Question

Write the equation of the plane which
makes intercepts 1, -2, -3 on the axes.​

Answer 1

Question:

Write the equation of the plane which makes intercepts 1, - 2, - 3 on the axes.

Given:

x-intercept = 1

y-intercept = -2

z-intercept = -3

To find:

Equation of the plane .

Answer:

• Required equation of plan in intercept form; x/1 + y/(-2) + z/(-3) = 1 .

• Required equation of plan in general form; 6x - 3y - 2z - 6 = 0

Note:

• The intercept form of the plane is given by; x/a + y/b + z/c = 1,

where,

a is x-intercept,

b is y-intercept,

c is z-intercept.

Also,

Point of intersection on x-axis is "a",

Point of intersection on y-axis is "b",

Point of intersection on z-axis is "c".

• The general form of the plane is given by; Ax + By + Cz + D = 0.

Solution:

Here, it is given that;

x-intercept,(a) = 1

y-intercept,(b) = -2

z-intercept,(c) = -3

We know that ,

The intercept form of the plane is given by; x/a + y/b + z/c = 1,

where,

a is x-intercept,

b is y-intercept,

c is z-intercept.

Thus,

The required equation of the plane will be given by ;

=> x/1 + y/(-2) + z/(-3) = 1

=> x - y/2 - z/3 = 1

=> (6x - 3y - 2z)/6 = 1

=> 6x - 3y - 2z = 6

=> 6x - 3y - 2z - 6 = 0

Hence,

The required equation of the plane is;

x/1 + y/(-2) + z/(-3) = 1

OR

6x - 3y - 2z - 6 = 0

Answer 2

Answer:

$\bold\red{6x-3y-2z-6=0}$

Step-by-step explanation:

If a plane makes intercept of lengths a, b, c with the x - axis, y- axis and z - axis respectively,

then

the equation of plane is

$\bold{\frac{x}{a} + \frac{y}{b} + \frac{z}{c} = 1}$

Now,

In the Given Question,

Here,

a = 1

b = -2

c = -3

Therefore,

the Equation of the plane is

$= > \frac{x}{1} + \frac{y}{ - 2} + \frac{z}{ - 3} = 1 \\ \\ = > \frac{x}{1} - \frac{y}{2} - \frac{z}{3} = 1$

Now,

taking LCM of Denominator and solving,

we get,

$= > \frac{6x - 3y - 2z}{6} = 1 \\ \\ = > 6x - 3y - 2z = 6 \\ \\ = > 6x - 3y - 2z - 6 = 0$

Hence,

The required equation of plane is

$\bold{6x-3y-2z-6=0}$