Let P _ 1 : 2x + y - z = 3and P _ 2 : x + 2y + z = 2 be two planes....

Let

{P}_{1}: 2x + y - z = 3and{P}_{2}: x + 2y + z = 2

be two planes. Then, which of the following statement(s) is (are) TRUE?

Options

AThe line of intersection of ${P}_{1}, {P}_{2}$ has direction ratios $1, 2, - 1$
BThe line $\dfrac{3x - 4}{9} = \dfrac{1 - 3y}{9} = \dfrac{z}{3}$ is perpendicular to the line of intersection of ${P}_{1}, {P}_{2}$
CThe acute angle between ${P}_{1}, {P}_{2}$ is ${60}^{o}$
DIf ${P}_{3}$ is the plane passing through the point \left(4, 2, - 2\right) $and perpendicular to the line of intersection of$ {P}_{1}, {P}_{2} $, then the distance of the point \left(2,1, 1\right)$ from the plane ${P}_{3}is\dfrac{2}{\sqrt{3}}$

Correct answer

C. The acute angle between ${P}_{1}, {P}_{2}$ is ${60}^{o}$

Step-by-step solution

D.C of line intersection

\left(a, b, c\right)

⇒ 2a + b - c = 0

a + 2b + c = 0

\dfrac{a}{1 + 2} = \dfrac{b}{- 1 - 2} = \dfrac{c}{4 - 1}

∴ D . C\left(1, - 1, 1\right)

\left(B\right)

\dfrac{3x - 4}{9} = \dfrac{1 - 3y}{9} = \dfrac{z}{3}

⇒ \dfrac{x - \dfrac{4}{3}}{3} = \dfrac{y - \dfrac{1}{3}}{- 3} = \dfrac{z}{3}

⇒

lines are parallel

( C )

Acute angle between

{P}_{1}, {P}_{2} = \cos^{- 1}

\left(\dfrac{2 \times 1 + 1 \times 2 - 1 \times 1}{\sqrt{6}\sqrt{6}}\right)

<br />

= \cos^{- 1}\left(\dfrac{3}{6}\right) = \cos^{- 1}\left(\dfrac{1}{2}\right) = 60 °

<br />\left(D\right)

Plane is given by \left(x - 4\right) - \left(y - 2\right) + \left(z + 2\right) = 0

<br />

⇒ x - y + z = 0

<br />Distance of\left(2, 2, 1\right)

from plane

= \dfrac{2 - 1 + 1}{\sqrt{3}} = \dfrac{2}{\sqrt{3}}

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