find the equation of plane passing through points (1,3,1) (2,5,7) and (4,1,5).
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Answer:
Answer
Vector equation of a plane passing through three points with position vectors
a
,
b
,
c
is
(
r
−
a
).[(
b
−
a
)×(
c
−
a
)]=0
Now, the plane passes through the points
A(2,3,4),B(−3,5,1) and C(4,−1,2)
b
−
a
=−3
i
^
+5
j
^
+
k
^
−2
i
^
−3
j
^
−4
k
^
=−5
i
^
+2
j
^
−3
k
^
c
−
a
=4
i
^
−
j
^
+2
k
^
−2
i
^
−3
j
^
−4
k
^
=2
i
^
−4
j
^
−2
k
^
(
b
−
a
)×(
c
−
a
)=
∣
∣
∣
∣
∣
∣
∣
∣
i
^
−5
2
j
^
2
−4
k
^
−3
−2
∣
∣
∣
∣
∣
∣
∣
∣
=(−4−12)
i
^
−(10+6)
j
^
+(20−4)
k
^
=−16
i
^
−16
j
^
+16
k
^
∴vector equation of plane is (x
i
^
+y
j
^
+z
k
^
−2
i
^
−3
j
^
−4
k
^
)(8
i
^
−16
j
^
−16
k
^
)=0
⇒[(x−2)
i
^
+(y−3)
j
^
+(z−4)
k
^
](8
i
^
−16
j
^
−16
k
^
)=0
⇒8x−16−16y+48−16z+64=0
⇒8x−16y−16z+96=0
or
x−2y−2
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