If a plane passes through the point (1 , 1 , 1) and is perpendicular to the line (x - 1/3)=(y - 1/0)=(z - 1/4) , then its perpendicular distance from the origin is-

Question

If a plane passes through the point (1 , 1 , 1) and is perpendicular to the line (x - 1/3)=(y - 1/0)=(z - 1/4) , then its perpendicular distance from the origin is- (A) (3/4) (B) (4/3) (C) (7/5) (D) 1

Tardigrade Admin · Accepted Answer

Equation of the plane which passes through (1 , 1 , 1) and dr&rsquo;s of its normal are 3,0,4 is- 3(x - 1)+0(y - 1)+4(z - 1)=0 3x+4z-7=0 Perpendicular distance from the origin is- =|(- 7/5)|=(7/5)

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

$\frac{3}{4}$

$\frac{4}{3}$

$\frac{7}{5}$

$1$

Equation of the plane which passes through $(1, 1, 1)$ and dr’s of its normal are $3, 0, 4$ is-
$3 (x - 1) + 0 (y - 1) + 4 (z - 1) = 0$
$3 x + 4 z - 7 = 0$
Perpendicular distance from the origin is-
$= ∣ ∣ \frac{- 7}{5} ∣ ∣ = \frac{7}{5}$

Q. If a plane passes through the point (1,1,1) and is perpendicular to the line 3x−1​=0y−1​=4z−1​ , then its perpendicular distance from the origin is-

43​

34​

57​

1

Equation of the plane which passes through (1,1,1) and dr’s of its normal are 3,0,4 is- 3(x−1)+0(y−1)+4(z−1)=0 3x+4z−7=0 Perpendicular distance from the origin is- =∣∣​5−7​∣∣​=57​

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

$\frac{3}{4}$

$\frac{4}{3}$

$\frac{7}{5}$

$1$

Equation of the plane which passes through $(1, 1, 1)$ and dr’s of its normal are $3, 0, 4$ is-
$3 (x - 1) + 0 (y - 1) + 4 (z - 1) = 0$
$3 x + 4 z - 7 = 0$
Perpendicular distance from the origin is-
$= ∣ ∣ \frac{- 7}{5} ∣ ∣ = \frac{7}{5}$