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
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.
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