Solving Vector Equations for Unknown Components

Why Solve a Vector Equation Component by Component?

A vector equation can look like a single statement, but in three dimensions it contains information about three coordinates. When some components are unknown, the equation can be translated into ordinary scalar equations that are solved one at a time.

This approach is useful whether the equation involves vector addition, scalar multiplication, or both. The key is to keep corresponding components aligned: first with first, second with second, and third with third.

Definition: A vector equation in \(\mathbb{R}^3\) is an equality between vectors whose components may include known or unknown real numbers. It is true exactly when the corresponding first, second, and third components are equal.

We write a vector in component form as \((a,b,c)\). Thus, solving a vector equation means finding values for its unknowns so that all three coordinate equalities hold at once.

The Componentwise Principle

Suppose two vectors in \(\mathbb{R}^3\) are equal. Their first components must match, their second components must match, and their third components must match. Therefore, a vector equation can be expanded into three scalar equations.

(a_1,a_2,a_3)=(b_1,b_2,b_3) \quad\Longleftrightarrow\quad a_1=b_1,\qquad a_2=b_2,\qquad a_3=b_3.

This principle is the reason the method works: equality of vectors is defined component by component. It is not enough for two coordinates to agree; all three must agree.

Theorem: If the two sides of a vector equation in \(\mathbb{R}^3\) are written in component form, the vector equation is equivalent to the three scalar equations obtained by equating corresponding components.

Justification. If the vectors are equal, they are the same ordered triple, so each coordinate in one triple equals the coordinate in the same position in the other. Conversely, if all three corresponding coordinates are equal, the ordered triples have exactly the same entries, so they are equal vectors. This proves both directions.

Vector addition and scalar multiplication are also defined componentwise. For vectors \((a,b,c)\), \((d,e,f)\) and a real scalar \(k\),

(a,b,c)+(d,e,f)=(a+d,b+e,c+f), \qquad k(a,b,c)=(ka,kb,kc).

These rules let us expand an equation before solving it. If the scalar multiplying a vector is known, it multiplies each of that vector’s components. If the scalar itself is unknown, the same unknown scalar appears in each coordinate equation.

Solving an Equation with Vector Addition

Worked Example: Find the Unknown Components in a Sum

Solve for \(x\), \(y\), and \(z\) in

$$ (x,2,y)+(3,z,-1)=(5,7,4). $$

Step 1: Match corresponding components. The first coordinate on the left is \(x+3\). The second is \(2+z\), and the third is \(y-1\). Thus the vector equation is equivalent to

x+3=5,\qquad 2+z=7,\qquad y-1=4.

Step 2: Solve each scalar equation.

x=5-3=2,\qquad z=7-2=5,\qquad y=4+1=5.

Therefore, \(x=2\), \(y=5\), and \(z=5\). The order in which the equations are solved does not have to match the order in which the unknowns were named; each equation determines the variable in its coordinate.

Check. Substitute the values into the original equation:

$$ (2,2,5)+(3,5,-1) =(2+3,2+5,5-1) =(5,7,4). $$

This is the required right-hand vector, so the solution is correct.

Including Scalar Multiplication

A scalar multiplying a vector scales every component, not just one of them. For instance, \(3(x,-2,z)\) means \((3x,-6,3z)\). After expanding, compare the resulting coordinates with the corresponding coordinates on the other side.

Worked Example: Solve a Scalar-Multiplication Equation

Find \(x\) and \(z\) in

$$ 3(x,-2,z)=(12,-6,15). $$

Step 1: Multiply every component by 3.

$$ 3(x,-2,z)=(3x,-6,3z). $$

Step 2: Equate the corresponding components.

3x=12,\qquad -6=-6,\qquad 3z=15.

The second equation is already true. The first and third give

x=\frac{12}{3}=4,\qquad z=\frac{15}{3}=5.

Check. Substituting the values yields

$$ 3(4,-2,5)=(12,-6,15). $$

All three components match, including the middle component, which serves as a consistency check even though it did not contain an unknown.

Combining Addition and Scalar Multiplication

When both operations occur, expand the scalar multiplication first and then perform the componentwise addition or subtraction. Each coordinate of the resulting vector equation gives one scalar equation. Those equations may contain different unknowns or may involve the same unknowns together.

Worked Example: Solve a Mixed Vector Equation

Solve for \(x\), \(y\), and \(z\):

$$ 2(x,y,-1)-(3,1,z)=(5,7,-4). $$

Step 1: Expand the scalar multiplication.

$$ 2(x,y,-1)=(2x,2y,-2). $$

Step 2: Subtract corresponding components.

$$ (2x,2y,-2)-(3,1,z) =(2x-3,2y-1,-2-z). $$

Equating this vector with the right-hand side gives

2x-3=5,\qquad 2y-1=7,\qquad -2-z=-4.

Step 3: Solve the three equations.

2x=8 \quad\Longrightarrow\quad x=4, $$ $$ 2y=8 \quad\Longrightarrow\quad y=4, $$ $$ -z=-2 \quad\Longrightarrow\quad z=2.

Check. Substitute \(x=4\), \(y=4\), and \(z=2\) into the original equation:

$$ 2(4,4,-1)-(3,1,2) =(8,8,-2)-(3,1,2) =(8-3,8-1,-2-2) =(5,7,-4). $$

The result equals the right-hand side, so the solution satisfies the original vector equation.

When the Unknown Is a Scalar

Sometimes the unknown is a single scalar that appears in all three coordinates. The coordinate equations must then give the same value for that scalar. This provides a useful test: values obtained from different coordinates must agree.

Worked Example: Determine a Common Scalar

Find \(\lambda\) such that

\lambda(2,-1,3)+(1,4,-2)=(9,0,10).

Step 1: Expand the scalar multiplication and add.

(2\lambda,-\lambda,3\lambda)+(1,4,-2) =(2\lambda+1,4-\lambda,3\lambda-2).

Step 2: Equate components.

2\lambda+1=9,\qquad 4-\lambda=0,\qquad 3\lambda-2=10.

Solving each equation gives

2\lambda=8 \quad\Longrightarrow\quad \lambda=4, $$ $$ -\lambda=-4 \quad\Longrightarrow\quad \lambda=4, $$ $$ 3\lambda=12 \quad\Longrightarrow\quad \lambda=4.

All three coordinates agree on \(\lambda=4\). Check:

$$ 4(2,-1,3)+(1,4,-2) =(8,-4,12)+(1,4,-2) =(9,0,10). $$

Thus \(\lambda=4\) satisfies the vector equation.

Consistency and No-Solution Cases

A vector equation may have no solution if the coordinate equations contradict one another. This can happen especially when one unknown scalar appears in multiple coordinates. In such a case, matching one or two components does not establish a solution; the remaining component must also match.

For example, the equation \(\lambda(2,-1,3)=(6,-3,8)\) would require \(2\lambda=6\), so \(\lambda=3\). The second coordinate would require \(-\lambda=-3\), which also gives \(\lambda=3\). But the third coordinate would require \(3\lambda=8\), hence \(\lambda=\frac{8}{3}\). Since \(3\ne\frac{8}{3}\), no single value of \(\lambda\) satisfies all three equations, and the vector equation has no solution.

When each unknown appears in only one coordinate, as in the first example, the equations can be solved separately. When the same unknown appears in several coordinates, compare the resulting conditions. They must all be compatible.

A Common Component-Matching Mistake

A frequent error is to match components that happen to look convenient rather than components in the same position. For example, consider

$$ (x,2,1)+(3,y,-2)=(7,5,-1). $$

An incorrect calculation might pair the \(2\) in the first vector with the \(3\) in the second and write \(2+3=5\), concluding that the middle coordinate is already satisfied and that \(y\) is irrelevant. That calculation adds a second component to a first component, so it does not follow the definition of vector addition.

The correct expansion adds first to first, second to second, and third to third:

$$ (x+3,2+y,1-2)=(7,5,-1). $$

Therefore, the scalar equations are

x+3=7,\qquad 2+y=5,\qquad 1-2=-1.

They give \(x=4\), \(y=3\), and a true third-coordinate equation. Substitution confirms the result:

$$ (4,2,1)+(3,3,-2)=(7,5,-1). $$

Key takeaway: Expand operations using the component rules, equate entries in the same position, solve all resulting scalar equations, and check the proposed values in the original vector equation.

Check Your Understanding

For each question, write the coordinate equations before solving. Check any proposed solution by substituting it into the vector equation.

Solve for \(a\), \(b\), and \(c\): \((a,1,b)+(2,c,4)=(7,6,3)\).
Find \(p\) and \(q\): \(2(p,-3,q)=(10,-6,14)\).
Solve for \(x\) and \(y\): \(3(x,y,2)+(1,-2,0)=(10,4,6)\).
Does a real number \(k\) satisfy \(k(1,2,3)=(4,8,12)\)? Explain by comparing all three coordinates.
Can a real number \(t\) satisfy \(t(2,1,-1)=(6,3,4)\)? Identify which coordinate equations are incompatible, if any.