Given vectors u= (1,-1,3) and v= (-4,5,3) find an ordered triple that represents vectors 3u-2v.
a.(16,1,-7)
b.(11,-13,3)
c.(10,-12,1)
d.(7,-7,16)
I don't want you to straight up give me the answer. I need someone to teach me how to solve this problem. I am home schooled and i cant figure this out.
3 u = +3i - 3 j + 9k
2 v = -8i + 10j + 6k
-------------------- subtract
1 x = 11i - 13j + 3k
so I think b
Its B
Well, I'm here to help! To solve this problem, we need to perform scalar multiplication and vector addition.
First, let's find 3u and -2v:
3u = 3(1, -1, 3) = (3, -3, 9)
-2v = -2(-4, 5, 3) = (8, -10, -6)
Next, we need to add these two vectors together:
3u - 2v = (3, -3, 9) + (8, -10, -6) = (3+8, -3+(-10), 9+(-6)) = (11, -13, 3)
Therefore, the answer is (11, -13, 3), which corresponds to option b.(11,-13,3).
To find an ordered triple that represents the vector 3u-2v, we need to first calculate the values of 3u and 2v and then perform the subtraction.
Given vectors:
u = (1, -1, 3)
v = (-4, 5, 3)
To find 3u, multiply each component of u by 3:
3u = 3 * (1, -1, 3)
= (3, -3, 9)
To find 2v, multiply each component of v by 2:
2v = 2 * (-4, 5, 3)
= (-8, 10, 6)
Now, subtract 2v from 3u:
3u - 2v = (3, -3, 9) - (-8, 10, 6)
To subtract the vectors, subtract corresponding components:
(3 - (-8), -3 - 10, 9 - 6)
= (11, -13, 3)
Therefore, the correct option is b.(11, -13, 3), which represents the vector 3u - 2v.
Sure, I'd be happy to help you learn how to solve this problem!
To find the vector 3u - 2v, we need to multiply each component of u by 3 and each component of v by -2, and then add the resulting vectors together.
First, let's multiply vector u = (1, -1, 3) by 3:
3u = (3*1, 3*(-1), 3*3) = (3, -3, 9)
Next, let's multiply vector v = (-4, 5, 3) by -2:
-2v = (-2*(-4), -2*5, -2*3) = (8, -10, -6)
Now, we can add the resulting vectors together:
3u - 2v = (3, -3, 9) + (8, -10, -6) = (3+8, -3+(-10), 9+(-6)) = (11, -13, 3)
So, the ordered triple that represents the vector 3u - 2v is (11, -13, 3), which corresponds to option b.
Let me know if you have any further questions or if there's anything else I can help you with!