The Foot of the parpendicular drawn from t the origin to a plane is M (2,1,- 2) find Vector Equation of the th plane.
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Answer:
Let equation of the given point lying in the plane be
r
0
.
Let equation of the normal be
n
.
∴
n
=
0
−(4
i
^
−2
j
^
−5
k
^
)=−4
i
^
+2
j
^
+5
k
^
Now, equation of plane with normal
n
and point
n
0
is
n
⋅(
r
−
r
0
)=0,
where
r
=x
i
^
+y
j
^
+z
k
^
is any general point in the plane.
⟹(−4
i
^
+2
j
^
+5
k
^
)⋅(
r
−(4
i
^
−2
j
^
−5
k
^
))=0
⟹−4x+2y+5z+16+4+25=0
⟹4x−2y−5z=45
This is the required equation of the plane.
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