Suppose that p, q and r are three non-coplanar vectors in R³. Let the components of a vector s along

Q: Wait, why are we allowed to just equate the coefficients of $\vec{p}, \vec{q},$ and $\vec{r}$? Is that mathematically legal?

That is a fantastic question! Yes, it is absolutely legal because $\vec{p}, \vec{q},$ and $\vec{r}$ are non-coplanar. In linear algebra, this means they are linearly independent and form a basis for $\mathbb{R}^3$. If a set of vectors forms a basis, every vector in that space has a unique representation as a linear combination of those basis vectors. If you have two expressions for the same vector $\vec{s}$ in the same basis, the coefficients must be identical. It is the same principle as saying if $a\hat{i} + b\hat{j} + c\hat{k} = x\hat{i} + y\hat{j} + z\hat{k}$, then $a=x, b=y,$ and $c=z$.

Q: Is there a faster way to solve this than setting up a system of equations? Maybe using matrices?

You are thinking like a mathematician! Yes, you could represent this as a matrix equation $A\mathbf{X} = \mathbf{B}$, where $A$ is the matrix of coefficients of the new basis vectors, $\mathbf{X}$ is the column vector $[x, y, z]^T$, and $\mathbf{B}$ is $[4, 3, 5]^T$. You could then solve for $\mathbf{X}$ by finding $A^{-1}\mathbf{B}$. However, for a $3 \times 3$ system like this, the algebraic elimination method we used is often faster and less prone to calculation errors than inverting a matrix by hand during an exam.

Q: What would happen if the vectors $\vec{p}, \vec{q},$ and $\vec{r}$ were coplanar? Would the problem still work?

If the vectors were coplanar, they would be linearly dependent. This means one vector could be written as a combination of the other two. In that case, the representation of $\vec{s}$ would not be unique. You would have infinitely many solutions or no solution at all, depending on the vector $\vec{s}$. The problem would essentially break because you wouldn't have a valid 3D basis. The condition 'non-coplanar' is the bedrock that makes this entire solution possible!

Question

Suppose that $\vec{p}, \vec{q}$ and $\vec{r}$ are three non-coplanar vectors in $\mathbb{R}^3$. Let the components of a vector $\vec{s}$ along $\vec{p}, \vec{q}$ and $\vec{r}$ be 4, 3 and 5, respectively. If the components of this vector $\vec{s}$ along $(-\vec{p} + \vec{q} + \vec{r}), (\vec{p} - \vec{q} + \vec{r})$ and $(-\vec{p} - \vec{q} + \vec{r})$ are $x, y$ and $z$, respectively, then the value of $2x + y + z$ is .........

Sigma Percentile · Accepted Answer

Welcome, future engineer. Today, we are going to peel back the layers of a classic vector problem that tests not just your calculation skills, but your fundamental understanding of what a 'basis' really is.

Imagine you are standing in a room. To describe your position, you use the standard axes: forward, left, and up. These are your $\hat{i}, \hat{j}, \hat{k}$ vectors. But what if the world was skewed? What if your axes were tilted, stretched, and rotated? That is exactly what we are dealing with here.

We have three vectors, $\vec{p}, \vec{q},$ and $\vec{r}$. They are non-coplanar, which is our mathematical way of saying they are not trapped on a single flat sheet of paper. They span the entire 3D space and serve as our custom-made coordinate system.

### The First Perspective: The Given Reality

We are told that a vector $\vec{s}$ exists in this space. We don't know what $\vec{s}$ looks like, but we know its 'coordinates' in our custom system. It is composed of 4 units of $\vec{p}$, 3 units of $\vec{q}$, and 5 units of $\vec{r}$.

Mathematically, we write this as:
$$\vec{s} = 4\vec{p} + 3\vec{q} + 5\vec{r}$$

This is our anchor. This is the truth of what $\vec{s}$ is.

### The Second Perspective: The Transformation

Now, the problem introduces a new set of basis vectors: $(-\vec{p} + \vec{q} + \vec{r})$, $(\vec{p} - \vec{q} + \vec{r})$, and $(-\vec{p} - \vec{q} + \vec{r})$. It asks us to express $\vec{s}$ in terms of these new vectors with unknown components $x, y,$ and $z$.

So, we write:
$$\vec{s} = x(-\vec{p} + \vec{q} + \vec{r}) + y(\vec{p} - \vec{q} + \vec{r}) + z(-\vec{p} - \vec{q} + \vec{r})$$

I know, it looks like a mess of variables. But take a deep breath. This is just a change of perspective; the vector $\vec{s}$ hasn't changed, only the way we are describing it has.

### The Bridge: Uniqueness

Here is the magic moment. Because $\vec{p}, \vec{q},$ and $\vec{r}$ are non-coplanar, they form a basis. In any basis, a vector has a unique representation.

This means if we expand our second equation and group the terms by $\vec{p}, \vec{q},$ and $\vec{r}$, the coefficients must match our original equation. Expanding the terms, we get:
$$\vec{s} = (-x + y - z)\vec{p} + (x - y - z)\vec{q} + (x + y + z)\vec{r}$$

Now, compare this to $\vec{s} = 4\vec{p} + 3\vec{q} + 5\vec{r}$. The coefficient of $\vec{p}$ must be 4, the coefficient of $\vec{q}$ must be 3, and the coefficient of $\vec{r}$ must be 5.

### The Symphony of Equations

We have arrived at a system of three linear equations:
1) $-x + y - z = 4$
2) $x - y - z = 3$
3) $x + y + z = 5$

This is where the beauty of the problem reveals itself. Look at the symmetry! If we add equation (2) and equation (3), the $y$ and $z$ terms vanish instantly:
$$(x - y - z) + (x + y + z) = 3 + 5 \Rightarrow 2x = 8 \Rightarrow x = 4$$

Similarly, adding equation (1) and equation (3) makes the $x$ and $z$ terms vanish:
$$(-x + y - z) + (x + y + z) = 4 + 5 \Rightarrow 2y = 9 \Rightarrow y = \frac{9}{2}$$

Finally, substituting these into equation (3), we find:
$$4 + \frac{9}{2} + z = 5 \Rightarrow z = 5 - 4 - \frac{9}{2} = 1 - \frac{9}{2} = -\frac{7}{2}$$

### The Final Celebration

The question asks for the value of $2x + y + z$. We have our values: $x = 4, y = \frac{9}{2}, z = -\frac{7}{2}$.

Plugging them in:
$$2(4) + \frac{9}{2} - \frac{7}{2} = 8 + \frac{2}{2} = 8 + 1 = 9$$

And there it is. A complex-looking vector transformation collapses into a clean, elegant integer. The final answer is **9**.

Enter Numerical Value:

Visualized Solution

Defining the Vector $s$

Changing the Basis

Grouping the Terms

Uniqueness of Representation

Forming the Equations

Solving for $x$

Solving for $y$

Solving for $z$

Final Calculation

The Sigma Insight: Components of a Vector

The First Perspective

The Second Perspective

The Bridge

The Symphony of Equations

The Final Celebration

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