Given the vectors u=(1,−3,3), w=(2,1,−2), and the equation 3u−2v+4w=(−1,2,−3), we need to solve the following:
Part (a) - Finding the vector v
We start with the equation:
3u−2v+4w=(−1,2,−3)
Substitute u=(1,−3,3) and w=(2,1,−2):
3(1,−3,3)−2v+4(2,1,−2)=(−1,2,−3)
First, calculate 3u and 4w:
3u=3(1,−3,3)=(3,−9,9)
4w=4(2,1,−2)=(8,4,−8)
Now substitute these into the equation:
(3,−9,9)−2v+(8,4,−8)=(−1,2,−3)
Add the vectors 3u and 4w:
(3+8,−9+4,9−8)=(11,−5,1)
So the equation becomes:
11−2v=(−1,2,−3)
Now, isolate 2v by subtracting (11,−5,1) from both sides:
−2v=(−1,2,−3)−(11,−5,1)
−2v=(−1−11,2+5,−3−1)=(−12,7,−4)
Finally, solve for v by dividing each component by -2:
v=(−2−12,−27,−2−4)=(6,−27,2)
So, the vector v is:
v=(6,−27,2)
Part (b) - Finding the linear combination 3u−5v+2w
Now we calculate the linear combination 3u−5v+2w.
Substitute the values of u, v, and w:
3u=(3,−9,9)
5v=5(6,−27,2)=(30,−235,10)
2w=2(2,1,−2)=(4,2,−4)
Now, compute the linear combination:
3u−5v+2w=(3,−9,9)−(30,−235,10)+(4,2,−4)
Subtract 5v from 3u:
(3−30,−9+235,9−10)=(−27,−218+235,−1)=(−27,217,−1)
Finally, add 2w:
(−27+4,217+2,−1−4)=(−23,217+24,−5)=(−23,221,−5)
So the linear combination 3u−5v+2w is:
(−23,221,−5)
Would you like more details or have any questions?
Here are five related questions to explore:
- How would you find 2u+v−w given the vectors u, v, and w?
- What is the dot product of vectors u and v?
- How do you determine if vectors u and w are orthogonal?
- Can you find the magnitude of the vector v found in part (a)?
- How would you calculate the cross product of vectors u and w?
Tip: When solving vector equations, it's helpful to work systematically through each operation to avoid errors and ensure clarity.