Practicing Success

Target Exam

CUET

Subject

-- Mathematics - Section B1

Chapter

Three-dimensional Geometry

Question:

The equation of the plane whose intercepts on the coordinate axes are -4, 2 and 3, is

Options:

$3x + 6y + 4z = 12 $

$-3x + 6y + 4z = 12 $

$-3x - 6y - 4z = 12 $

none of these

Correct Answer:

$-3x + 6y + 4z = 12 $

Explanation:

We know that the equation of a plane whose intercepts on the coordinate axes are a, b and c respectively, is $\frac{x}{a}+\frac{y}{b}+\frac{z}{c}=1$

Here, a = -4, b = 2, and c = 3

So, the equation of the required plane is

$\frac{x}{-4}+\frac{y}{2}+\frac{z}{3}= 1 $ or , $ -3x + 6y + 4 z = 12 $