Find the equation of line (vector and cartesian both) which is p

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 Multiple Choice QuestionsLong Answer Type

31. The projection of a line on the co-ordinate axes are 6, 2, 3. Find the length of the line and its direction cosines. Also, find the projection of the line segment joining (2, – 3, 1) to (4, 2, 3) on this line.
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32. A (– 1, 2, – 3), B (5, 0, – 6), C (0, 4, – 1) are three points. Show that the direction cosines of the bisectors of     are proportional to 25, 8, 5 and -11, AB space equals space square root of left parenthesis 5 plus 1 right parenthesis squared plus left parenthesis 0 minus 2 right parenthesis squared plus left parenthesis negative 6 plus 3 right parenthesis squared end root
space space space space space space equals space square root of 36 plus 4 plus 9 end root space equals space square root of 49 space equals space 7
AC space equals space square root of left parenthesis 0 plus 1 right parenthesis squared plus left parenthesis 4 minus 2 right parenthesis squared plus left parenthesis negative 1 plus 3 right parenthesis squared end root
space space space space space space space space equals space square root of 1 plus 4 plus 4 end root space equals space square root of 9 space equals space 3
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 Multiple Choice QuestionsShort Answer Type

33. The cartesian equations of a line are fraction numerator straight x minus 5 over denominator 3 end fraction space equals space fraction numerator straight y plus 4 over denominator 7 end fraction space equals fraction numerator straight z minus 6 over denominator 2 end fraction. Find a vector equation for the line.
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34.

The Cartesian equation of a line is fraction numerator straight x plus 3 over denominator 2 end fraction space equals space fraction numerator straight y minus 5 over denominator 4 end fraction space equals fraction numerator straight z plus 6 over denominator 2 end fraction. Find the vector equation for the line. 

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35.

Find the cartesian equation of the line which passes through the point (-2, 4, -5) and is parallel to the line given by fraction numerator straight x plus 3 over denominator 3 end fraction space equals space fraction numerator straight y minus 4 over denominator 5 end fraction space equals fraction numerator straight z plus 8 over denominator 6 end fraction

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36.

Find the equations of a line which is parallel to the line fraction numerator straight x minus 2 over denominator 3 end fraction space equals fraction numerator straight y plus 1 over denominator 1 end fraction space equals space fraction numerator straight z minus 7 over denominator 9 end fraction and which passes through the point (3, 0, 5).

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 Multiple Choice QuestionsLong Answer Type

37.

For the cartesian and vector equation of a line which passes through the point (1, 2, 3) and is parallel to the line fraction numerator negative straight x minus 2 over denominator 1 end fraction space equals space fraction numerator straight y plus 3 over denominator 7 end fraction space equals space fraction numerator 2 straight z minus 6 over denominator 3 end fraction.

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 Multiple Choice QuestionsShort Answer Type

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38.

Find the equation of line (vector and cartesian both) which is parallel to the vector 2 straight i with hat on top space minus straight j with hat on top plus 3 space straight k with hat on top and which passes through the point (5, -2, 4).


We know that the equation of a straight line passing through a fixed point with position vector straight a with rightwards arrow on top and parallel to the vector straight m with rightwards arrow on top is 
                              straight r with rightwards arrow on top space equals straight a with rightwards arrow on top plus straight lambda straight m with rightwards arrow on top                         ...(1)
where λ is a parameter.
Here straight a with rightwards arrow on top space equals space 5 space straight i with hat on top space minus space 2 space straight j with hat on top space plus space 4 space straight k with hat on top  and  straight m with rightwards arrow on top space equals space 2 space straight i with hat on top space minus space straight j with hat on top space plus space 3 space straight k with hat on top
∴ from (1), the vector equation of line is
straight r with rightwards arrow on top space equals space open parentheses 5 space straight i with hat on top space minus space 2 stack space straight j with hat on top space plus space 4 space straight k with hat on top close parentheses space space plus space straight lambda space open parentheses 2 space straight i with overparenthesis on top space minus space straight j with overparenthesis on top plus space 3 space straight k with overparenthesis on top close parentheses                    ...(2)
Now  straight r with rightwards arrow on top space equals space straight x straight i with hat on top space plus space straight y space straight j with hat on top space plus straight z space straight k with hat on top space space as space straight r with rightwards arrow on top is the position vector of (x, y, z).
∴ from (2), we get,
straight x straight i with hat on top space plus space straight y space straight j with hat on top plus space straight z space straight k with hat on top space equals space left parenthesis 2 straight lambda plus 5 right parenthesis space straight i with hat on top space minus space left parenthesis straight lambda plus 2 right parenthesis space straight j with hat on top plus left parenthesis 3 space straight lambda space plus space 4 right parenthesis space straight k with overparenthesis on top
Comparing the coefficients of straight i with hat on top comma space straight j with hat on top comma space stack straight k comma with hat on top space we space get comma
x = 2 λ + 5, y = – (λ + 2), z = 3 λ + 4
This is the parametric form of the equation.
Again, fraction numerator straight x minus 5 over denominator 2 end fraction space equals fraction numerator straight y plus 2 over denominator negative 1 end fraction equals fraction numerator straight z minus 4 over denominator 3 end fraction space equals straight lambda
∴ cartesian form of the line is   fraction numerator straight x minus 5 over denominator 2 end fraction space equals fraction numerator straight y plus 2 over denominator negative 1 end fraction space equals fraction numerator straight z minus 4 over denominator 3 end fraction
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39. Find the vector equation of the line passing through the point (2, –1, –1) which is parallel to the lines 6 x – 2 = 3 y + 1 = 2 z – 2.
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40.

The equation of a line is given by fraction numerator 4 minus straight x over denominator 2 end fraction space equals space fraction numerator straight y plus 3 over denominator 3 end fraction space equals fraction numerator straight z plus 2 over denominator 6 end fraction. Write the direction cosines of a line parallel to the above line. 

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