Question

The Cartesian equations of a line are \[3x - 1 = 6y + 2 = 1 - z\] . Find the fixed point through which it passes, its direction ratios and also its vector equation.

Answer 1

Hint: In order to solve this question, first of all we will rewrite the given line in the standard form of line i.e., \[\dfrac{{x - a}}{l} = \dfrac{{y - b}}{m} = \dfrac{{z - c}}{n}\] . Here, for this line \[\left( {a,b,c} \right)\] is the fixed point through which it passes and \[\left( {l,m,n} \right)\] are its direction ratios. Then we will find the vector equation of a line which is given by \[\overrightarrow r = a\hat i + b\hat j + c\hat k + \lambda \left( {l\hat i + m\hat j + n\hat k} \right)\] . And hence, we will get the required results.

Complete answer:
We are given a cartesian equation of a line as \[3x - 1 = 6y + 2 = 1 - z\]
We have to find the fixed point through which it passes, its direction ratios and its vector equation.
Now first of all, let’s rewrite the above equation in the standard form of line
i.e., \[\dfrac{{x - a}}{l} = \dfrac{{y - b}}{m} = \dfrac{{z - c}}{n}{\text{ }} - - - \left( i \right)\]
For this line \[\left( {a,b,c} \right)\] is the fixed point through which it passes and \[\left( {l,m,n} \right)\] are its direction ratios.
Now taking out \[3,6, - 1\] common form the first, second and third term of the given equation respectively, we have
\[3\left( {x - \dfrac{1}{3}} \right) = 6\left( {y + \dfrac{2}{6}} \right) = - \left( {z - 1} \right)\]
\[ \Rightarrow \dfrac{{\left( {x - \dfrac{1}{3}} \right)}}{{\dfrac{1}{3}}} = \dfrac{{\left( {y + \dfrac{2}{6}} \right)}}{{\dfrac{2}{6}}} = \dfrac{{\left( {z - 1} \right)}}{{ - 1}}{\text{ }} - - - \left( {ii} \right)\]
Now on comparing equation \[\left( i \right)\] and \[\left( {ii} \right)\] we get
\[a = \dfrac{1}{3},{\text{ }}b = \dfrac{2}{6},{\text{ }}c = 1\]
Therefore, the point through which the given line passes is \[\left( {\dfrac{1}{3},\dfrac{2}{6},1} \right)\]
Also, from the equation \[\left( i \right)\] and \[\left( {ii} \right)\] we get
\[l = \dfrac{1}{3},{\text{ }}m = \dfrac{2}{6},{\text{ }}n = - 1\]
Therefore, the direction ratios of the given line are \[\left( {\dfrac{1}{3},\dfrac{2}{6}, - 1} \right)\]
Now we know that if a line passes through points \[\left( {a,b,c} \right)\] and its direction ratios are \[\left( {l,m,n} \right)\] then its vector equation is given by \[\overrightarrow r = a\hat i + b\hat j + c\hat k + \lambda \left( {l\hat i + m\hat j + n\hat k} \right)\]
So, on substituting the values, we have
\[\overrightarrow r = \dfrac{1}{3}\hat i + \dfrac{2}{6}\hat j + \hat k + \lambda \left( {\dfrac{1}{3}\hat i + \dfrac{2}{6}\hat j - \hat k} \right)\]
which is the required vector equation of the given line.
Hence, we get the results as
The fixed point through which the line passes is \[\left( {\dfrac{1}{3},\dfrac{2}{6},1} \right)\]
The direction ratios of the given line is \[\left( {\dfrac{1}{3},\dfrac{2}{6}, - 1} \right)\]
The vector equation of the line is \[\overrightarrow r = \dfrac{1}{3}\hat i + \dfrac{2}{6}\hat j + \hat k + \lambda \left( {\dfrac{1}{3}\hat i + \dfrac{2}{6}\hat j - \hat k} \right)\]

Note:
In these type of questions, it is better to convert the given line into a standard form of line other than remembering multiple formulas to calculate the direction ratios etc. So be careful while taking out the common terms and putting the signs in the number. Also remember the vector equation of a line is used to identify the position vector of every point along the line.