Question

Find the equation of the plane whose intercepts on X, Y, Z-axes are 1, 2, 4 respectively.​

Answer 1

Answer:

Equation of plane with intercepts a,b and c on x,y and z-axis respectively is given as-

a

x

+

b

y

+

c

z

=1

Now,

Intercept on x-axis (a)=1

Intercept on y-axis (b)=2

Intercept on z-axis (c)=4

Therefore,

Equation of plane-

1

x

+

2

y

+

4

z

=1

4

4x+2y+z

=1

⇒4x+2y+z=4

Hence the equation of plane whose intercept on x,y and z-axis are 1,2 and 4 respectively is-

4x+2y+z=4

Answer 2

$\color{green}{•••Answer•••} \\equation \: of \: plane \: is \: given \: by \\ \frac{x}{1} + \frac{y}{2} + \frac{z}{4} = 1 \\ 4x + 2y + z = 4 \\ \\ \\ \color{green}{•••hope \: it \: helps \: you•••}$