Is this question about vectors faulty?

Question

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

This is a question from an entrance exam (JEE Advanced 2015). In the official answer key, all candidates were awarded the points for this question which led me to guess that the question is somehow incorrect or has information missing. The only problem I can see is that $\mathbf{p}$, $\mathbf{q}$ and $\mathbf{r}$ are not mentioned to be mutually perpendicular, otherwise we could have written $\mathbf{s}=4\dfrac{\mathbf{p}}{|\mathbf{p}|}+3\dfrac{\mathbf{q}}{|\mathbf{q}|}+5\dfrac{\mathbf{r}}{|\mathbf{r}|}$, but we would still not be able to proceed because $-\mathbf{p}+\mathbf{q}+\mathbf{r}$, $\mathbf{p}-\mathbf{q}+\mathbf{r}$ and $-\mathbf{p}-\mathbf{q}+\mathbf{r}$ aren't guaranteed to be mutually perpendicular.

My question is, I need a mathematical explanation that either:

• The problem is inconsistent.
• The problem has more than one correct answer.
• A crucial piece of information is missing.
  Basically, I need to know why this question is faulty.
  EDIT 1: I accepted the answer of user @DanielCunha because it was understandable at my level and the answer offered a pretty reasonable alternative to what could have been. However I feel that the bounty should be awarded to user @mweiss because they put more time and effort into answering the problem and making it easier for me to understand, as is evidenced by this post: Reference request: A primer on the difference between "Vectors" in linear algebra vs. "Vectors" in ohysics/vector calculus

Answer 1

For given linearly independent vectors $\mathbf{p}$, $\mathbf{q}$ and $\mathbf{r}$ $\in \mathbb{R}^3$, we have:

$\frac{p_1}{\|\mathbf{p}\|}\,s_1 + \frac{p_2}{\|\mathbf{p}\|}\,s_2 + \frac{p_3}{\|\mathbf{p}\|}\,s_3 = 4$

$\frac{q_1}{\|\mathbf{q}\|}\,s_1 + \frac{q_2}{\|\mathbf{q}\|}\,s_2 + \frac{q_3}{\|\mathbf{q}\|}\,s_3 = 3$

$\frac{r_1}{\|\mathbf{r}\|}\,s_1 + \frac{r_2}{\|\mathbf{r}\|}\,s_2 + \frac{r_3}{\|\mathbf{r}\|}\,s_3 = 5$

This system has a unique solution, so $\mathbf{s}$ is well defined in the problem, it is a known function of the unit vectors in the directions $\mathbf{p}$, $\mathbf{q}$ and $\mathbf{r}$.

However, this does not mean that $2\,x + y + z$ can be obtained as a scalar value, independent of $\mathbf{p}$, $\mathbf{q}$ and $\mathbf{r}$. As it is, if there is a solution, the resolution seems to be cumbersome.

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Since they removed the question, I agree that there must be something wrong. One possibility is that the people who proposed this question were considering that everything was defined as projections of the other vectors along $\mathbf{s}$, instead of projections of $\mathbf{s}$ along the other vectors ($\mathbf{p}$, $\mathbf{q}$, $\mathbf{r}$, $-\mathbf{p}+\mathbf{q}+\mathbf{r}$, $\mathbf{p}-\mathbf{q}+\mathbf{r}$ and $-\mathbf{p}-\mathbf{q}+\mathbf{r}$).

In this case, assuming valid vectors for $\mathbf{p}$, $\mathbf{q}$ and $\mathbf{r}$, the resolution is very direct:

$p_1\,s_1 + p_2\,s_2 + p_3\,s_3 = 4\,\|\mathbf{s}\|$

$q_1\,s_1 + q_2\,s_2 + q_3\,s_3 = 3\,\|\mathbf{s}\|$

$r_1\,s_1 + r_2\,s_2 + r_3\,s_3 = 5\,\|\mathbf{s}\|$

Therefore

$x = -4+3+5 = 4$

$y = 4-3+5 = 6$

$z = -4-3+5 = -2$

$2\,x + y + z = 8 + 6 - 2 = 12$

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Note that there exist valid solutions for this, e.g.:

$\mathbf{p} = [4\,\sqrt{3},0,0]$

$\mathbf{q} = [0,3\,\sqrt{3},0]$

$\mathbf{r} = [0,0,5\,\sqrt{3}]$

$\mathbf{s} = [1,1,1]$

Answer 2

TL;DR:

The OP is (perhaps reasonably) interpreting "component" in the vector calculus sense, in which "the component of $\mathbf{v}$ in the direction $\mathbf{w}$" means $\frac{\mathbf{v} \cdot \mathbf{w}}{\| \mathbf{w} \|}$, but I think the intent of the problem is to think of "component" in the linear algebra sense, i.e. as the coefficients of a vector relative to a basis (which need not be orthogonal nor orthonormal). Probably the problem was discarded precisely because of this ambiguity.

Full explanation:

Perpendicularity has nothing to do with this problem. The key concept is linear independence. The statement

Suppose that $\mathbf{p}$, $\mathbf{q}$ and $\mathbf{r}$ are three non-coplanar vectors in $\mathbb{R}^3$

is, I can only assume, a somewhat informal and imprecise way of saying that these three vectors are linearly independent. (See below for more on this...)

Because of the linear independence, any vector $\mathbf{v} \in \mathbb R^3$ can be written uniquely as a linear combination $a\mathbf{p} + b\mathbf{q} + c\mathbf{r}$. We are told

Let the components of a vector $\mathbf{s}$ along $\mathbf{p}$, $\mathbf{q}$ and $\mathbf{r}$ be $4$, $3$ and $5$, respectively

which means that $\mathbf{s} = 4\mathbf{p} + 3\mathbf{q} + 5\mathbf{r}$.

Now it is an interesting fact, which can be proven (consider this a side quest, if you will) that if $\mathbf{p}$, $\mathbf{q}$ and $\mathbf{r}$ are linearly independent, then so too are the three vectors $$\mathbf{w_1} = -\mathbf{p}+\mathbf{q}+\mathbf{r}$$ $$\mathbf{w_2} = \mathbf{p}-\mathbf{q}+\mathbf{r}$$ $$\mathbf{w_3} = -\mathbf{p}-\mathbf{q}+\mathbf{r}$$ which means that any vector $\mathbf{v}$ can also be written uniquely as a linear combination of $\mathbf{w_1}, \mathbf{w_2}$ and $\mathbf{w_3}$. We are told that the components of $\mathbf{s}$ along these vectors are $x, y, z$, which means that $\mathbf{s} = x\mathbf{w_1} + y\mathbf{w_2} + z\mathbf{w_3}$.

Equating these two representations together, we get

$$4\mathbf{p} + 3\mathbf{q} + 5\mathbf{r} = x(-\mathbf{p}+\mathbf{q}+\mathbf{r}) + y(\mathbf{p}-\mathbf{q}+\mathbf{r}) + z(-\mathbf{p}-\mathbf{q}+\mathbf{r})$$

Collecting like terms on the right hand side, we end up with the equation $$4\mathbf{p} + 3\mathbf{q} + 5\mathbf{r} = (-x+y-z)\mathbf{p} + (x-y-z)\mathbf{q} + (x+y+z)\mathbf{r}$$

Now because the vectors are linearly independent, the components on the left must exactly match the components on the right; that is, the values of $x, y, z$ must satisfy the system of equations $$-x + y - z = 4$$ $$x - y - z = 3$$ $$x + y + z = 5$$

This system has a unique solution, so we can unambiguously determine the values of $x, y$ and $z$, and hence calculate $2x + y + z$ as well.

So as far as I can tell, there is no fundamental problem with the question.

HOWEVER: As I briefly alluded to at the start of the problem, referring to the three vectors as "non-coplanar" (instead of "linearly independent", which is what the problem posers presumably intended) is both not standard and potentially misleading. It is commonplace to identify vectors with points, and (as user @DanielWainfleet observed in the comments under the OP) "any 3 points in $\mathbb R^3$ lie in a plane". So if you interpret the problem that way, the question makes no sense from the very beginning. I can only assume that this is the reason why they decided to accept all answers as correct: they realized, belatedly, that in choosing the informal phrase "non-coplanar" over the more precise "linearly independent", they were opening themselves up to an unintended second interpretation of the question, one in which it had no solution because the premise was impossible.