3D Geometry
Equation of Plane
Grade 12
Question:
<p>The equation of the plane with intercepts <i>2, 3</i> and <i>4</i> on the <i>X, Y</i> and <i>Z</i>-axes respectively, is</p>
<p>(a) <i>2x + 3y + 4z = 12</i></p>
<p>(b) <i>6x + 4y + 3z = 12</i></p>
<p>(c) <i>x + 3y + 2z = 6</i></p>
<p>(d) None of these</p>
Step-by-Step Solution
Key Concept: The intercept form of a plane with intercepts a, b, c on the coordinate axes is x/a + y/b + z/c = 1.
Solution: Let the equation of the plane be $\frac{x}{a} + \frac{y}{b} + \frac{z}{c} = 1$ Here, $a = 2, b = 3, c = 4$ Substituting the values of a, b and c , we get: $\frac{x}{2} + \frac{y}{3} + \frac{z}{4} = 1$ Multiplying throughout by 12 : $6x + 4y + 3z = 12$ ∴ Answer is (b).
Correct Answer: B