Q.Two vectors $\vec{u}$ and $\vec{v}$ are drawn in the $XY$-plane. Both lie to the right of the $Y$-axis, so both have positive $x$-components. $\vec{u}$ is directed upward and to the right (it points above the horizontal, so its $y$-component is positive), while $\vec{v}$ is directed downward and to the right (it slopes below the horizontal, so its $y$-component is negative). If $\vec{u} = a\hat{i} + b\hat{j}$ and $\vec{v} = p\hat{i} + q\hat{j}$, which of the following is correct?

A component is positive when the vector points along the $+$ axis. $\vec{u}$ (up-right) gives $a>0,b>0$; $\vec{v}$ (down-right) gives $p>0,q<0$.

Concept

For a vector $\vec{A}=A_x\hat{i}+A_y\hat{j}$, the sign of each component is fixed by direction: $A_x>0$ if it points toward $+x$ (right), $A_x<0$ if toward $-x$ (left); $A_y>0$ if it points toward $+y$ (up), $A_y<0$ if down.

Steps

1. Vector $\vec{u}$: it lies in the first quadrant, pointing up and to the right. Rightward $\Rightarrow a>0$; upward $\Rightarrow b>0$.
2. Vector $\vec{v}$: it points to the right but slopes downward (below the horizontal). Rightward $\Rightarrow p>0$; downward $\Rightarrow q<0$.
3. Collecting: $a>0,\ b>0,\ p>0,\ q<0$.

Why the others fail

• (A) claims $b<0$, but $\vec{u}$ points upward, so $b>0$. Wrong.
• (C) claims $p<0$, but $\vec{v}$ points to the right, so $p>0$. Wrong.
• (D) claims all positive, but $\vec{v}$ points downward, so $q<0$. Wrong.

Concept: Instantiate Concrete Numbers That Match the Picture, Then Just Read Off the Signs

Method: Pick an Explicit Numeric Vector for $\vec u$ and $\vec v$ (Not Abstract Quadrant/Sign Reasoning)

Both existing solutions reason abstractly about which quadrant each vector points into and what sign that implies for each component. This method instead writes down one concrete, fully numeric pair of vectors that matches the pictorial description exactly, computes their components directly, and checks each answer option by substitution — turning an abstract sign argument into simple arithmetic.

Step 1 — Build a concrete $\vec u$ matching "up and to the right"

Any vector with both components positive points up-and-to-the-right. Pick a simple example:

Both are chosen positive, exactly matching "lies to the right of the $Y$-axis" ($x$-component positive) and "directed upward" ($y$-component positive).

Step 2 — Build a concrete $\vec v$ matching "down and to the right"

The $x$-component is positive (still to the right of the $Y$-axis), but the $y$-component is negative (sloping below the horizontal, downward).

Step 3 — Read off the actual signs

Step 4 — Test every option by direct substitution

• (a) claims $a,p>0$ and $b,q<0$: but $b=4>0$ here, contradicting the claim — fails.
• (b) claims $a,p,b>0$ and $q<0$: matches exactly ($3>0,\,3>0,\,4>0,\,-4<0$) — holds.
• (c) claims $a,q,b>0$ and $p<0$: but $p=3>0$ here, contradicting the claim — fails.
• (d) claims all four positive: but $q=-4<0$ here, contradicting the claim — fails.

Why using concrete numbers instead of abstract reasoning is a genuine cross-check

The abstract argument ("up-right means both components positive") and this numeric instantiation reach the same conclusion, but the numeric version leaves no room for a sign-convention slip: once real numbers are written down, checking each option is pure substitution, not a second round of directional reasoning that could itself go wrong. Since $\vec u,\vec v$ were only required to match the two given directions (not any specific magnitude), any other numeric choice with the same signs — e.g. $\vec u=1\hat i+2\hat j$, $\vec v=5\hat i-1\hat j$ — would confirm the identical conclusion.

Final Answer

With $\vec u=3\hat i+4\hat j$ (up-right) and $\vec v=3\hat i-4\hat j$ (down-right): $a,b,p>0$ and $q<0$, matching option (b) exactly — $\boxed{a,\,p,\,b\text{ positive},\ q\text{ negative}}$.