Discussion
The equation of a line passing through (-1,2,-4) and parallel to the straight line \(\frac{-x - 1}{4} = \frac{2y + 1}{-1} = \frac{-z + 4}{3}\) , is
Answer: C
The line is \(\vec{r} = (-\hat{i} + 2\hat{j} - 4\hat{k}) + t(8\hat{i} + \hat{j} + 6\hat{k})\).
Rewrite the given line in standard form \(\frac{x - x_1}{l} = \frac{y - y_1}{m} = \frac{z - z_1}{n}\):
\[ \begin{aligned} \frac{x + 1}{-4} &= \frac{y + \frac{1}{2}}{-\frac{1}{2}} \\ &= \frac{z - 4}{-3} \end{aligned} \]
So its direction ratios are \((-4, -\frac{1}{2}, -3)\), proportional to \((8, 1, 6)\). A parallel line through \((-1, 2, -4)\) is therefore \(\vec{r} = (-\hat{i} + 2\hat{j} - 4\hat{k}) + t(8\hat{i} + \hat{j} + 6\hat{k})\).
The last option has the right direction but passes through \((-1, 2, -6)\). The others read the denominators without first making the coefficients of \(x\), \(y\), \(z\) equal to 1.
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