Three vectors A,B,and C are related to one another such that
5A + 6B = 60C. A fourth vector D
exists such that 30C + D = A.
Determine an expression for D in terms of the vectors A and B.
1st: equation1: 5A+6B=60C equation2: 30C+D=A
2nd: divide equation1 by2-->((5A+6B=60C)/2)=5/2A+6/2B=30C now reduce itto 5/2A+3B=30C
3rd: subtract30C on both side of the equal sign on equation2 and get: D=A-30C
4th:subtitute for equation1 into the Nnew equation2 30C--> D=A-[5/2A+3B]
5th: distribute the negative sign and get-->D=A-5/2A-3B-->2/2A-5/2A-3B
6th: answer: -3/2A-3B
To determine an expression for vector D in terms of vectors A and B, let's solve the given equations step by step.
We are given the equation: 5A + 6B = 60C
Rearranging the equation, we have: C = (5A + 6B) / 60
Now, we are also given the equation: 30C + D = A
Substituting the value of C from the previous equation, we get: 30(5A + 6B) / 60 + D = A
Simplifying the equation, we have: (5A + 6B) / 2 + D = A
To isolate D, we can move the other terms to the other side of the equation: D = A - (5A + 6B) / 2
Expanding the right side of the equation, we get: D = A - (5A/2) - (6B/2)
Simplifying further, we have: D = A - (5/2)A - (3B)
Therefore, the expression for vector D in terms of vectors A and B is: D = A - (5/2)A - (3B)
To determine an expression for D in terms of the vectors A and B, we need to use the information provided in the given equations and rearrange them.
From the equation 5A + 6B = 60C, we can rewrite it as:
A = (60/5)C - (6/5)B
Now, let's substitute this expression for A into the equation 30C + D = A:
30C + D = (60/5)C - (6/5)B
To isolate D, we can move (60/5)C to the right side of the equation by subtracting it from both sides:
D = (60/5)C - (6/5)B - 30C
Simplifying further, we can combine like terms:
D = (12C - 30C)/5 - (6/5)B
Finally, we can simplify it even more:
D = -18C/5 - (6/5)B
Therefore, the expression for D in terms of the vectors A and B is -18C/5 - (6/5)B.