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Straight line in space — Maths STD 12 Science — Question

Maharashtra BoardEnglish MediumSTD 12 ScienceMathsStraight line in space2 Marks
Question
The equations of a line are given by $\frac{4-\text{x}}{3}=\frac{\text{y}+3}{3}=\frac{\text{z}+2}{6}.$ Write the direction cosines of a line parallel to this line.
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Answer

We have$\frac{4-\text{x}}{3}=\frac{\text{y}+3}{3}=\frac{\text{z}+2}{6}$
The equation of the given line can be re-written as.
$\frac{\text{x}-4}{-3}=\frac{\text{y}+3}{3}=\frac{\text{z}+2}{6}$
The direction ratios of the line parallel to the given line are proportional to -3, 3, 6.
Hence, the direction cosines of the line parallel to the given line are proportional to
$\frac{-3}{\sqrt{(-3)^2+3^2+6^2}},\frac{-3}{\sqrt{(3)^2+3^2+6^2}},\frac{6}{\sqrt{(-3)^2+3^2+6^2}}$
$=\frac{-1}{\sqrt{6}},\frac{1}{\sqrt{6}},\frac{2}{\sqrt{6}}$

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