3D Geometry
Equation of plane / locus
Grade 12
Question:
<p>The equation of a plane is \(\dfrac{x}{a} + \dfrac{y}{b} + \dfrac{z}{c} = 1\). The plane passes through the point \((3, 2, 1)\) and meets the axes at \(A(a, 0, 0)\), \(B(0, b, 0)\) and \(C(0, 0, c)\). The locus of the point of intersection of planes through \(A\), \(B\) and \(C\) parallel to the \(yz\)-, \(zx\)- and \(xy\)-planes respectively is:</p>
<p>\(\dfrac{3}{x} + \dfrac{1}{y} + \dfrac{1}{z} = 1\)</p>
<p>\(\dfrac{3}{x} + \dfrac{2}{y} + \dfrac{1}{z} = 1\)</p>
<p>\(\dfrac{1}{x} + \dfrac{2}{y} + \dfrac{1}{z} = 1\)</p>
<p>\(\dfrac{3}{x} + \dfrac{2}{y} + \dfrac{3}{z} = 1\)</p>
Step-by-Step Solution
Key Concept: The planes through A, B, C parallel to coordinate planes are x = a, y = b, z = c respectively. Their intersection point is (a, b, c), and since the original plane passes through (3, 2, 1), we use the intercept form to find the locus relation.
Step 1: Identify the three planes and their intersection point. • Plane through A(a, 0, 0) parallel to yz-plane: x = a • Plane through B(0, b, 0) parallel to zx-plane: y = b • Plane through C(0, 0, c) parallel to xy-plane: z = c The intersection point of these three planes is P(a, b, c) . Step 2: Use the constraint that the original plane passes through (3, 2, 1). Since the plane $\frac{x}{a} + \frac{y}{b} + \frac{z}{c} = 1$ passes through (3, 2, 1): $\frac{3}{a} + \frac{2}{b} + \frac{1}{c} = 1$ Step 3: Express the locus in terms of the intersection point. Let the intersection point be (h, k, l), so a = h, b = k, c = l. Substituting into the constraint equation: $\frac{3}{h} + \frac{2}{k} + \frac{1}{l} = 1$ Step 4: Write the locus equation. Replacing (h, k, l) with (x, y, z): $\frac{3}{x} + \frac{2}{y} + \frac{1}{z} = 1$ ∴ Answer: $\frac{3}{x} + \frac{2}{y} + \frac{1}{z} = 1$
Correct Answer: B