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Three Dimensional Geometry

Question
CBSEENMA12033260

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

Solution
The cartesian equations of line are
                          3 straight x plus 1 space equals space 6 straight y minus 2 space equals space 1 minus straight z
or             3 open parentheses straight x plus 1 third close parentheses space equals space 6 space open parentheses straight y minus 1 third close parentheses space equals space minus left parenthesis straight z minus 1 right parenthesis
or          fraction numerator straight x minus open parentheses negative begin display style 1 third end style close parentheses over denominator begin display style 1 third end style end fraction space equals space fraction numerator straight y minus begin display style 1 third end style over denominator begin display style 1 over 6 end style end fraction space equals space fraction numerator straight z minus 1 over denominator negative 1 end fraction.
or              fraction numerator straight x minus open parentheses negative begin display style 1 third end style close parentheses over denominator 2 end fraction space equals space fraction numerator straight y minus begin display style 1 third end style over denominator 1 end fraction space equals space fraction numerator straight z minus 1 over denominator negative 6 end fraction
This is symmetric form of the given line.
This line passes through fixed point open parentheses negative 1 third comma space 1 third comma space 1 close parentheses and has direction ratios 2, 1, -6.
Here straight a with rightwards arrow on top space equals space minus 1 third straight i with overparenthesis on top space plus space 1 third straight j with hat on top plus straight k with hat on top comma space space space space space space straight m with rightwards arrow on top space equals space 2 straight i with hat on top plus straight j with hat on top space minus space 6 straight k with hat on top
therefore    vector equation of line is
                                     straight r with rightwards arrow on top space equals space straight a with rightwards arrow on top plus straight lambda straight m with rightwards arrow on top
or                                   straight r with rightwards arrow on top space equals space open parentheses negative 1 third straight i with hat on top plus 1 third straight j with overparenthesis on top plus straight k with hat on top close parentheses space plus space straight lambda space left parenthesis 2 straight i with hat on top plus straight j with hat on top minus 6 straight k with hat on top right parenthesis

Some More Questions From Three Dimensional Geometry Chapter

Find the direction cosines of x, y and z-axis.