Problem 7 - Advanced Calculus

Question

Solve for \(P\) in each of the following equations: \((a)(2,1,-3)+P=(0,2,4)\) \((b)(1,-1,4)+2 P=3 P+(2,0,5)\)

Step-by-Step Solution

Answer

The solution for vector \( P \) in equation (a) is (-2, 1, 7) and in equation (b) is (1, 1, 1)

Step 1: Understanding and Rearranging Equation (a)

The equation is \( (2,1,-3) + P = (0,2,4) \). Because addition in vectors is commutative, this equation can be rewritten to solve for \(P\) such that \( P = (0,2,4) - (2,1,-3) \) which implies that the addition or subtraction is done component-wise.

Step 2: Solving for \(P\) in Equation (a)

Therefore, by subtracting component-wise, the vector \( P \) is calculated as follows: \( P = (0 - 2, 2 - 1, 4 - (-3)) = (-2, 1, 7) \).

Step 3: Understanding and Rearranging Equation (b)

The equation is \( (1,-1,4) + 2P = 3P + (2,0,5) \). We can rearrange the equation by moving terms with \(P\) to one side and constants to the other side. The rearranged equation will be \( 2P - 3P = (2,0,5) - (1,-1,4) \). This reduces to \( P = (2,0,5) - (1,-1,4) \).

Step 4: Solving for \(P\) in Equation (b)

Therefore, by subtracting component-wise, the vector \(P\) is calculated as follows: \( P = (2 - 1, 0 - (-1), 5 - 4) = (1, 1, 1) \).

Key Concepts

Vector Addition

When dealing with vectors, addition is performed by adding corresponding components. Imagine you have two vectors, for example, \((x_1, y_1, z_1)\) and \((x_2, y_2, z_2)\). Vector addition would be performed as follows:

• Add the first components: \(x_1 + x_2\)
• Add the second components: \(y_1 + y_2\)
• Add the third components: \(z_1 + z_2\)

This gives you a new vector \((x_1 + x_2, y_1 + y_2, z_1 + z_2)\). Vector addition is used extensively in physics and engineering to find resultant vectors.
It's important to understand that like vector addition, subtraction is also done component-wise. So, subtract one vector from the other by subtracting each of its components.

Component-wise Operations

Component-wise operations mean handling each part, or component, of a vector separately. For example, to solve an equation like \(P = (0,2,4) - (2,1,-3)\), you'd perform the following steps:

• Subtract the first components: \(0 - 2 = -2\)
• Subtract the second components: \(2 - 1 = 1\)
• Subtract the third components: \(4 - (-3) = 4 + 3 = 7\)

Thus, the solution \( P = (-2, 1, 7)\) is reached by performing arithmetic separately on each corresponding pair of components. This approach allows us to solve vector equations in a straightforward manner without needing specialized software or calculations.

Equation Solving

Solving equations that involve vectors requires rearranging terms, much like in algebraic equations. Consider equation \( (1,-1,4) + 2P = 3P + (2,0,5) \).
To solve, first rearrange the terms so that all terms involving the unknown vector \(P\) are on one side. This brings us to \( 2P - 3P = (2,0,5) - (1,-1,4) \).
We aim to isolate \(P\) by combining like terms. Here, \(2P - 3P\) simplifies to \(-P\), so the equation becomes \(-P = (1,1,1)\). Multiply through by \(-1\) to solve for \(P\), thereby giving you \(P = (1,1,1)\).
Solving vector equations involves many familiar steps from basic algebra, such as combining like terms and rearranging.

Algebraic Manipulation

In vector algebra, manipulating equations often involves careful rearrangement to isolate the unknowns. This process is similar to manipulating algebraic equations but requires attention to vector components because each operation is carried out on every corresponding pair.
Let's revisit the equation \( (1,-1,4) + 2P = 3P + (2,0,5) \). By shifting all the \(P\)-related terms to one side, the equation becomes \( 2P - 3P = (2,0,5) - (1,-1,4) \).
Simplifying this yields \(-P = (1,1,1)\).
Multiply through by \(-1\) to resolve this into a simpler form: \( P = (1,1,1)\).
This algebraic manipulation allows us to solve the equation by aligning all the components clearly and consistently, ensuring each vector's structure is maintained throughout the process.