Chapter 11 Three Dimensional Geometry Worksheet Page 20
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MATHEMATICS
6
Substituting these in the equation of the plane, we get k =
.
29
−
12
18 24
,
,
Hence, the foot of the perpendicular is
.
29 29 29
A
Note
If d is the distance from the origin and l, m, n are the direction cosines of
the normal to the plane through the origin, then the foot of the perpendicular is
(ld, md, nd).
11.6.2 Equation of a plane perpendicular to a
given vector and passing through a given point
In the space, there can be many planes that are
perpendicular to the given vector, but through a given
point P(x
, y
, z
), only one such plane exists (see
1
1
1
Fig 11.12).
Let a plane pass through a point A with position
KH
H
Fig 11.12
vector
and perpendicular to the vector
.
a
N
H
Let
be the position vector of any point P(x, y, z) in the plane. (Fig 11.13).
r
Then the point P lies in the plane if and only if
KKK H
KH
KKK H
KH
AP
is perpendicular to
N
. i.e.,
AP
.
N
= 0. But
KKK H H H
H
H H
= −
−
⋅ =
. Therefore,
(
r a
) N 0
… (1)
AP r a
This is the vector equation of the plane.
Cartesian form
Let the given point A be (x
, y
, z
), P be (x, y, z)
KH
1
1
1
Fig 11.13
N
and direction ratios of
are A, B and C. Then,
H
H
H
=
+
+
ˆ
=
+
+
=
+
+
ˆ
ˆ
ˆ
ˆ
ˆ
ˆ
ˆ
ˆ
and
N A
i
B
j
C
k
a x i y j z k r xi y j z k
,
1
1
1
H
H H
r a ⋅
Now
( – ) N = 0
(
) (
) (
)
−
+
−
+
−
⋅
+
+
=
ˆ
ˆ
ˆ
ˆ
ˆ
ˆ
So
x x i
y y j
z z k
(A
i
B
j
C ) 0
k
1
1
1
i.e.
A (x – x
) + B (y – y
) + C (z – z
) = 0
1
1
1
Example 17
Find the vector and cartesian equations of the plane which passes through
the point (5, 2, – 4) and perpendicular to the line with direction ratios 2, 3, – 1.
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