Question

find the equation of plane passing through points (1,3,1) (2,5,7) and (4,1,5).​

Answer 1

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