Answer:
Let
the co-ordinates of the foot of the perpendicular from the original to the
plane is P(x1, y1, z1).
direction ratios
of line OP are x1 ? 0, y1 ? 0, z1 ? 0, i.e., x1,
y1, z1,
(a)
Equation of plane is 2x + 3y + 4z = 12
Divide
it by
Its
direction cosines are
As
direction ratios area proportional to direction cosines
?
(1)
Also
P(x1, y1, z1) lies on plane
2x
+ 3y + 4z =12
Put
foot of
perpendicular is
(b)
Equation of plane is 3y + 4z ? 6 = 0
Dividing i.e. 5
Its
direction cosines are
As
direction ratios are proportional to direction cosines.
..
(1)
Also
P(x1, y1, z1) lies on the plane
3y
+ 4z ? 6 = 0
Put
foot of
perpendicular is
(c)
Equation of plane is x + y + z = 1
Divide
it by
Its
direction cosines are
As
direction ratios are proportional to direction cosines.
Also
P(x1, y1, z1) lies on the plane
x
+ y + z = 1
Put
, we get
foot of
perpendicular is
(d) Equation of plane
is 5y + 8 = 0, divide it by
Its direction cosines
are 0, ?1, 0.
As direction ratios
are proportional to direction cosines
Also P(x1,
y1, z1) lies on the plane 5y + 8 = 0
Put K = we get
foot of
perpendicular is
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