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Three Dimensional Geometry

Short Answer Type

41. The equation of a line is

fraction numerator 2 straight x minus 5 over denominator 4 end fraction space equals space fraction numerator straight y plus 4 over denominator 3 end fraction space equals fraction numerator 6 minus straight z over denominator 6 end fraction

. Find the direction cosines of a line parallel to this line.

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Long Answer Type

42. 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.

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

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

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

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

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Short Answer Type

43. Find the equation of the line which passes through the point (3, 4, 5) and is parallel to the vector

2 straight i with hat on top plus 2 straight j with hat on top minus 3 straight k with hat on top.

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44. Find the vector and Cartesian equations of the line through the point (5, 2, – 4) and which is parallel to the vector

3 straight i with overparenthesis on top space plus space 2 straight j with overparenthesis on top space minus space 8 space straight k with overparenthesis on top.

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45.

Find the equation of the line which passes through the point (1, 2, 3) and is parallel to the vector $3 straight i with overparenthesis on top space plus space 2 straight j with overparenthesis on top space minus space 2 straight k with overparenthesis on top.$