Find the equation of line passing through the point of intersecation of the lines L1 : 2x – y = 14 and L2 : 2x + y = 10 and perpendicular to the L3 : 3x – y + 6 = 0
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Answer:
Normal to plane=2x−2y+3z−2=0 is
n
1
=2
i
^
−2
j
^
−3
k
^
Normal to plane=x−y+z+1=0 is
n
2
=
i
^
−
j
^
−
k
^
Normal to plane=x+2y−z−3=0 is
n
3
=
i
^
+2
j
^
−
k
^
Normal to plane=3x−y+2z−1=0 is
n
4
=3
i
^
−
j
^
+2
k
^
So line of intersection of planes 1 and 2 is along
n
1
×
n
2
=
i
^
+
j
^
:
L
1
similarly the line of intersection lof planes 3 and 4 is along
n
3
×
n
4
=3
i
^
−5
j
^
−7
k
^
:
L
2
The normal to the plane containing
L
1
and
L
2
is along (
i
+
j
)×(3
i
^
−5
j
^
−7
k
^
)=−7
i
^
+7
j
^
−3
k
^
By inspection
L
1
passes through(0,5,4)
so the equation of the required plane is=−7x+7y−8z−3=0
and its distance from origin is
(7
2
+7
2
+8
2
)
2
1
3
=
162
3
=
3
2
1
.
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