3D Geometry
Plane Equation
Grade 12
Question:
<p>Let <i>(l, 2, 1)</i> be a point on the plane which passes through the point <i>(4, -2, 2)</i>. If the plane is perpendicular to the line joining the points <i>(-2, -21, 29)</i> and <i>(-1, -16, 23)</i>, then \(\left(\frac{l}{11}\right)^2 - \frac{4l}{11} - 4\) is equal to</p>
Step-by-Step Solution
Key Concept: A plane perpendicular to a line has the direction vector of the line as its normal vector. Use this to form the plane equation.
Step 1: Find the direction vector of the line: $\vec{d} = (-1-(-2), -16-(-21), 23-29) = (1, 5, -6)$ Step 2: Since the plane is perpendicular to this line, the normal vector to the plane is $\vec{n} = (1, 5, -6)$ Step 3: The plane equation is $1(x-4) + 5(y+2) - 6(z-2) = 0$, which simplifies to $x + 5y - 6z + 12 = 0$ Step 4: Since point (l, 2, 1) lies on the plane: $l + 5(2) - 6(1) + 12 = 0$ Step 5: $l + 10 - 6 + 12 = 0$ ⟹ $l = -16$ Step 6: Calculate $\left(\frac{-16}{11}\right)^2 - \frac{4(-16)}{11} - 4 = \frac{256}{121} + \frac{64}{11} - 4 = \frac{256 + 704 - 484}{121} = \frac{476}{121} = \frac{44}{11} = 4$ (Check computation) ∴ Answer is (d).
Correct Answer: d