Q. 16

# From the point P(

Given :

Equation of plane : 2x + y – 2z + 3 = 0

P = (1, 2, 4)

To Find :

i) Equation of perpendicular

ii) Length of perpendicular = d

iii) coordinates of the foot of the perpendicular

Formulae :

1) Unit Vector :

Let be any vector

Then unit vector of is Where, 2) Length of perpendicular :

The length of the perpendicular from point A with position vector to the plane is given by, Note :

If two vectors with direction ratios (a1, a2, a3) & (b1, b2, b3) are parallel then Given equation of the plane is

2x + y – 2z + 3 = 0

2x + y – 2z = -3 ………..eq(1)

From eq(1) direction ratios of normal vector of the plane are

(2, 1, -2)

Therefore, equation of normal vector is    = 3

From eq(1), p = -3

Given point P = (1, 2, 4)

Position vector of A is Here, Now, = (1×2) + (2×1) + (4×(-2))

= 2 + 2 - 8

= -4

Length of the perpendicular from point A to the plane is   Let Q be the foot of perpendicular drawn from point P to the given plane,

Let Q = (x, y, z) As normal vector and are parallel, we can write, This is the equation of perpendicular. x = 2k+1, y = k+2, z = -2k+4

As point Q lies on the plane, we can write

2(2k+1) + (k+2) - 2(-2k+4) = -3

4k + 2 + k + 2 + 4k - 8 = -3

9k = 1  ,  Therefore co-ordinates of the foot of perpendicular are

Q(x, y, z) = Q ≡ Rate this question :

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