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Three Dimensional Geometry

Long Answer Type

61. Find the vector equation of a line passing through a point with position vector

2 straight i with hat on top space minus space straight j with hat on top space plus space straight k with hat on top

and parallel to the line joining the points with position vectors

negative straight i with hat on top space plus space 4 space straight j with hat on top space plus space straight k with hat on top

and

straight i with hat on top space plus space 2 space straight j with hat on top space plus space 2 space straight k with hat on top

. Also, find the cartesian equivalent of this equations.

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Short Answer Type

62. The cartesian equations of a line are 6 x – 2 = – 3 y + 1 = 2 z = 2. Find the direction ratios of the line and write down the vector equation of the line through (2, – 1, –1) which is parallel to the given line.

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63. The points A (4, 5, 10), B (2, 3, 4) and C (1, 2, – 1) are three vertices of a parallelogram ABCD. Find vector and cartesian equations for the sides AB and BC and find the coordinates of D.

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Long Answer Type

64. A line passes through the point with position vector $2 space straight i with hat on top space minus space straight j with hat on top space plus space 4 space straight k with hat on top$ and is in direction $straight i with hat on top space plus space straight j with hat on top space minus space 2 space straight k with hat on top.$ Find equations for the line in vector and in cartesian form .

Here $straight a with rightwards arrow on top space equals space 2 space straight i with hat on top space minus space straight j with hat on top space space plus space 4 space straight k with hat on top comma space space space straight m with rightwards arrow on top space equals space straight i with hat on top space plus space straight j with hat on top space minus space 2 space straight k with hat on top$
Let $straight r with rightwards arrow on top$ be the position vector of any point P (x, y, z) on the line
∴ the vector equation of line is $straight r with rightwards arrow on top space equals space straight a with rightwards arrow on top space plus space straight lambda space straight m with rightwards arrow on top$
or    $straight r with rightwards arrow on top space equals space open parentheses 2 straight i with hat on top space minus space straight j with hat on top space plus space 4 space straight k with hat on top close parentheses space plus space straight lambda space open parentheses straight i with hat on top space plus space straight j with hat on top space minus space 2 space straight k with hat on top close parentheses$
This equation can be written as
   $straight x space straight i with hat on top space plus space straight y space straight j with hat on top space plus space straight z space straight k with hat on top space equals space left parenthesis 2 plus straight lambda right parenthesis space straight i with hat on top space plus space left parenthesis negative 1 plus space straight lambda right parenthesis space straight j with hat on top space plus space left parenthesis 4 minus 2 space straight lambda right parenthesis space straight k with hat on top$
Comparing coefficients of $straight i with hat on top comma space straight j with hat on top comma space straight k with hat on top comma$ we get,
$straight x space equals space 2 space plus space straight lambda comma space space space straight y space equals space minus 1 plus straight lambda comma space space straight z space equals space 4 minus 2 space straight lambda$
$rightwards double arrow space$    $straight x minus 2 space equals space straight lambda comma space space space straight y plus 1 space equals space straight lambda comma space space space fraction numerator straight z minus 4 over denominator negative 2 end fraction space equals space straight lambda$
Equationing values of λ, we get, $fraction numerator straight x minus 2 over denominator 1 end fraction space equals space fraction numerator straight y plus 1 over denominator 1 end fraction space equals space fraction numerator straight z minus 4 over denominator negative 2 end fraction$
which are cartesian equations of the line.