Question

A plane \( \pi \) is passing through the points \( A(1, -2, 3) \) and \( B(6, 4, 5) \). If the plane \( \pi \) is perpendicular to the plane \( 3x - y + z = 2 \), then the perpendicular distance from \( (0, 0, 0) \) to the plane \( \pi \) is

• A. \( \dfrac{63}{\sqrt{594}} \)
• B. \( \dfrac{32}{\sqrt{594}} \)
• C. \( \dfrac{72}{\sqrt{435}} \)
• D. \( \dfrac{23}{\sqrt{135}} \)

Hint:

To find the perpendicular distance from a point to a plane, use the formula:
\[ \frac{|Ax + By + Cz + D|}{\sqrt{A^2 + B^2 + C^2}} for plane } Ax + By + Cz + D = 0 \]

Answer 1

The Correct Option is A

Step 1: Direction vector of line through points A and B 
Let \( \vec{AB} = \langle 6-1,\ 4 - (-2),\ 5 - 3 \rangle = \langle 5,\ 6,\ 2 \rangle \) 
Step 2: Normal vector of plane \( \pi \) 
Plane \( \pi \) is perpendicular to the given plane \( 3x - y + z = 2 \), so the normal vector of the given plane is: 
\[ \vec{n}_1 = \langle 3,\ -1,\ 1 \rangle \] Since \( \pi \) is perpendicular to this plane, its normal vector \( \vec{n}_\pi \) is perpendicular to \( \vec{n}_1 \), and also perpendicular to vector \( \vec{AB} \). So \( \vec{n}_\pi = \vec{AB} \times \vec{n}_1 \) 


Step 4: Equation of plane \( \pi \) 
Using point \( A(1, -2, 3) \), the equation of the plane is: 
\[ 8(x - 1) + 1(y + 2) - 23(z - 3) = 0 \Rightarrow 8x + y - 23z = -8 + (-2) + 69 = 59 \] \[ \Rightarrow 8x + y - 23z = 59 \] Step 5: Perpendicular distance from origin to this plane 
Use distance formula: 
\[ D = \frac{|8 \cdot 0 + 0 \cdot 1 - 23 \cdot 0 - 59|}{\sqrt{8^2 + 1^2 + (-23)^2}} = \frac{59}{\sqrt{64 + 1 + 529}} = \frac{59}{\sqrt{594}} \] Wait! The options suggest the answer is \( \dfrac{63}{\sqrt{594}} \), so let’s re-check Step 4’s constant term. 
Backtrack: 
\[ 8(x - 1) + 1(y + 2) - 23(z - 3) = 0 \Rightarrow 8x - 8 + y + 2 -23z + 69 = 0 \Rightarrow 8x + y - 23z + 63 = 0 \Rightarrow 8x + y - 23z = -63 \] Correct plane: \( 8x + y - 23z = -63 \) 
Step 6: Recalculate distance from origin to plane 
\[ D = \frac{|8(0) + 1(0) - 23(0) + 63|}{\sqrt{8^2 + 1^2 + 23^2}} = \frac{63}{\sqrt{64 + 1 + 529}} = \frac{63}{\sqrt{594}} \] Therefore, the perpendicular distance is \( \boxed{\dfrac{63}{\sqrt{594}}} \)