Math Problem Statement

. Let u = (1, −3, 3), w = (2, 1, −2), and 3u − 2v + 4w = (−1, 2, −3). Find (a) the vector v. (b) the linear combination: 3u − 5v + 2w

Solution

Given the vectors u=(1,3,3)u = (1, -3, 3), w=(2,1,2)w = (2, 1, -2), and the equation 3u2v+4w=(1,2,3)3u - 2v + 4w = (-1, 2, -3), we need to solve the following:

Part (a) - Finding the vector vv

We start with the equation: 3u2v+4w=(1,2,3)3u - 2v + 4w = (-1, 2, -3)

Substitute u=(1,3,3)u = (1, -3, 3) and w=(2,1,2)w = (2, 1, -2):

3(1,3,3)2v+4(2,1,2)=(1,2,3)3(1, -3, 3) - 2v + 4(2, 1, -2) = (-1, 2, -3)

First, calculate 3u3u and 4w4w:

3u=3(1,3,3)=(3,9,9)3u = 3(1, -3, 3) = (3, -9, 9)

4w=4(2,1,2)=(8,4,8)4w = 4(2, 1, -2) = (8, 4, -8)

Now substitute these into the equation:

(3,9,9)2v+(8,4,8)=(1,2,3)(3, -9, 9) - 2v + (8, 4, -8) = (-1, 2, -3)

Add the vectors 3u3u and 4w4w:

(3+8,9+4,98)=(11,5,1)(3 + 8, -9 + 4, 9 - 8) = (11, -5, 1)

So the equation becomes:

112v=(1,2,3)11 - 2v = (-1, 2, -3)

Now, isolate 2v2v by subtracting (11,5,1)(11, -5, 1) from both sides:

2v=(1,2,3)(11,5,1)-2v = (-1, 2, -3) - (11, -5, 1)

2v=(111,2+5,31)=(12,7,4)-2v = (-1 - 11, 2 + 5, -3 - 1) = (-12, 7, -4)

Finally, solve for vv by dividing each component by -2:

v=(122,72,42)=(6,72,2)v = \left(\frac{-12}{-2}, \frac{7}{-2}, \frac{-4}{-2}\right) = (6, -\frac{7}{2}, 2)

So, the vector vv is:

v=(6,72,2)\boxed{v = \left(6, -\frac{7}{2}, 2\right)}

Part (b) - Finding the linear combination 3u5v+2w3u - 5v + 2w

Now we calculate the linear combination 3u5v+2w3u - 5v + 2w.

Substitute the values of uu, vv, and ww:

3u=(3,9,9)3u = (3, -9, 9)

5v=5(6,72,2)=(30,352,10)5v = 5\left(6, -\frac{7}{2}, 2\right) = \left(30, -\frac{35}{2}, 10\right)

2w=2(2,1,2)=(4,2,4)2w = 2(2, 1, -2) = (4, 2, -4)

Now, compute the linear combination:

3u5v+2w=(3,9,9)(30,352,10)+(4,2,4)3u - 5v + 2w = (3, -9, 9) - \left(30, -\frac{35}{2}, 10\right) + (4, 2, -4)

Subtract 5v5v from 3u3u:

(330,9+352,910)=(27,182+352,1)=(27,172,1)(3 - 30, -9 + \frac{35}{2}, 9 - 10) = \left(-27, -\frac{18}{2} + \frac{35}{2}, -1\right) = \left(-27, \frac{17}{2}, -1\right)

Finally, add 2w2w:

(27+4,172+2,14)=(23,172+42,5)=(23,212,5)(-27 + 4, \frac{17}{2} + 2, -1 - 4) = \left(-23, \frac{17}{2} + \frac{4}{2}, -5\right) = \left(-23, \frac{21}{2}, -5\right)

So the linear combination 3u5v+2w3u - 5v + 2w is:

(23,212,5)\boxed{\left(-23, \frac{21}{2}, -5\right)}

Would you like more details or have any questions?


Here are five related questions to explore:

  1. How would you find 2u+vw2u + v - w given the vectors uu, vv, and ww?
  2. What is the dot product of vectors uu and vv?
  3. How do you determine if vectors uu and ww are orthogonal?
  4. Can you find the magnitude of the vector vv found in part (a)?
  5. How would you calculate the cross product of vectors uu and ww?

Tip: When solving vector equations, it's helpful to work systematically through each operation to avoid errors and ensure clarity.

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Math Problem Analysis

Mathematical Concepts

Vectors
Linear combinations

Formulas

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Theorems

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Suitable Grade Level

Advanced High School