To express the vector \( R \) in terms of the vectors \( A \), \( B \), and \( C \), we start with the given definitions of these vectors:
1. \( A = 2p + 3q - r \)
2. \( B = p - 2q + 2r \)
3. \( C = -2p + q - 2r \)
4. \( R = 3p - q + 2r \)
We want to express \( R \) as a linear combination of \( A \), \( B \), and \( C \):
\[
R = xA + yB + zC
\]
where \( x \), \( y \), and \( z \) are the coefficients we need to determine.
### Step 1: Substitute \( A \), \( B \), and \( C \) into the equation
Substituting the expressions for \( A \), \( B \), and \( C \) into the equation for \( R \):
\[
R = x(2p + 3q - r) + y(p - 2q + 2r) + z(-2p + q - 2r)
\]
### Step 2: Expand the equation
Expanding the right-hand side:
\[
R = (2x + y - 2z)p + (3x - 2y + z)q + (-x + 2y - 2z)r
\]
### Step 3: Set up the equations by comparing coefficients
Now we compare the coefficients of \( p \), \( q \), and \( r \) from both sides of the equation:
1. Coefficient of \( p \): \( 2x + y - 2z = 3 \)
2. Coefficient of \( q \): \( 3x - 2y + z = -1 \)
3. Coefficient of \( r \): \( -x + 2y - 2z = 2 \)
### Step 4: Solve the system of equations
We now have a system of three equations:
1. \( 2x + y - 2z = 3 \) (Equation 1)
2. \( 3x - 2y + z = -1 \) (Equation 2)
3. \( -x + 2y - 2z = 2 \) (Equation 3)
#### Step 4a: Eliminate \( z \)
From Equation 1, we can express \( z \):
\[
z = \frac{2x + y - 3}{2}
\]
Substituting this into Equations 2 and 3 to eliminate \( z \).
#### Step 4b: Substitute \( z \) into Equation 2
Substituting \( z \) into Equation 2:
\[
3x - 2y + \frac{2x + y - 3}{2} = -1
\]
Multiplying through by 2 to eliminate the fraction:
\[
6x - 4y + 2x + y - 3 = -2
\]
Combining like terms:
\[
8x - 3y = 1 \quad \text{(Equation 4)}
\]
#### Step 4c: Substitute \( z \) into Equation 3
Substituting \( z \) into Equation 3:
\[
-x + 2y - 2\left(\frac{2x + y - 3}{2}\right) = 2
\]
This simplifies to:
\[
-x + 2y - (2x + y - 3) = 2
\]
Combining like terms:
\[
-x - 2x + 2y - y + 3 = 2
\]
This simplifies to:
\[
-3x + y = -1 \quad \text{(Equation 5)}
\]
### Step 5: Solve Equations 4 and 5
Now we solve the system of equations 4 and 5:
1. \( 8x - 3y = 1 \) (Equation 4)
2. \( -3x + y = -1 \) (Equation 5)
From Equation 5, we can express \( y \):
\[
y = 3x - 1
\]
Substituting \( y \) back into Equation 4:
\[
8x - 3(3x - 1) = 1
\]
This simplifies to:
\[
8x - 9x + 3 = 1
\]
Thus:
\[
-x + 3 = 1 \implies -x = -2 \implies x = 2
\]
Now substituting \( x = 2 \) back into Equation 5 to find \( y \):
\[
y = 3(2) - 1 = 6 - 1 = 5
\]
Now substituting \( x \) and \( y \) into the expression for \( z \):
\[
z = \frac{2(2) + 5 - 3}{2} = \frac{4 + 5 - 3}{2} = \frac{6}{2} = 3
\]
### Final Result
Thus, we have:
\[
x = 2, \quad y = 5, \quad z = 3
\]
Therefore, we can express \( R \) in terms of \( A \), \( B \), and \( C \):
\[
R = 2A + 5B + 3C
\]
