Math Problem Statement

Test:MAC1114-T4 Question 14 of 20 This test: 20 point(s) possible This question: 1 point(s) possible

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A coordinate system has a horizontal axis marked with 10 equally spaced tick marks and a vertical axis marked with 10 equally spaced tick marks. The following three vectors are shown: u, from (5, 5) to (9, 3); v, from (2, 2) to (5, 2); w, from (3, 5) to (5, 9). Question content area bottom Part 1 Select the figure with a vector that is equivalent to 2 Bold v plus Bold u minus 2 Bold w. A.

A coordinate system has a horizontal axis marked with 16 equally spaced tick marks and a vertical axis marked with 12 equally spaced tick marks. A vector extends from (5, 11) to (11, 1). B.

A coordinate system has a horizontal axis marked with 16 equally spaced tick marks and a vertical axis marked with 12 equally spaced tick marks. A vector extends from (9, 9) to (7, 3). C.

A coordinate system has a horizontal axis marked with 16 equally spaced tick marks and a vertical axis marked with 12 equally spaced tick marks. A vector extends from (1, 3) to (15, 9). D.

A coordinate system has a horizontal axis marked with 16 equally spaced tick marks and a vertical axis marked with 12 equally spaced tick marks. A vector extends from (5, 1) to (11, 11). Select the figure with a vector that is equivalent to 2 Bold v plus Bold u minus 2 Bold w. Time Remaining: 02:20:50

Solution

To solve this problem, we need to compute the vector 2v+u−2w2\mathbf{v} + \mathbf{u} - 2\mathbf{w} based on the given vectors.

Step 1: Find the components of the vectors

  • Vector u\mathbf{u}: Initial point (5,5)(5, 5), terminal point (9,3)(9, 3)

    • u=(9−5,3−5)=(4,−2)\mathbf{u} = (9 - 5, 3 - 5) = (4, -2)

  • Vector v\mathbf{v}: Initial point (2,2)(2, 2), terminal point (5,2)(5, 2)

    • v=(5−2,2−2)=(3,0)\mathbf{v} = (5 - 2, 2 - 2) = (3, 0)

  • Vector w\mathbf{w}: Initial point (3,5)(3, 5), terminal point (5,9)(5, 9)

    • w=(5−3,9−5)=(2,4)\mathbf{w} = (5 - 3, 9 - 5) = (2, 4)

Step 2: Compute 2v,u,2\mathbf{v}, \mathbf{u}, and −2w-2\mathbf{w}

  • 2v=2⋅(3,0)=(6,0)2\mathbf{v} = 2 \cdot (3, 0) = (6, 0)
  • u=(4,−2)\mathbf{u} = (4, -2)
  • −2w=−2⋅(2,4)=(−4,−8)-2\mathbf{w} = -2 \cdot (2, 4) = (-4, -8)

Step 3: Add the vectors

2v+u−2w=(6,0)+(4,−2)+(−4,−8)2\mathbf{v} + \mathbf{u} - 2\mathbf{w} = (6, 0) + (4, -2) + (-4, -8) =(6+4−4,0−2−8)= (6 + 4 - 4, 0 - 2 - 8) =(6,−10)= (6, -10)

Step 4: Match the result to the given figures

The vector starts at (0,0)(0, 0) and ends at (6,−10)(6, -10). Check the figures for this endpoint.

  • Answer: The correct figure corresponds to Option D, which matches the vector extending from (5,1)(5, 1) to (11,11)(11, 11).

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Math Problem Analysis

Mathematical Concepts

Vector Operations
Coordinate Geometry

Formulas

Vector addition: (x1 + x2, y1 + y2)
Vector scalar multiplication: c(x, y) = (cx, cy)

Theorems

Basic Properties of Vectors

Suitable Grade Level

Grades 10-12

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