The general equation ax + by + cz + d = 0 of first degree in x, y, z represents a plane.

**Intercept form of the equation of a plane**

Let the plane * *â‹… * *= *q
*meets the coordinate axes at A,B,C respectively such that the intercepts on the axes are *OA *= *a*, *OB *= *b*, *OC *= *c
*. Now position vector of the point *A *is *ai*Ë†. Since *A
*lies on the given plane, we have *ai*Ë†â‹… = *q *which
gives .

Similarly, since the vectors bË†*j* and cË†*k* lie on the given plane, we have . Substituting
= *x*Ë†*i* +* y*Ë†*j + z*Ë†*k* in * *â‹… = q
, we get

Dividing by q, we get, . This is called the **intercept form **of equation of the plane
having intercepts a, b, c on the x, y, z axes respectively.

The general equation *ax *+ *by *+ *cz *+ *d *=
0 of first degree in *x*, *y*, *z *represents a plane.

The equation *ax *+ *by *+ *cz *+ *d *= 0
can be written in the vector form as follows

Since this is the vector form of the equation of a plane in
standard form, the given equation *ax *+ *by *+ *cz *+ *d *= 0 represents a plane. Here
= *ai*Ë† + *b*Ë†*j *+ *ck*Ë†. is a vector normal to the plane.

In the general equation *ax *+ *by *+ *cz *+ *d
*= 0 of a plane, *a*, *b*, *c *are direction ratios of the normal
to the plane.

Find the vector and Cartesian form of the equations of a plane
which is at a distance of 12 units from the origin and perpendicular to 6*i*Ë†
+ 2 Ë†*j *âˆ’ 3*k*Ë† .

Let * *= 6*i*Ë† + 2 Ë†*j *âˆ’ 3*k*Ë†
and P =12.

If *d*Ë† is the unit normal vector in the direction
of the vector 6*i*Ë† + 2Ë†*j *âˆ’ 3*k*Ë† , then

If is the position vector of an arbitrary point (*x, y, z*) on the plane, then using . = p , the vector equation of the plane in normal form is

Substituting = *x*Ë†*i + y*Ë†*j + z*Ë†*k* in the above
equation, we get (*x*Ë†*i + y*Ë†*j + z*Ë†*k *) . 1/7 (6Ë†*i + 2*Ë†* j - 3*Ë†*k* ) = 12 .

Applying dot product in the above equation and simplifying, we get
6*x *+ 2 *y *âˆ’ 3*z *= 84, which is the the standard form.

**Example 6.39**

If the Cartesian equation of a plane is 3*x *- 4 *y *+ 3*z *= -8 , find the
vector equation of the plane in the standard form.

**Solution**

If * *= *xi *+ *yj *+ *zk *is the
position vector of an arbitrary point (*x*, *y*, *z*) on the
plane, then the given equation can be written as (*xi*Ë† + *y*Ë†*j *+
*zk*Ë†) â‹… (3*i*Ë† âˆ’ 4 Ë†*j *+ 3*k*Ë†) = âˆ’8 or
(*xi*Ë† + *y*Ë†*j *+ *zk*Ë†) â‹… (âˆ’3*i*Ë† + 4 Ë†*j *âˆ’
3*k*Ë†) = 8 . That is, * *â‹… (âˆ’3Ë†*i *+ 4Ë†*j *âˆ’
3Ë†*k *) = 8 which is the vector equation of the given plane in standard
form.

**Example 6.40**

Find the direction cosines of the normal to the plane and length
of the perpendicular from the origin to the plane * *â‹… (3Ë†*i *âˆ’ 4Ë†
*j *+12Ë†*k *) = 5.

**Solution**

Let = 3*i*Ë† âˆ’ 4 Ë†*j *+12*k*Ë† and
*q *= 5 .

If *d*Ë† is the unit vector in the direction of the vector 3*i*Ë†
âˆ’ 4 Ë†*j *+12*k*Ë† , then *d*Ë† = 1/13 (3*i*Ë† âˆ’ 4 Ë†*j *+12*k*Ë†)

Now, dividing the given equation by 13 , we get

which is the equation of the plane in the normal form .Ë†*d* = *p*

From this equation, we infer that is a unit
vector normal to the plane from the origin. Therefore, the direction cosines of
*d*Ë† are and the length of the perpendicular from the origin to the
plane is 5/13.

Find the vector and Cartesian equations of the plane passing
through the point with position vector 4*i*Ë† + 2 Ë†*j *âˆ’ 3*k*Ë†
and normal to vector 2*i*Ë† âˆ’ Ë†*j *+ *k*Ë† .

If the position vector of the given point is * *= 4*i
*+ 2 *j *âˆ’ 3*k *and * *= 2*i *âˆ’ *j *+ *k *,
then the equation of the plane passing through a point and normal to a vector
is given by (* *âˆ’ ) â‹… * *= 0
or * *â‹… = * *â‹… .

Substituting * *= 4*i *+ 2 *j *âˆ’ 3*k *and * *= 2*i *âˆ’ *j *+ *k *in the
above equation, we get

= (4*i *+ 2 *j *âˆ’ 3*k *). (2*i *âˆ’ *j *+ *k *)

Thus, the required vector equation of the plane is * *â‹…(2Ë†*i *âˆ’ Ë†*j *+ Ë†*k *) = 3 . If * *=
*x*Ë†*i *+ *y*Ë†*j *+ *z*Ë†*k *then we get the Cartesian
equation of the plane 2*x *âˆ’ *y *+ *z *= 3 .

A variable plane moves in such a way that the sum of the
reciprocals of its intercepts on the coordinate axes is a constant. Show that the
plane passes through a fixed point

The equation of the plane having intercepts *a*, *b*, *c *on
the *x*, *y*, *z *axes respectively is .

Since the sum of the reciprocals of the intercepts on the
coordinate axes is a constant, we have where *k *is a
constant, and which can be written as

This shows that the plane passes through the fixed point

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12th Mathematics : UNIT 6 : Applications of Vector Algebra : Intercept form of the equation of a plane |

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