Find the vector and Cartesian forms of the equations of the plane containing the two lines
and .
.
Given : Equations of lines -
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To Find : Equation of plane.
Formulae :
1) Cross Product :
If
are two vectors
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then,

2) Dot Product :
If
are two vectors
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![]()
then,
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3) Equation of plane :
If two lines
are coplanar then equation of the plane containing them is
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Where,
Given equations of lines are
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Let, ![]()
Where,
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Now,

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Therefore,
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= 6 – 56 – 48
= - 98
………eq(1)
Equation of plane containing lines
is
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Now,
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From eq(1)
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Therefore, equation of required plane is
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This vector equation of plane.
As ![]()
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= 6x – 28y + 12z
Therefore, equation of plane is
6x – 28y + 12z = -98
6x – 28y + 12z + 98 = 0
This Cartesian equation of plane.
Couldn't generate an explanation.
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