Question

Which of the following statements are true?
(A) The equations of the plane passing through the point (1, -1, 2) having 2, 3, 2 as direction ratios of normal to the plane is 2x + 3y + 2z = 3
(B) Angle between the normal to the plane 2x - y + z = 6 and x + y + 2z = 7 is \(\frac{\pi}{3}\)
(C) The angle at which the normal vectors to the plane 4x + 8y + z = 5 is inclined to the z-axis is \( \sin^{-1}(\frac{1}{9}) \)
(D) The equation of the plane passing through the point (3, -3, 1) and normal to the line joining the points (3, 4, -1) and (2, -1, 5) is x + 5y + 6z = -18
(E) A normal vector to the plane 2x - y + 2z = 5 is \( \frac{1}{3}(2\vec{i} - \vec{j} + 2\vec{k}) \)
Choose the correct answer from the options given below:

• A. (A), (B), (C) and (E) only
• B. (B), (C), (D) and (E) only
• C. (A), (B), (D) and (E) only
• D. (A), (B) and (E) only

Hint:

The coefficients of x, y, and z in the equation of a plane \(ax+by+cz=d\) directly give the direction ratios (a, b, c) of the normal vector. This is the starting point for most problems involving angles with planes.

Answer 1

The Correct Option is D

Step 1: Understanding the Concept:
This question tests fundamental concepts of planes and lines in 3D geometry, including finding the equation of a plane, the angle between two planes, the angle between a line and an axis, and normal vectors.
Step 2: Detailed Explanation:
(A): The equation of a plane passing through \((x_0, y_0, z_0)\) with normal direction ratios (a, b, c) is \(a(x-x_0) + b(y-y_0) + c(z-z_0) = 0\).
Given point (1, -1, 2) and normal (2, 3, 2).
\(2(x-1) + 3(y-(-1)) + 2(z-2) = 0 \implies 2x-2+3y+3+2z-4=0 \implies 2x+3y+2z-3=0 \implies 2x+3y+2z=3\).
Statement (A) is true.

(B): Angle \(\theta\) between two planes is the angle between their normals. \(\vec{n_1} = (2, -1, 1)\), \(\vec{n_2} = (1, 1, 2)\).
\(\cos\theta = \frac{|\vec{n_1} \cdot \vec{n_2}|}{|\vec{n_1}||\vec{n_2}|} = \frac{|2(1)+(-1)(1)+1(2)|}{\sqrt{2^2+(-1)^2+1^2}\sqrt{1^2+1^2+2^2}} = \frac{|2-1+2|}{\sqrt{6}\sqrt{6}} = \frac{3}{6} = \frac{1}{2}\).
So, \(\theta = \cos^{-1}(1/2) = \pi/3\). Statement (B) is true.

(C): The normal vector to the plane is \(\vec{n} = (4, 8, 1)\). The z-axis is represented by the vector \(\vec{k} = (0, 0, 1)\).
The angle \(\alpha\) between \(\vec{n}\) and the z-axis is given by:
\(\cos\alpha = \frac{|\vec{n} \cdot \vec{k}|}{|\vec{n}||\vec{k}|} = \frac{|4(0)+8(0)+1(1)|}{\sqrt{4^2+8^2+1^2}\sqrt{1}} = \frac{1}{\sqrt{16+64+1}} = \frac{1}{\sqrt{81}} = \frac{1}{9}\).
So, \(\alpha = \cos^{-1}(1/9)\). The statement says \(\sin^{-1}(1/9)\). Statement (C) is false.

(D): The normal vector to the plane is parallel to the line joining the points.
\(\vec{n} = (3-2, 4-(-1), -1-5) = (1, 5, -6)\).
The plane passes through (3, -3, 1).
Equation: \(1(x-3) + 5(y-(-3)) - 6(z-1) = 0 \implies x-3+5y+15-6z+6=0 \implies x+5y-6z+18=0\).
The statement gives \(x+5y+6z=-18\). The sign of the z-term is wrong. Statement (D) is false.

(E): A normal vector to the plane \(2x - y + 2z = 5\) is \(\vec{n} = 2\vec{i} - \vec{j} + 2\vec{k}\). Any scalar multiple of this vector is also a normal vector.
\(\tfrac{1}{3}(2\vec{i} - \vec{j} + 2\vec{k})\) is a scalar multiple (specifically, it's the unit normal vector \(\hat{n}\)). So, it is a normal vector. Statement (E) is true.

Step 3: Final Answer:
The true statements are (A), (B), and (E). This corresponds to option (D).