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:
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Step-by-step explanation:
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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