Question

The foot of the perpendicular drawn from the origin to the plane is $(4,-2,5)$, then the Cartesian equation of the plane is

• A. $4x-2y+5z=45$
• B. $-4x+2y+5z=45$
• C. $4x-2y+5z+45=0$
• D. $4x+2y-5z+45=0$

Hint:

Logic Tip: A blazing fast shortcut is to recognize that since the plane must pass through the point $(4,-2,5)$, you can simply plug these coordinates into the given options. Option A: $4(4) - 2(-2) + 5(5) = 16 + 4 + 25 = 45$. This matches perfectly, and it's much faster during an exam!

Answer 1

The Correct Option is A

Concept:
If a perpendicular is drawn from the origin $(0,0,0)$ to a plane and it meets the plane at a point $P(x_1, y_1, z_1)$, then the vector $\vec{OP}$ acts as the normal vector $\vec{n}$ to the plane. The equation of the plane passing through point $P$ with normal vector $\vec{n}$ is given by $(\vec{r} - \vec{p}) \cdot \vec{n} = 0$, which translates to $a(x-x_1) + b(y-y_1) + c(z-z_1) = 0$ in Cartesian form.
Step 1: Identify the normal vector to the plane.
The foot of the perpendicular from the origin $O(0,0,0)$ is given as $P(4, -2, 5)$. The normal vector $\vec{n}$ to the plane is the vector $\vec{OP}$: $$\vec{n} = (4-0)\hat{i} + (-2-0)\hat{j} + (5-0)\hat{k}$$ $$\vec{n} = 4\hat{i} - 2\hat{j} + 5\hat{k}$$ Thus, the direction ratios of the normal are $a=4$, $b=-2$, and $c=5$.
Step 2: Write the Cartesian equation of the plane.
The plane passes through the point $P(4, -2, 5)$. Using the point-normal form: $$a(x-x_1) + b(y-y_1) + c(z-z_1) = 0$$ Substitute the values: $$4(x - 4) - 2(y - (-2)) + 5(z - 5) = 0$$ $$4(x - 4) - 2(y + 2) + 5(z - 5) = 0$$
Step 3: Simplify the equation.
Expand the terms: $$4x - 16 - 2y - 4 + 5z - 25 = 0$$ Group the variables and constants: $$4x - 2y + 5z - (16 + 4 + 25) = 0$$ $$4x - 2y + 5z - 45 = 0$$ $$4x - 2y + 5z = 45$$