11. Find unit vector perpendicular to the plane passing through the points (1, 2, 3),
(2,-1, 1) and (1, 2,-4).
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Answer:
Let the points be A(1,2,3),B(2,−1,1),C(1,2,−4)
Let,
x
1
=1,y
1
=2,z
1
=3
x
2
=2,y
2
=−1,z
2
=1
x
3
=1,y
3
=2,z
3
=−4
Equation of plane passing through A,B and C is
∣
∣
∣
∣
∣
∣
∣
∣
x−x
1
x
2
−x
1
x
3
−x
1
y−y
1
y
2
−y
1
y
3
−y
1
z−z
1
z
2
−z
1
z
3
−z
1
∣
∣
∣
∣
∣
∣
∣
∣
=0
∣
∣
∣
∣
∣
∣
∣
∣
x−1
1
0
y−2
−3
0
z−3
−2
−7
∣
∣
∣
∣
∣
∣
∣
∣
=0
Expanding along R
3
⇒−7[−3(x−1)−(y−2)]=0
⇒3x+y−5=0
Directional ratios of normal to the plane are 3,1,0
Therefore, let a vector perpendicular to the plane is
P
=3
i
^
+
j
^
+0
k
^
∣
P
∣=
(3)
2
+(1)
2
+(0)
2
=
10
Hence a unit vector perpendicular to the plane =
∣
P
∣
P
=
10
3
i
^
+
j
^
+0
k
^
=
10
3
i
^
+
10
1
j
^
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