Find the vector equation of a plane which is at 42 units from the origin and which is normal to the vector 2 i+ j— 2 k
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Answer:
If
n
^
is a unit vector along the normal and p i the length of the perpendicular from origin to the plane, then the vector equation of the plane
r
^
.
n
^
=p
Hence,
n
ˉ
=2
i
^
+
j
^
−2
k
^
and p=42
∴∣
n
ˉ
∣=
2
2
+1
2
+(−2)
2
=
9
=3
n
^
=
∣N∣
n
ˉ
=
3
1
(2
i
^
+
j
^
−2
k
^
)
∴ the vector equation of the required plane is
r
ˉ
[
3
1
(2
i
^
+
j
^
−2
k
^
)]=42
i.e.,
r
ˉ
(2
i
^
+
j
^
=2
k
^
)=126
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