Question

Find the vector and the cartesian equations of the line that passes through the points(3,-2,-5),(3,-2,6).

Answer 1

Let the line passing through the points P (3,-2,-5) and Q(3,-2,6), be PQ.

Since PQ passes through P (3,-2,-5), its position vector is given by, 

\(\overrightarrow a=3\widehat{i}-2\widehat j-5\widehat k\)
The direction ratios of PQ are given by, (3-3)=0, (-2+2)=0, (6+5)=11
The equation of the vector in the direction of PQ is 

\(\overrightarrow b=0 \widehat i-0. \widehat j+11 \widehat k=11 \widehat k\)

The equation of PQ in vector form is given by, \(\overrightarrow r= \overrightarrow a+ \lambda \overrightarrow b, \lambda \in R\)
\(\Rightarrow \overrightarrow r=(3 \widehat i-2\widehat j-5\widehat k)+11 \lambda \widehat k\)

The equation of PQ in cartesian form is

\(\frac{x-x_1}{a}=\frac{y-y_1}{b}=\frac{z-z_1}{c} i.e.,\frac{x-3}{0}=\frac{y+2}{0}=\frac{z+5}{11}\)