Question

If $A$ and $B$ are the foot of the perpendicular drawn from the point $Q(a, b, c)$ to the planes $YZ$ and $ZX$ respectively, then the equation of the plane through the points $A$, $B$, and the origin $O$ is

• A. $\frac{x}{a} - \frac{y}{b} - \frac{z}{c} = 0$
• B. $\frac{x}{a} + \frac{y}{b} - \frac{z}{c} = 0$
• C. $\frac{x}{a} - \frac{y}{b} + \frac{z}{c} = 0$
• D. $\frac{x}{a} + \frac{y}{b} + \frac{z}{c} = 0$

Hint:

To bypass developing determinants, test your derived coordinates directly within the options! For point $A(0,b,c)$, substituting it into option (B) gives: $\frac{0}{a} + \frac{b}{b} - \frac{c}{c} = 0 + 1 - 1 = 0$. For point $B(a,0,c)$, it gives $\frac{a}{a} + \frac{0}{b} - \frac{c}{c} = 1 + 0 - 1 = 0$. Since it satisfies all conditions, option (B) is confirmed immediately.

Answer 1

The Correct Option is B

Step 1: Understanding the Question:
We are given a coordinate point $Q(a, b, c)$. Points $A$ and $B$ represent the spatial feet of the perpendiculars dropped from $Q$ onto the $YZ$-plane and $ZX$-plane respectively. We need to construct the equation of the unique plane that passes through points $A$, $B$, and the coordinate origin $O(0, 0, 0)$.

Step 2: Key Formula or Approach:
The coordinates for projection feet onto primary reference planes are found by nullifying the excluded axis index:

• Foot on $YZ$-plane ($x=0$): $A \equiv (0, b, c)$

• Foot on $ZX$-plane ($y=0$): $B \equiv (a, 0, c)$
The Cartesian equation of a plane passing through three points $(x_1, y_1, z_1)$, $(x_2, y_2, z_2)$, and $(x_3, y_3, z_3)$ is evaluated using the coplanar determinant constraint: $$\begin{vmatrix} x - x_1 & y - y_1 & z - z_1 \\ x_2 - x_1 & y_2 - y_1 & z_2 - z_1 \\ x_3 - x_1 & y_3 - y_1 & z_3 - z_1 \end{vmatrix} = 0$$

Step 3: Detailed Explanation:
Let's select the coordinate origin $O(0,0,0)$ as our primary reference point $(x_1, y_1, z_1)$. Substituting points $O(0,0,0)$, $A(0,b,c)$, and $B(a,0,c)$ into the structural determinant gives: $$\begin{vmatrix} x - 0 & y - 0 & z - 0 \\ 0 - 0 & b - 0 & c - 0 \\ a - 0 & 0 - 0 & c - 0 \end{vmatrix} = 0 \implies \begin{vmatrix} x & y & z \\ 0 & b & c \\ a & 0 & c \end{vmatrix} = 0$$ Expand this determinant expression along the first row: $$x(b \cdot c - 0 \cdot c) - y(0 \cdot c - a \cdot c) + z(0 \cdot 0 - a \cdot b) = 0$$ $$bcx + acy - abz = 0$$ To clean up coefficients and match the option layouts, divide the entire equation by the non-zero constant product $abc$: $$\frac{bcx}{abc} + \frac{acy}{abc} - \frac{abz}{abc} = 0 \implies \frac{x}{a} + \frac{y}{b} - \frac{z}{c} = 0$$ This matches the formula given in option (B).

Step 4: Final Answer:
The equation of the plane is $\frac{x}{a} + \frac{y}{b} - \frac{z}{c} = 0$, which corresponds to option (B).