Question

Find the equation of the plane through the line of intersection of the planes 3x – 4y + 5z = 10, 2x + 2y – 3z = 4 and parallel to the line x = 2y = 3z.

Solution

The equation of plane through the line of intersection of the planes
3x – 4y + 5z – 10 = 0 and 2x + 2y – 3z – 4 = 0 is
(3x – 4 y + 5 z – 10) + k (2x + 2y – 3z – 4) = 0    ...(1)
or (2k + 3) x + (2k – 4) y + (–3 k + 5) z – (4k + 10) = 0
Direction ratios of normal to the plane are
2k + 3, 2k – 4, –3k + 5
Consider the line x = 2y = 3z
or    $straight x over 1 space equals space fraction numerator straight y over denominator begin display style 1 half end style end fraction space equals space fraction numerator straight z over denominator begin display style 1 third end style end fraction$
or    $straight x over 6 space equals space straight y over 3 space equals space straight z over 2$
Its direction ratios are 1, $1 half comma space 1 third$
Since this line is parallel to plane (1).
∴ this line is perpendicular to the normal to the plane (1).
$therefore space space space left parenthesis 2 straight k plus 3 right parenthesis thin space left parenthesis 1 right parenthesis space plus space left parenthesis 2 straight k minus 4 right parenthesis space open parentheses 1 half close parentheses plus left parenthesis negative 3 straight k plus 5 right parenthesis space open parentheses 1 third close parentheses space equals space 0 therefore space space 2 straight k space plus space 3 space plus space straight k space minus space 2 space minus straight k space plus space 5 over 3 space equals space 0 therefore space space space space space space space space space 2 straight k space equals space minus 8 over 3 space space space space space space space space space space space space space space rightwards double arrow space space space space straight k space equals space minus 4 over 3$
Putting $straight k space equals space minus 4 over 3 space in space left parenthesis 1 right parenthesis comma space we space get comma$
$left parenthesis 3 straight x minus 4 straight y plus 5 straight z minus 10 right parenthesis space minus space 4 over 3 left parenthesis 2 straight x plus 2 straight y minus 3 straight z minus 4 right parenthesis space equals space 0$

or 3 (3x – 4y + 5z – 10) – 4 (2x + 2 y – 3z – 4) = 0
or 9x – 12y + 15z – 30 – 8x – 8y + 12z + 16 = 0
or x – 20y + 27z = 14
which is required equation of plane.

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Three Dimensional Geometry

Find the equation of the plane through the line of intersection of the planes 3x – 4y + 5z = 10, 2x + 2y – 3z = 4 and parallel to the line x = 2y = 3z.

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