Math Problem Statement

(a) Find the components of a vector u which has initial point (−5,−1) , terminal point (x,0) for some x∈R and which points in the same direction as (1,3)

Use exact values as vector entries, so use fractions rather than decimal approximations, for example (1/7,6). Preview

(b) Find the components of a vector w which has initial point (0,y) for some y∈R , terminal point (4,2) and which points in the same direction as (3,1) .

Solution

Let's solve each part of the problem step by step.

Part (a)

Given:

  • Initial point of vector u\mathbf{u} is (5,1)(-5, -1)
  • Terminal point of vector u\mathbf{u} is (x,0)(x, 0)
  • The vector u\mathbf{u} points in the same direction as v=(1,3)\mathbf{v} = (1, 3)

To find the components of vector u\mathbf{u}, we first express it as: u=(x(5),0(1))=(x+5,1)\mathbf{u} = (x - (-5), 0 - (-1)) = (x + 5, 1) This vector must be a scalar multiple of v=(1,3)\mathbf{v} = (1, 3), so: u=k(1,3)=(k,3k)\mathbf{u} = k(1, 3) = (k, 3k) Therefore, we have the equations: x+5=kand1=3kx + 5 = k \quad \text{and} \quad 1 = 3k

Solving for kk in the second equation: k=13k = \frac{1}{3}

Now, substitute k=13k = \frac{1}{3} into the first equation: x+5=13x + 5 = \frac{1}{3}

Solving for xx: x=135=13153=143x = \frac{1}{3} - 5 = \frac{1}{3} - \frac{15}{3} = \frac{-14}{3}

So the vector u\mathbf{u} is: u=(13,1)\mathbf{u} = \left(\frac{1}{3}, 1\right)

Part (b)

Given:

  • Initial point of vector w\mathbf{w} is (0,y)(0, y)
  • Terminal point of vector w\mathbf{w} is (4,2)(4, 2)
  • The vector w\mathbf{w} points in the same direction as v=(3,1)\mathbf{v} = (3, 1)

To find the components of vector w\mathbf{w}, we express it as: w=(40,2y)=(4,2y)\mathbf{w} = (4 - 0, 2 - y) = (4, 2 - y) This vector must be a scalar multiple of v=(3,1)\mathbf{v} = (3, 1), so: w=k(3,1)=(3k,k)\mathbf{w} = k(3, 1) = (3k, k) Therefore, we have the equations: 4=3kand2y=k4 = 3k \quad \text{and} \quad 2 - y = k

Solving for kk in the first equation: k=43k = \frac{4}{3}

Substitute k=43k = \frac{4}{3} into the second equation: 2y=432 - y = \frac{4}{3}

Solving for yy: y=243=6343=23y = 2 - \frac{4}{3} = \frac{6}{3} - \frac{4}{3} = \frac{2}{3}

So the vector w\mathbf{w} is: w=(4,23)\mathbf{w} = \left(4, \frac{2}{3}\right)

Summary of Results

  • Part (a): The components of vector u\mathbf{u} are (13,1)\left(\frac{1}{3}, 1\right).
  • Part (b): The components of vector w\mathbf{w} are (4,23)\left(4, \frac{2}{3}\right).

Would you like more details or have any questions? Here are some related questions to consider:

  1. How do you determine the direction of a vector?
  2. What is the significance of scalar multiplication in vectors?
  3. How would you find the magnitude of the vectors u\mathbf{u} and w\mathbf{w}?
  4. What happens if the initial or terminal points of the vector are modified?
  5. How do you normalize a vector, and why might it be useful?

Tip: When solving vector problems, always check if the direction and magnitude make sense in the context of the problem.

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Math Problem Analysis

Mathematical Concepts

Vectors
Scalar multiplication
Direction of vectors

Formulas

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Theorems

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Suitable Grade Level

High School