Question

Find the perpendicular distance of P (1, 1, 1) from the plane 2x + 2y – z = 6.​

Answer 1

Answer:

perpendicular dustance if plane from tne point (1,1,1,) 2x+2y-z+k =o

1

1

1

1

1

=2

2

+2

2

+1

2

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

1

1

1

1

1

=3

1

1

1

1

1

3=k+2 =9

k =-11 ,7

k=3

Answer 2

Answer:

Let the foot of the perpendicular to the plane drawn from (x, y, z) be the point (x, y, z) (1, 1, 1). As a result, the vector (x-1, y-1, z-1) is perpendicular to the plane.

The plane's equation is, which indicates that the vector is perpendicular to the plane.

⇒2x + 2y – z = 6

⇒ (x, y, z) . ( 2, 2, -1) = 6

Because they're perpendicular to the plane, the vectors (x-1, y-1, z-1) and   ( 2, 2, -1) are now parallel to each other.

⇒ ( x-1, y-1, z-1 ) = k ( 2, 2 -1 )

So, where k is a real number, we can write:-

⇒ x = 2k + 1 ( equation 1)

⇒ y = 2k + 1 ( equation 2)

⇒ z = 1 - k ( equation 3)

Since the point (x, y, z) is on the plane,

⇒ 2x + 2y – z = 6.

⇒ 2 ( 2k + 1 ) + 2( 2k + 1 ) + ( 1 - k ) = 6

⇒ 9k + 3 = 6

⇒ k = 1/3

Put k = 1/3 in equation 1, 2 and 3.

⇒ x =  5/3

⇒ y = 5/3

⇒z = 2/3

Now, the perpendicular distance is

⇒ d = $\sqrt (1- \frac{5}{3})^{2} +( 1 - \frac{5}{3})^{2} + ( 1 - \frac{2}{3})^{2}$

⇒ d = $\sqrt \frac{4}{9} + \frac{4}{9} + \frac{1}{9}$

⇒ d = 1

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