1
You visited us 1 times! Enjoying our articles? Unlock Full Access!
Question

# Find the vector equation of the plane with intercepts 3, –4 and 2 on x, y and z-axis respectively.

Open in App
Solution

## The equation of the plane in the intercept form is $\frac{x}{a}+\frac{y}{b}+\frac{z}{c}=1$, where a, b and c are the intercepts on the x, y and z-axis, respectively. It is given that the intercepts made by the plane on the x, y and z-axis are 3, –4 and 2, respectively. ∴ a = 3, b = −4, c = 2 Thus, the equation of the plane is $\frac{x}{3}+\frac{y}{\left(-4\right)}+\frac{z}{2}=1\phantom{\rule{0ex}{0ex}}⇒4x-3y+6z=12$ $⇒\left(x\stackrel{^}{i}+y\stackrel{^}{j}+z\stackrel{^}{k}\right).\left(4\stackrel{^}{i}-3\stackrel{^}{j}+6\stackrel{^}{k}\right)=12\phantom{\rule{0ex}{0ex}}⇒\stackrel{\to }{r}.\left(4\stackrel{^}{i}-3\stackrel{^}{j}+6\stackrel{^}{k}\right)=12$ This is the vector form of the equation of the given plane.

Suggest Corrections
1
Join BYJU'S Learning Program
Related Videos
Equation of a Plane: Three Point Form and Intercept Form
MATHEMATICS
Watch in App
Join BYJU'S Learning Program