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प्रश्न
Find the direction cosines of the line \[\frac{4 - x}{2} = \frac{y}{6} = \frac{1 - z}{3} .\] Also, reduce it to vector form.
उत्तर
The cartesian equation of the given line is
\[\frac{4 - x}{2} = \frac{y}{6} = \frac{1 - z}{3}\]
It can be re-written as
\[\frac{x - 4}{- 2} = \frac{y - 0}{6} = \frac{z - 1}{- 3}\]
This shows that the given line passes through the point (4,0,1) and its direction ratios are proportional to -2,6,-3
So, its direction cosines are
\[\frac{- 2}{\sqrt{\left( - 2 \right)^2 + 6^2 + \left( - 3 \right)^2}}, \frac{6}{\sqrt{\left( - 2 \right)^2 + 6^2 + \left( - 3 \right)^2}}, \frac{- 3}{\sqrt{\left( - 2 \right)^2 + 6^2 + \left( - 3 \right)^2}}\]
\[ = \frac{- 2}{7}, \frac{6}{7}, \frac{- 3}{7} \]
Thus, the given line passes through the point having position vector \[\overrightarrow{a} = 4 \hat{i} + \hat{k} \] and is parallel to the vector \[\overrightarrow{b} = - 2 \hat{i} + 6 \hat{j} - 3 \hat{k}\]
We know that the vector equation of a line passing through a point with position vector `vec a` and parallel to the vector `vec b` is \[\overrightarrow{r} = \overrightarrow{a} + \lambda \overrightarrow{b}\] Here,
\[\overrightarrow{a} = 4 \hat{i} + \hat{k} \]
\[ \overrightarrow{b} = - 2 \hat{i} + 6 \hat{j} - 3 \hat{k} \]
Vector equation of the required line is
\[\overrightarrow{r} = \left( 4 \hat{i} + 0 \hat{j}+ \hat{k} \right) + \lambda \left( - 2 \hat{i} + 6 \hat{j} - 3 \hat{k} \right) \]
\[\text{ Here } , \lambda \text{ is a parameter } . \]
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