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: Find the feet of the perpendiculars.
Dropping a perpendicular from $Q(a,b,c)$ to the $YZ$-plane kills the $x$-coordinate, giving $A(0,b,c)$. To the $ZX$-plane it kills the $y$-coordinate, giving $B(a,0,c)$.
Step 2: Assume a plane through the origin.
Since the plane passes through $O(0,0,0)$, it has the form $\alpha x + \beta y + \gamma z = 0$.
Step 3: Apply point $A$.
Substituting $A(0,b,c)$: $\beta b + \gamma c = 0$.
Step 4: Apply point $B$.
Substituting $B(a,0,c)$: $\alpha a + \gamma c = 0$.
Step 5: Express in terms of $\gamma$.
From these, $\alpha = -\dfrac{\gamma c}{a}$ and $\beta = -\dfrac{\gamma c}{b}$. Choosing $\gamma = -\dfrac{1}{c}$ gives $\alpha = \dfrac{1}{a}$ and $\beta = \dfrac{1}{b}$.
Step 6: Write the plane.
The equation becomes $\dfrac{x}{a} + \dfrac{y}{b} - \dfrac{z}{c} = 0$, the same as the determinant method but obtained by solving for the coefficients directly.
\[ \boxed{\dfrac{x}{a} + \dfrac{y}{b} - \dfrac{z}{c} = 0} \]