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प्रश्न
The cartesian equations of a line are \[\frac{x - 5}{3} = \frac{y + 4}{7} = \frac{z - 6}{2} .\] Find a vector equation for the line.
उत्तर
The cartesian equation of the given line is \[\frac{x - 5}{3} = \frac{y + 4}{7} = \frac{z - 6}{2}\]
It can be re-written as
\[\frac{x - 5}{3} = \frac{y - \left( - 4 \right)}{7} = \frac{z - 6}{2}\]
Thus, the given line passes through the point having position vector \[\vec{a} = 5 \hat{i} - 4 \hat{j} + 6 \hat{k}\] and is parallel to the vector \[\vec{b} = 3 \hat{i} + 7 \hat{j}+ 2 \hat{k} \]
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 \[\vec{r} = \vec{a} + \lambda \vec{b}\]
Vector equation of the required line is
\[\vec{r} = \left( 5 \hat{i} - 4 \hat{j} + 6 \hat{k} \right) + \lambda \left( 3 \hat{i} + 7 \hat{j} + 2 \hat{k} \right)\]
\[\text{ Here} , \lambda \text{ is a parameter } . \]
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