Determine whether the following pair of lines intersect:

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Three Dimensional Geometry

Long Answer Type

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131. Determine whether the following pair of lines intersect:
$fraction numerator straight x minus 1 over denominator 2 end fraction space equals fraction numerator straight y plus 1 over denominator 3 end fraction space equals straight z$

$fraction numerator straight x plus 5 over denominator 5 end fraction space equals space fraction numerator straight y minus 2 over denominator 1 end fraction comma space space straight z space equals space 2$

The equations of given lines are:

fraction numerator straight x minus 1 over denominator 2 end fraction space equals fraction numerator straight y plus 1 over denominator 3 end fraction space equals straight z comma space space fraction numerator straight x plus 1 over denominator 5 end fraction space equals space fraction numerator straight y minus 2 over denominator 1 end fraction semicolon space space straight z space equals space 2

or

fraction numerator straight x minus 1 over denominator 2 end fraction space equals space fraction numerator straight y plus 1 over denominator 3 end fraction space space equals fraction numerator straight z minus 0 over denominator 1 end fraction

and

fraction numerator straight x plus 1 over denominator 5 end fraction space equals space fraction numerator straight y minus 2 over denominator 1 end fraction space equals space fraction numerator straight z minus 2 over denominator 0 end fraction

The equations of two lines are

straight r with rightwards arrow on top space equals space straight i with hat on top space minus space straight j with hat on top space plus space straight lambda open parentheses 2 straight i with hat on top space plus space 3 space straight j with hat on top space plus space straight k with hat on top close parentheses

...(1)
and

straight r with rightwards arrow on top space equals space minus 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 space plus space straight mu open parentheses 5 space straight i with hat on top space plus space straight j with hat on top close parentheses

...(2)
Comparing these equations with

straight r with rightwards arrow on top space equals space stack straight a subscript 1 with rightwards arrow on top space plus space straight lambda space stack straight b subscript 1 with rightwards arrow on top space and space straight r with rightwards arrow on top space equals space stack straight a subscript 2 with rightwards arrow on top space plus space straight mu space stack straight b subscript 2 with rightwards arrow on top comma space space we space get comma stack straight a subscript 2 with rightwards arrow on top space equals space minus 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 comma space space stack straight b subscript 2 with rightwards arrow on top space equals space 5 straight i with hat on top space plus space straight j with hat on top

Let S be points on line (1) with position vector

stack straight a subscript 1 with rightwards arrow on top

and T be point on line (2) with position vector

stack straight alpha subscript 2 with rightwards arrow on top

so that

ST with rightwards arrow on top space equals space stack straight a subscript 2 with rightwards arrow on top space minus space stack straight a subscript 1 with rightwards arrow on top space equals space minus 2 space straight i with hat on top space plus space 3 space straight j with hat on top space plus space 2 space straight k with hat on top

stack straight b subscript 1 with rightwards arrow on top space cross times space stack straight b subscript 2 with rightwards arrow on top space equals space open vertical bar table row cell straight i with hat on top end cell cell straight j with hat on top end cell cell straight k with hat on top end cell row 2 3 1 row 5 1 0 end table close vertical bar space equals space left parenthesis 0 minus 1 right parenthesis space straight i with hat on top space minus space left parenthesis 0 minus 5 right parenthesis space straight j with hat on top space plus space left parenthesis 2 minus 15 right parenthesis space straight k with hat on top space equals space minus straight i with hat on top space plus space 5 space straight j with hat on top space minus space 13 space straight k with hat on top

therefore space space space space space open vertical bar stack straight b subscript 1 with rightwards arrow on top cross times stack straight b subscript 2 with rightwards arrow on top close vertical bar space equals space square root of 1 plus 25 plus 169 end root space equals space square root of 195

Let

PQ with rightwards arrow on top

be the S.D. vector between given lines. Therefore, it is parallel to

stack straight b subscript 1 with rightwards arrow on top cross times stack straight b subscript 2 with rightwards arrow on top.

If

straight n with rightwards arrow on top

is a unit vector along

PQ with rightwards arrow on top

, then

straight n with rightwards arrow on top space equals space fraction numerator stack straight b subscript 1 with rightwards arrow on top cross times stack straight b subscript 2 with rightwards arrow on top over denominator open vertical bar stack straight b subscript 1 with rightwards arrow on top cross times stack straight b subscript 2 with rightwards arrow on top close vertical bar end fraction space equals space fraction numerator 1 over denominator square root of 195 end fraction left parenthesis negative straight i with hat on top space plus space 5 space straight j with hat on top space minus space 13 space space straight k with hat on top right parenthesis

Now S.D. = Projection of

ST with rightwards arrow on top space on space PQ with rightwards arrow on top space equals space Projection space of space ST with rightwards arrow on top space on space straight n with rightwards arrow on top space equals space ST with rightwards arrow on top space. space straight n with rightwards arrow on top

equals space open parentheses negative 2 straight i with hat on top space plus space 3 space straight j with hat on top space plus space 2 space straight k with hat on top close parentheses. space fraction numerator 1 over denominator square root of 195 end fraction left parenthesis negative space straight i with hat on top space plus space 5 space straight j with hat on top space minus space 13 space straight k with hat on top right parenthesis space equals space fraction numerator 1 over denominator square root of 195 end fraction left parenthesis 2 plus 15 minus 26 right parenthesis space equals negative fraction numerator 9 over denominator square root of 195 end fraction space equals space fraction numerator 9 over denominator square root of 195 end fraction left parenthesis in space magnitude right parenthesis space not equal to 0

∴ given lines do not intersect.

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Short Answer Type

132. Find the intercepts cut off by the plane x + 2y – 2z = 9 with the axes.

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133. Find the equation of the plane with intercepts 2, 3 and 4 on the x, y and z-axis respectively.

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134. Find the intercepts cut off by the plane 2x + y – z = 5.

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135. Find the intercept form of the equation of the plane x + 3 y – 7 z + 2 = 0.

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136. Find the equation of the plane with intercept 3 on the y-axis and parallel to ZOX plane..

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137. Find a unit normal vector to the plane x + 2y + 3z – 6 = 0.

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138. Find the normal form of the equation of the plane 3x– 4y + z + 5 = 0.

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139. Find the value of k for which the planes 3 x– 6 y– 2 z = 7 and 2x + y'– k z = 5 are perpendicular to each other.

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140.

The foot of the perpendicular drawn from the origin to the plane is (4, 3, 2). Find the equation of the plane.

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