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Question: If a plane passes through the point (1,1,1) and is perpendicular to the line \(\frac{x - 1}{3} = \fr...

If a plane passes through the point (1,1,1) and is perpendicular to the line x−13=y−10=z−14\frac{x - 1}{3} = \frac{y - 1}{0} = \frac{z - 1}{4}, then its perpendicular distance from the origin is.

A

34\frac{3}{4}

B

43\frac{4}{3}

C

75\frac{7}{5}

D

1

Answer

75\frac{7}{5}

Explanation

Solution

According to direction ratio of plane are respectively (3, 0, 4).

Equation of plane passing through point (1, 1, 1) is

⇒A(x−x1)+B(y−y1)+C(z−z1)=0\Rightarrow A \left( x - x _ { 1 } \right) + B \left( y - y _ { 1 } \right) + C \left( z - z _ { 1 } \right) = 0

⇒3(x−1)+0(y−1)+4(z−1)=0\Rightarrow 3 ( x - 1 ) + 0 ( y - 1 ) + 4 ( z - 1 ) = 0 ⇒3x+4z−7=0\Rightarrow 3 x + 4 z - 7 = 0

Normal form of plane is, 3x5+4z5=75\frac { 3 x } { 5 } + \frac { 4 z } { 5 } = \frac { 7 } { 5 }

∴\therefore Perpendicular distance from (0,0,0)=75( 0,0,0 ) = \frac { 7 } { 5 }.