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.
Answers
hey mate...
here is your 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→=i^+j^+2k^
Let C be the midpoint of AB.
(x1+x22,y1+y22,z1+z22)
=C(2+42,3+52,4+82)
=C(62,82,122)=(3,4,6)
Let this be a→=3i^+4j^+6k^
Hence the vector equation of the plane passing through C and ⊥ AB is
(r→−(3i^+4j^+6k^)).(i^+j^+2k^)=0
We know r→=xi^+yj^+zk^
=>[(xi^+yj^+zk^).(3i^+4j^+6k^)].(i^+j^+2k^)=0
On simplifying we get,
(x−3)i^+(y−4)j^+(z−6)k^).(i^+j^+2k^)=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
hope it helps...
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→=i^+j^+2k^
Let C be the midpoint of AB.
(x1+x22,y1+y22,z1+z22)
=C(2+42,3+52,4+82)
=C(62,82,122)=(3,4,6)
Let this be a→=3i^+4j^+6k^
Hence the vector equation of the plane passing through C and ⊥ AB is
(r→−(3i^+4j^+6k^)).(i^+j^+2k^)=0
We know r→=xi^+yj^+zk^
=>[(xi^+yj^+zk^).(3i^+4j^+6k^)].(i^+j^+2k^)=0
On simplifying we get,
(x−3)i^+(y−4)j^+(z−6)k^).(i^+j^+2k^)=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