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I'm struggling with this problem:

Find the position vector a of a point on the line described by the equation that follows, and also find a vector b parallel to the line $$\frac{x + 1}3 = \frac{y + 4}{-3} = \frac z2$$

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1 Answer

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From

$$\frac{x + 1}3 = \frac{y + 4}{-3} = \frac z2$$

let $x=t\in \mathbb{R}$ and obtain the parametric equation for the line

$$x=t\implies y=-t-5 \quad z=\frac23(t+1)$$

thus the position vector is

$$P=\left(t,-t-5,\frac23(t+1)\right)=\left(0,-5,\frac23\right)+t\left(1,-1,\frac23\right)$$

thus a parallel vector is

$$b=\left(1,-1,\frac23\right)$$

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