Math, asked by bznshvvss, 10 months ago

Find the equation of a plane which bisects perpendicularly the line joining the points A(2,3,4) and B(4,5,8) at right angles.​

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Answered by Anonymous
7

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Answered by Anonymous
0

Answer:

The given points are A(2,3,4)andB(4,5,8)

The line segement AB is given by (x2−x1),(y2−y1),(z2−z1)

(ie)(4−2),(5−3),(8−4)

=(2,2,4)=(1,1,2)

Since the plane bisects AB at rightangles,

AB

is the normal to the plane. which is

n

Therefore

n

=

ˆ

i

+

ˆ

j

+2

ˆ

k

Let C be the midpoint of AB.

(

x1+x2

2

,

y1+y2

2

,

z1+z2

2

)

=C(

2+4

2

,

3+5

2

,

4+8

2

)

=C(

6

2

,

8

2

,

12

2

)=(3,4,6)

Let this be

a

=3

ˆ

i

+4

ˆ

j

+6

ˆ

k

Hence the vector equation of the plane passing through C and ⊥ AB is

(

r

−(3

ˆ

i

+4

ˆ

j

+6

ˆ

k

)).(

ˆ

i

+

ˆ

j

+2

ˆ

k

)=0

We know

r

=x

ˆ

i

+y

ˆ

j

+z

ˆ

k

=>[(x

ˆ

i

+y

ˆ

j

+z

ˆ

k

).(3

ˆ

i

+4

ˆ

j

+6

ˆ

k

)].(

ˆ

i

+

ˆ

j

+2

ˆ

k

)=0

On simplifying we get,

(x−3)

ˆ

i

+(y−4)

ˆ

j

+(z−6)

ˆ

k

).(

ˆ

i

+

ˆ

j

+2

ˆ

k

)=0

Applying the product rule we get,

(x−3)+(y−4)+2(z−6)=0

On simplifying we get,

x+y+2z=19

This is the required equation of the plane

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