -i+j-k. The angle bisector theorem for vectors might be useful here, but we definitely need a more efficient approach than just brute force calculations. Let’s set up the vectors and explore how the theorem applies to find c’s direction. We can probably solve it through vector relationships.

Refining the solution

Alright, from the question, we need to find the unit vector in the direction of vector c, which is unknown at the moment, with the condition that the vector -i + j - k bisects the angle between c and 3i + 4j.

Using the angle bisector property, we know the bisector (q) is proportional to the sum of the unit vectors of c and b. This gives us:
q = k * (c + (3i + 4j) / 5)
Now, we can set up a system of equations with a scalar λ to find c. This gives us c + (3i + 4j) / 5 = λ (-i + j - k).

Finalizing the solution

Let's denote the unknown unit vector c as (c1, c2, c3). We're given that the angle bisector vector q = -i + j - k bisects the angle between vector c and 3i + 4j. Using the angle bisector property, we can express c as:
c = λ (-1, 1, -1) - (3/5, 4/5, 0).

After equating the magnitude to 1, we find λ = 2/15. Calculating c, we get (-11/15, -10/15, -2/15). The magnitude checks out as 1.

For a more efficient solution, we use the property of an angle bisector in vector form, which simplifies the problem!

Reasoned about vector angle bisector for 1m 2s

Answer

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

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

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

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

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

cc+bb=λu.\frac{\mathbf{c}}{\|\mathbf{c}\|} + \frac{\mathbf{b}}{\|\mathbf{b}\|} = \lambda\,\mathbf{u}.

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

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

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

b=32+42=5.\|\mathbf{b}\| = \sqrt{3^2 + 4^2} = 5.

Thus, our “smart” step is to write

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

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

Rearrange the equation:

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

Writing this componentwise gives

c=(λ35)i+(λ45)jλk.\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 c\mathbf{c}:

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

(λ35)2+(λ45)2+(λ)2=1.\left(-\lambda - \frac{3}{5}\right)^2 + \left(\lambda - \frac{4}{5}\right)^2 + (-\lambda)^2 = 1.

Let’s compute each term:

  • (λ35)2=λ2+65λ+925\left(-\lambda - \frac{3}{5}\right)^2 = \lambda^2 + \frac{6}{5}\lambda + \frac{9}{25},
  • (λ45)2=λ285λ+1625\left(\lambda - \frac{4}{5}\right)^2 = \lambda^2 - \frac{8}{5}\lambda + \frac{16}{25},
  • (λ)2=λ2(-\lambda)^2 = \lambda^2.

Summing them up we get:

λ2+65λ+925+λ285λ+1625+λ2=3λ225λ+2525.\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 2525=1\frac{25}{25} = 1, the norm condition becomes:

3λ225λ+1=1.3\lambda^2 - \frac{2}{5}\lambda + 1 = 1.

Subtract 1 from both sides:

3λ225λ=0,3\lambda^2 - \frac{2}{5}\lambda = 0,

or

λ(3λ25)=0.\lambda\left(3\lambda - \frac{2}{5}\right) = 0.

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

3λ25=0λ=215.3\lambda - \frac{2}{5} = 0 \quad\Longrightarrow\quad \lambda = \frac{2}{15}.

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

Substitute λ=215\lambda = \frac{2}{15} into the expression for c\mathbf{c}:

c=215(i+jk)15(3i+4j).\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:

c=(215915)i+(2151215)j215k.\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,

c=1115i1015j215k.\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 c\mathbf{c} is

c=11i10j2k15.\boxed{\mathbf{c} = \frac{-11\,\mathbf{i} - 10\,\mathbf{j} - 2\,\mathbf{k}}{15}}.

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 c\mathbf{c} and b\mathbf{b}.
  2. Set up the equation: Write c+15(3i+4j)=λ(i+jk),\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 c\mathbf{c} to the known bisector.
  3. Solve with the unit vector condition: Express c\mathbf{c} in terms of λ\lambda and impose c=1\|\mathbf{c}\|=1 to solve for λ\lambda.
  4. Substitute back: Finally, substitute back to obtain the components of c\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:

11i10j2k15.\boxed{\frac{-11\,\mathbf{i} - 10\,\mathbf{j} - 2\,\mathbf{k}}{15}}.

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