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:

Write the given line as the intersection of two simple coordinate planes, then use the family of planes through that line, $P_1+\lambda P_2=0$, to represent every plane containing it. Fix the unknown constant by setting the dot product of the normal vector of this family with the normal of the given plane to zero, since perpendicular planes have perpendicular normals. Once you have the equation of plane $P$, apply the standard formulas for distance between parallel planes, distance from a point, and angle between planes to test each option.

Answer 1

The Correct Option is A

Step 1: Find the normal vector of plane \(P\). 
Direction ratios of the given line: \[ (2,3,1) \] Normal vector of plane \[ x+2y+3z=4 \] is: \[ (1,2,3) \] Since plane \(P\) contains the line and is perpendicular to the given plane, its normal vector is perpendicular to both \[ (2,3,1) \] and \[ (1,2,3). \] Thus, \[ \vec n= \begin{vmatrix} \hat i & \hat j & \hat k \\ 2 & 3 & 1 \\ 1 & 2 & 3 \end{vmatrix} \] \[ \vec n=(7,-5,1) \] 

Step 2: Find equation of plane \(P\). 
Plane passes through point: \[ (1,3,-2) \] Using: \[ 7(x-1)-5(y-3)+(z+2)=0 \] \[ 7x-7-5y+15+z+2=0 \] \[ 7x-5y+z+10=0 \] \[ 7x-5y+z=-10 \] Therefore: \[ \Rightarrow \mathrm{Option\ (A)\ is\ Correct} \] 

Step 3: Check Option (B). 
Plane \(P_1\) parallel to \(P\) through: \[ (4,2,2) \] Equation: \[ 7(x-4)-5(y-2)+(z-2)=0 \] \[ 7x-5y+z=20 \] Distance between planes: \[ \frac{|20-(-10)|}{\sqrt{7^2+(-5)^2+1^2}} \] \[ =\frac{30}{\sqrt{75}} \] \[ =\frac{30}{5\sqrt{3}} \] \[ =2\sqrt{3} \] Thus: \[ \Rightarrow \mathrm{Option\ (B)\ is\ Incorrect} \] 

Step 4: Check Option (C). 
Distance of plane \[ 7x-5y+z+10=0 \] from origin: \[ \frac{|10|}{\sqrt{75}} \] \[ =\frac{2}{\sqrt{3}} \] not \[ 2\sqrt{3} \] Therefore: \[ \Rightarrow \mathrm{Option\ (C)\ is\ Incorrect} \] 

Step 5: Check Option (D). 
Normal vector of \[ 2x+2y+z=3 \] is: \[ (2,2,1) \] Angle between planes equals angle between normals. Thus: \[ \cos\theta= \frac{|(7)(2)+(-5)(2)+(1)(1)|} {\sqrt{75}\sqrt{9}} \] \[ = \frac{|14-10+1|} {5\sqrt{3}\cdot3} \] \[ = \frac{5}{15\sqrt{3}} \] \[ = \frac{1}{3\sqrt{3}} \] Hence: \[ \theta= \cos^{-1}\left(\frac{1}{3\sqrt{3}}\right) \] Therefore: \[ \Rightarrow \mathrm{Option\ (D)\ is\ Correct} \] 

Step 6: Identify the correct options. 
Hence: \[ \boxed{\mathrm{(A)\ and\ (D)}} \]

Answer 2

Concept:

• A straight line can be written as the intersection of two planes. Any plane containing that line then belongs to the family $P_1 + \lambda P_2 = 0$, for some constant $\lambda$.
• Two planes are perpendicular when the dot product of their normal vectors is zero. This condition fixes the value of $\lambda$.
• Once the equation of plane $P$ is known, the standard formulas for distance between parallel planes, distance of a plane from a point, and angle between two planes are used to test each option.

Step 1: Write the given line as the intersection of two planes.
From $\frac{x-1}{2}=\frac{y-3}{3}$: $3(x-1)=2(y-3) \Rightarrow 3x-2y+3=0$
From $\frac{y-3}{3}=\frac{z+2}{1}$: $(y-3)=3(z+2) \Rightarrow y-3z-9=0$
The given line is the intersection of these two planes.

Step 2: Form the family of planes through this line.
$(3x-2y+3)+\lambda(y-3z-9)=0$
$3x+(\lambda-2)y-3\lambda z+(3-9\lambda)=0$
This represents every plane that contains the given line, for some value of $\lambda$.

Step 3: Apply the perpendicularity condition with the plane $x+2y+3z=4$.
Normal of the family plane: $(3,\ \lambda-2,\ -3\lambda)$. Normal of the given plane: $(1,2,3)$.
For perpendicular planes, the dot product of normals is zero:
$3(1)+(\lambda-2)(2)+(-3\lambda)(3)=0$
$3+2\lambda-4-9\lambda=0$
$-1-7\lambda=0 \Rightarrow \lambda=-\frac{1}{7}$

Step 4: Substitute $\lambda$ back to get the equation of plane $P$.
$3x+\left(-\frac{1}{7}-2\right)y-3\left(-\frac{1}{7}\right)z+\left(3-9\left(-\frac{1}{7}\right)\right)=0$
$3x-\frac{15}{7}y+\frac{3}{7}z+3+\frac{9}{7}=0$
Multiplying throughout by $7$: $21x-15y+3z+21+9=0 \Rightarrow 21x-15y+3z+30=0$
Dividing by $3$: $7x-5y+z+10=0 \Rightarrow 7x-5y+z=-10$
This matches option (A), so option (A) is correct.

Step 5: Find plane $P_1$ and test option (B).
$P_1$ is parallel to $P$ and passes through $(4,2,2)$: $7(4)-5(2)+2=20$, so $P_1: 7x-5y+z=20$
Distance between the parallel planes $P$ and $P_1$:
$\frac{|20-(-10)|}{\sqrt{7^2+(-5)^2+1^2}}=\frac{30}{\sqrt{75}}=\frac{30}{5\sqrt3}=2\sqrt3$
This is not $30$, so option (B) is incorrect.

Step 6: Test option (C).
Distance of plane $7x-5y+z+10=0$ from the origin:
$\frac{|10|}{\sqrt{75}}=\frac{10}{5\sqrt3}=\frac{2}{\sqrt3}=\frac{2\sqrt3}{3}$
This is not $2\sqrt3$, so option (C) is incorrect.

Step 7: Test option (D).
Normal of $2x+2y+z=3$ is $(2,2,1)$. Angle between the planes equals the angle between their normals:
$\cos\theta=\frac{|(7)(2)+(-5)(2)+(1)(1)|}{\sqrt{75}\cdot\sqrt{9}}=\frac{|14-10+1|}{5\sqrt3\cdot3}=\frac{5}{15\sqrt3}=\frac{1}{3\sqrt3}$
$\theta=\cos^{-1}\left(\frac{1}{3\sqrt3}\right)$
This matches option (D), so option (D) is correct.

Final Answer: Options (A) and (D) are TRUE.