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4.12 G) Vector Simultaneous Equations
4.12 G) Vector Simultaneous Equations
The addition of two vectors can be made harder by adding in some unknowns. We are now going to have a look at an example where this is the case.
Example 1
Given that
Given that
Find the value of a and b.
The a and b that are on the outside of the vectors just mean a lots of the first vector and b lots of the second vector. These are scalar vectors. We are able to get these unknown inside the vectors by multiplying both the x component and the y component of the column vector by the unknown on the outside of the column vectors. When we do this, we get what is shown below.
The a and b that are on the outside of the vectors just mean a lots of the first vector and b lots of the second vector. These are scalar vectors. We are able to get these unknown inside the vectors by multiplying both the x component and the y component of the column vector by the unknown on the outside of the column vectors. When we do this, we get what is shown below.
I am now going to add the vectors on the left side of the equation. When we add column vectors, we add straight across; we add the x components together (the top components), and add the y components together (the bottom components). The outcome for the addition of the two vectors on the left side of the equation is shown below:
In order for this equation to work, the x components and y components of the vector on the left side of the equation must be equal to the x components and y components of the vector on the right side of the equation. This means that 3a + b must equal 10 (from the x components), and 5a + 4b must equal 12 (from the y components). This gives us the following two equations:
We now have simultaneous equations with two unknowns. The first step in solving simultaneous equations is to get the number of one of the unknowns the same (ignoring the sign). I am going to get the b’s the same. Currently there is 1b in the first equation and 4b in the second equation. We can make the b’s 4b in both of the equations by multiplying the first equation by 4 and keeping the second equation as it is. The working for multiplying the first equation by 4 is shown below.
Our two equations (the modified first equation and the original second equation) are:
The next step is to use STOP, which stands for Same Take, Opposite Plus. The signs for the 4b’s are the same (both positive) and this means that we take the equations (Same Take).
The taking of the equations has eliminated the b’s, which results in an equation with only one unknown; the equation only contains a. We solve the equation to find the value of a in the usual way.
Currently, we have 7a in the equation and we want to find the value of a. We are able to do this by dividing both sides of the equation by 7.
Currently, we have 7a in the equation and we want to find the value of a. We are able to do this by dividing both sides of the equation by 7.
This means that a is 4.
We find the value of b by subbing the value of a (4) into one of the original equations; it does not matter which equation we use. I am going to sub a as 4 into the first non-multiplied equation.
We find the value of b by subbing the value of a (4) into one of the original equations; it does not matter which equation we use. I am going to sub a as 4 into the first non-multiplied equation.
We find the value of b by moving the 12 from the left side of the equation to the right. We are able to do this by taking 12 from both sides of the equation.
This tells us that b is -2. We can check that we have obtained the correct values for a and b by subbing them into the other simultaneous equation (the second original simultaneous equation). The other simultaneous equation is shown below, and I am going to sub a in as 4 and b in as -2.
This equation works, which means that we have found the correct values for a and b; a is 4 and b is -2.