Question

Let \(P\) be the plane such that it contains the straight line \[ \frac{x-1}{2}=\frac{y-3}{3}=\frac{z+2}{1} \] and is perpendicular to the plane \[ x+2y+3z=4 \] Let \(P_1\) be the plane which passes through the point \((4,2,2)\) and is parallel to \(P\). Then which of the following statements is (are) TRUE?

• A. The equation of the plane \(P\) is 7x-5y+z=-10
• B. The distance between the planes \(P\) and \(P_1\) is \(30\)
• C. The distance of the plane \(P\) from the origin is \(2\sqrt3\)
• D. The acute angle between the plane \(P\) and the plane 2x+2y+z=3 is \[\cos^{-1}\left(\frac1{3\sqrt3}\right)\]

Hint:

If two planes are perpendicular, then: \[ \vec n_1\cdot\vec n_2=0 \] Distance between parallel planes: \[ \frac{|d_1-d_2|}{\sqrt{a^2+b^2+c^2}} \]

Answer 1

The Correct Option is A

To solve the problem, let's first break down the given pieces of information and solve it step by step:

1. \(P\) is a plane that contains the line \(\frac{x-1}{2}=\frac{y-3}{3}=\frac{z+2}{1}\) and is perpendicular to the plane \(x+2y+3z=4\).
   • The given line can be written in the vector form as: \(\textbf{r} = \begin{bmatrix} 1 \\ 3 \\ -2 \end{bmatrix} + \lambda \begin{bmatrix} 2 \\ 3 \\ 1 \end{bmatrix}\), where \(\lambda\) is a parameter.
   • Thus, the direction ratios of the line are \((2, 3, 1)\).
2. The plane \(P\) is perpendicular to the plane \(x + 2y + 3z = 4\).
   • This implies that the normal vector to plane \(P\) is a linear combination of the direction ratios of the line and the normal vector of the given plane. Thus, the normal vector of plane \(P\) can be calculated using the cross product of the two direction vectors.
   • The normal vector of the given plane is \((1, 2, 3)\).
   • The cross product of \((2, 3, 1)\) and \((1, 2, 3)\) is calculated as follows: \(\begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 2 & 3 & 1 \\ 1 & 2 & 3 \end{vmatrix} = \mathbf{i}(9 - 2) - \mathbf{j}(6 - 1) + \mathbf{k}(4 - 3) = 7\mathbf{i} - 5\mathbf{j} + \mathbf{k}.\)
   • Therefore, the equation of the plane \(P\) can be written as: \(7x - 5y + z = d\).
   • Since the plane \(P\) contains the point \((1, 3, -2)\): \(\Rightarrow 7(1) - 5(3) + (-2) = d \\ \Rightarrow 7 - 15 - 2 = d \\ \Rightarrow d = -10.\)
   • Thus, the equation of the plane \(P\) is: \(7x - 5y + z = -10\).