Question

The cartesian equation of line are 3x+1=6y-2=1-z. find its direction ratios and write down its equation in vector form

Answer 1

The given Cartesian equation of a line is:

$3x+1 = 6y-2 = 1-z$

$3(x+\frac{1}{3}) = 6(y-\frac{2}{6}) = -(z-1)$

$\frac{(x+\frac{1}{3})}{\frac{1}{3}} =\frac{(y-\frac{1}{3})}{\frac{1}{6}}= \frac{z-1}{-1}$

$\frac{(x-(\frac{-1}{3}))}{\frac{1}{3}} =\frac{(y-\frac{1}{3})}{\frac{1}{6}}= \frac{z-1}{-1}$

Multiplying the denominators of the given equation by '6', we get

$\frac{(x+(\frac{1}{3}))}{2} =\frac{(y-\frac{1}{3})}{1}= \frac{z-1}{-6}$

Therefore, the direction ratios of the given equation are (2,1,-6).

Now, we will determine the vector form of the given equation.

Let $\frac{(x+(\frac{1}{3}))}{2} =\frac{(y-\frac{1}{3})}{1}= \frac{z-1}{-6} = \lambda$

$x = 2 \lambda - \frac{1}{3}$ , $y= \lambda +\frac{1}{3}$ and $z= -6\lambda + 1$

So, the vector form of the given equation is:

$(2 \lambda - \frac{1}{3}) i + ( \lambda + \frac{1}{3}) j + (-6 \lambda+1) k$.