Which of the following cannot be the direction ratios of the straight line $\frac{x-3}{2}=\frac{2-y}{3}=\frac{z+4}{-1}$ ?

Direction ratios are the components of a vector that gives the direction of a line in 3D space. For a line in symmetric form like ours, the denominators directly give us the direction ratios.

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From our equation $\dfrac{x-3}{2} = \dfrac{2-y}{3} = \dfrac{z+4}{-1}$, let's rewrite the middle term to match the standard form:

$\dfrac{2-y}{3} = \dfrac{-1(y-2)}{3} = \dfrac{y-2}{-3}$

So our line can be written as: $\dfrac{x-3}{2} = \dfrac{y-2}{-3} = \dfrac{z+4}{-1}$

This means the direction ratios are proportional to $(2, -3, -1)$.

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A key concept to understand: Any scalar multiple of these ratios represents the same direction. This is because direction ratios define the orientation of the line, not its magnitude.

Think of it like this: If you have a vector pointing northeast, doubling or tripling its length doesn't change which way it points - it's still northeast.

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Now let's check each option:

Option 1: $(2, -3, -1)$

These are exactly our identified direction ratios, so they're valid.

Option 2: $(-2, 3, 1)$

These are $(-1) \times (2, -3, -1)$, so they're valid direction ratios.

Option 3: $(2, 3, -1)$

These numbers aren't proportional to our direction ratios.

Option 4: $(6, -9, -3)$

These are $3 \times (2, -3, -1)$, so they're valid direction ratios.

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Therefore, the option that cannot be the direction ratios of the given line is $(2, 3, -1)$.

Additional Learning: the direction ratios tell us how much we move in each coordinate direction as we travel along the line. If we move 2 units in the x-direction, we must move -3 units (not 3 units) in the y-direction to stay on our line. The middle value having the wrong sign means it points in the opposite direction from what our line requires.