Math Problem Statement

Find 2π˜‚, βˆ’3π˜ƒ, π˜‚ + π˜ƒ, and 3π˜‚ βˆ’ 4π˜ƒ for the given vectors π˜‚ and π˜ƒ.

π˜‚ = βˆ’6, 7,    π˜ƒ = 6, βˆ’9

2π˜‚

 =    βˆ’3π˜ƒ

 =    π˜‚ + π˜ƒ

 = 3π˜‚ βˆ’ 4π˜ƒ=

Solution

Given two vectors:

u=(βˆ’67),v=(6βˆ’9)\mathbf{u} = \begin{pmatrix} -6 \\ 7 \end{pmatrix}, \quad \mathbf{v} = \begin{pmatrix} 6 \\ -9 \end{pmatrix}

We are asked to find the following quantities:

  • 2u2\mathbf{u}
  • βˆ’3v-3\mathbf{v}
  • u+v\mathbf{u} + \mathbf{v}
  • 3uβˆ’4v3\mathbf{u} - 4\mathbf{v}

Step 1: Calculate 2u2\mathbf{u}

Multiply vector u\mathbf{u} by 2: 2u=2Γ—(βˆ’67)=(2Γ—βˆ’62Γ—7)=(βˆ’1214)2\mathbf{u} = 2 \times \begin{pmatrix} -6 \\ 7 \end{pmatrix} = \begin{pmatrix} 2 \times -6 \\ 2 \times 7 \end{pmatrix} = \begin{pmatrix} -12 \\ 14 \end{pmatrix}

Step 2: Calculate βˆ’3v-3\mathbf{v}

Multiply vector v\mathbf{v} by -3: βˆ’3v=βˆ’3Γ—(6βˆ’9)=(βˆ’3Γ—6βˆ’3Γ—βˆ’9)=(βˆ’1827)-3\mathbf{v} = -3 \times \begin{pmatrix} 6 \\ -9 \end{pmatrix} = \begin{pmatrix} -3 \times 6 \\ -3 \times -9 \end{pmatrix} = \begin{pmatrix} -18 \\ 27 \end{pmatrix}

Step 3: Calculate u+v\mathbf{u} + \mathbf{v}

Add vectors u\mathbf{u} and v\mathbf{v}: u+v=(βˆ’67)+(6βˆ’9)=(βˆ’6+67+βˆ’9)=(0βˆ’2)\mathbf{u} + \mathbf{v} = \begin{pmatrix} -6 \\ 7 \end{pmatrix} + \begin{pmatrix} 6 \\ -9 \end{pmatrix} = \begin{pmatrix} -6 + 6 \\ 7 + -9 \end{pmatrix} = \begin{pmatrix} 0 \\ -2 \end{pmatrix}

Step 4: Calculate 3uβˆ’4v3\mathbf{u} - 4\mathbf{v}

Multiply vector u\mathbf{u} by 3 and vector v\mathbf{v} by 4, then subtract: 3u=3Γ—(βˆ’67)=(βˆ’1821)3\mathbf{u} = 3 \times \begin{pmatrix} -6 \\ 7 \end{pmatrix} = \begin{pmatrix} -18 \\ 21 \end{pmatrix} 4v=4Γ—(6βˆ’9)=(24βˆ’36)4\mathbf{v} = 4 \times \begin{pmatrix} 6 \\ -9 \end{pmatrix} = \begin{pmatrix} 24 \\ -36 \end{pmatrix} Now subtract: 3uβˆ’4v=(βˆ’1821)βˆ’(24βˆ’36)=(βˆ’18βˆ’2421βˆ’(βˆ’36))=(βˆ’4257)3\mathbf{u} - 4\mathbf{v} = \begin{pmatrix} -18 \\ 21 \end{pmatrix} - \begin{pmatrix} 24 \\ -36 \end{pmatrix} = \begin{pmatrix} -18 - 24 \\ 21 - (-36) \end{pmatrix} = \begin{pmatrix} -42 \\ 57 \end{pmatrix}

Final Results:

  • 2u=(βˆ’1214)2\mathbf{u} = \begin{pmatrix} -12 \\ 14 \end{pmatrix}
  • βˆ’3v=(βˆ’1827)-3\mathbf{v} = \begin{pmatrix} -18 \\ 27 \end{pmatrix}
  • u+v=(0βˆ’2)\mathbf{u} + \mathbf{v} = \begin{pmatrix} 0 \\ -2 \end{pmatrix}
  • 3uβˆ’4v=(βˆ’4257)3\mathbf{u} - 4\mathbf{v} = \begin{pmatrix} -42 \\ 57 \end{pmatrix}

Do you have any questions or would you like more details?

5 Related Questions:

  1. How do you multiply a scalar with a vector?
  2. How is vector addition performed geometrically?
  3. What is the difference between scalar multiplication and vector subtraction?
  4. How do vectors behave under matrix transformations?
  5. Can vectors of different dimensions be added?

Tip:

When performing vector arithmetic, ensure that you are consistent with the dimensions of the vectors to avoid mistakes.

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Math Problem Analysis

Mathematical Concepts

Vector Arithmetic
Scalar Multiplication
Vector Addition
Vector Subtraction

Formulas

Scalar Multiplication: k * (a, b) = (k * a, k * b)
Vector Addition: (a, b) + (c, d) = (a + c, b + d)
Vector Subtraction: (a, b) - (c, d) = (a - c, b - d)

Theorems

Basic Properties of Vectors

Suitable Grade Level

Grades 9-12