Plane Exercise 10.1 Class 12 Basic Mathematics Solution [NEB UPDATED]

Syllabus: Content to Study

1. Equation of the plane in Intercept Form
2. Equation of the plane in normal form
3. Equation of plane through three given points
4. Plane through the intersection of two planes
5. Angle Between two planes
6. Angles between a line and a plane
7. Length of perpendicular from a given point on a given plane
   

Exercise 10

1. a) Find the intercepts made by the plane 2x + 3y+4z = 12 on the coordinate axes.

Solution:

Given plane is;

Or, 2x + 3y + 4z = 12

Or, $\frac{{\rm{x}}}{6} + \frac{{\rm{y}}}{4} + \frac{{\rm{z}}}{3}$ = 1.

Comparing with

Or, $\frac{{\rm{x}}}{{\rm{a}}} + \frac{{\rm{y}}}{{\rm{b}}} + \frac{{\rm{z}}}{{\rm{c}}}$ = 1.

x - intercept, a = 6 , y – intercept, b = 4 and z – intercept, c = 3.

So, required intercepts are 6,4,3. 

b. Reduce the equation 4x+8y + 8z = 9 to the normal form and also determine the direction cosines of the normal and the length of the perpendicular to it from the origin.

Solution:

Given plane is;

Or, 4x + 8y + 8z = 9.

Dividing by $\sqrt {{4^2} + {8^2} + {8^2}} $ = 12,

We have,

Or,$\frac{{\rm{x}}}{3} + \frac{{2{\rm{y}}}}{3} + \frac{{2{\rm{z}}}}{3}$ = $\frac{3}{4}$.

Which is the required normal form. The d.c.’s of the normal $\frac{1}{3},\frac{2}{3},\frac{2}{3}$ and the length of perpendicular from origin to the plane is, p = $\frac{3}{4}$.

 

2. a) Find the equation of a plane which makes intercepts 2, 3, 4 on x-axis, y-axis and z-axis respectively.

Solution:

Given, x – intercept, a = 2, y – intercept, b = 3 and z – intercept , c = 4.

So, the required plane is $\frac{{\rm{x}}}{{\rm{a}}} + \frac{{\rm{y}}}{{\rm{b}}} + \frac{{\rm{z}}}{{\rm{c}}}$ = 1.

So, $\frac{{\rm{x}}}{2} + \frac{{\rm{y}}}{3} + \frac{{\rm{z}}}{4}$ = 1.

So, 6x + 4y + 3x = 12.

 

b. Find the equation of the plane which makes equal intercepts on the axes and passes through the point (2, 3, 4).

Solution:

Let the equal intercept be a.

So, the required plane is $\frac{{\rm{x}}}{{\rm{a}}} + \frac{{\rm{y}}}{{\rm{b}}} + \frac{{\rm{z}}}{{\rm{c}}}$ = 1.

So, x + y + z = a …(i)

Since plane (i) passes through (2,3,4),

So, 2 + 3 + 4 = a

So, a = 9.

Thus, from (i), the required plane is, x + y + z = 9.

 

3. a) Find the equation of the plane passing through the points (1, 1, 0), (-2, 2, -1) and (1, 2, 1)

Solution:
The given points are A(1,1,0), B = (-2,2,-1) and C(1,2,1).

Now, any plane through A(1,1,0) is:

a(x – 1) + b(y – 1) + c(z – 0) = 0 …….(i)

The plant (i) also passes through B and C. so,

Or, a(- 2 – 1) + b(2 – 1) + c( - 1 – 0) = 0

So, - 3a + b – c = 0 …..(ii)

And, a(1 – 1) + b(2 – 1) + c(1 – 0) = 0.

So, a.0 + b + c = 0…… (iii)

From (ii) and (iii) by cross multiplication, we have,

Or, $\frac{{\rm{a}}}{{1 + 1}} = \frac{{\rm{b}}}{{0 + 3}} = \frac{{\rm{c}}}{{ - 3 - 0}}$.

Or, $\frac{{\rm{a}}}{2} = \frac{{\rm{b}}}{3} = \frac{{\rm{c}}}{{ - 3}}$   …….(iv)

Now from (i) and (ii), we have,

Or, 2(x – 1) + 3(y – 1) – 3(z – 0) = 0

So, 2x + 3y – 3z = 5.

 

b. Show that the four points (0,-1,-1), (4, 5, 1), (3, 9, 4) and (-4, 4, 4) are coplanar.

Solution:

Given points are A(0,-1,-1), B(4,5,1), C(3,9,4) and D(-4,4,4)

To show: A,B,C and D are coplanar.

Now, any plane through A(0,-1,-1) is:

Or, a(x – 0) + b(y + 1) + c(z + 1) = 0 …(i)

The plane (i) passes through B and C, so

Or, a(4 – 0) + b(5 + 1) + c(1 + 1) = 0

So, 4a + 6b + 2c = 0 …..(ii)

And, a(3 – 0) + b (9 + 1) + c(4 + 1) = 0 ……. (ii)

So, 3a + 10b + 5c = 0 …(iii).

From (ii) and (iii), we have,

Or, $\frac{{\rm{a}}}{{30 - 20}} = \frac{{\rm{b}}}{{6 - 20}} = \frac{{\rm{c}}}{{40 - 18}}$

So, $\frac{{\rm{a}}}{{10}} = \frac{{\rm{b}}}{{ - 14}} = \frac{{\rm{c}}}{{22}}$

So, $\frac{{\rm{a}}}{5} = \frac{{\rm{b}}}{{ - 7}} = \frac{{\rm{c}}}{{11}}$ …(iv)

Eliminating a,b,c from (i) and (iv), we have,

Or, 5(x – 0) – 7(y + 1) + 11(z + 1) = 0

So, 5x – 7y + 11z + 4 = 0 ….(v)

Is the plane through A,B and C.

Now, putting D(-4,4,4) in (v), we have,

Or, 5 * (-4) – 7 * 4 + 11 * 4 + 4 = - 20 – 28 + 44 + 4 = 0

Hence, the given points are coplanar and they lie on the plane,

5x – 7y + 11z + 4 = 0.

 

4. a) Find the equation of the plane through the point (3, 4, 5) and parallel to the plane 3x - 4y + 5z = 7

Solution:

Any plane through (3,-4,5) is:

Or, a(x – 3) + b(y + 4) + c(z – 5) = 0

The plane (i) is parallel to the plane,

Or, 3x – 4y + 5x = 7.

So, the normal’s plane (i) and (ii) are parallel.

So, $\frac{{\rm{a}}}{3} = \frac{{\rm{b}}}{{ - 4}} = \frac{{\rm{c}}}{5}$ ….(iii)

Now, from (i) and (iii) we have,

Or, 3(x – 3) – 4(y + 4) + 5(z – 5) = 0

So, 3x – 4y + 5z – 50 = 0.

 

b. Show that the equation of the plane through (alpha, - beta, gamma) and parallel to the plane ax + by + cz = 0 is ax + by + cz = aα+ bβ+ c γ.

Solution:

Given plane is ax + by + cz = 0.

Any plane to the plane (i) is ax + by + cz + k = 0 …(ii)

Where, k is to be determined,

The plane (ii) passes through (α,β,γ),so , aα + bβ + cγ+ k = 0.

So, k = - (aα + bβ + cγ)

Thus, from (ii) the required plane is, ax + by + cz = aα + bβ + cγ.

 

5. Find the angle between the two planes

i. x + 2y + z + 7 = 0 and 2x + y - z + 13 = 0

Solution:

The given plane are, x + 2y + z + 7 = 0 and 2x + y – z + 13 = 0.

Let θ be the angle between the planes,

So, cosθ = $\frac{{{{\rm{a}}_1}.{{\rm{a}}_2} + {{\rm{b}}_1}.{{\rm{b}}_2} + {{\rm{c}}_1}.{{\rm{c}}_2}}}{{\sqrt {({\rm{a}}_1^2 + {\rm{b}}_1^2 + {\rm{c}}_1^2)\left( {{\rm{a}}_2^2 + {\rm{b}}_2^2 + {\rm{c}}_2^2} \right)} .}}$.

= $\frac{{1.2 + 2.1 - 1.1}}{{\sqrt {\left( {{1^2} + {2^2} + {1^2}} \right)\left( {{2^2} + {1^2} + {{\left( { - 1} \right)}^2}} \right)} }}$ = $\frac{3}{6}$ = $\frac{1}{2}$.

So, θ = 60° = $\frac{{\rm{\pi }}}{3}$.

 

ii. x + 2y + 3z = 6 and 3x - 3y + z = 1

The given planes are, x + 2y + 3z = 6 and 3x – 3y + z = 1.

Let θ be the angle between the planes.

So, cosθ = $\frac{{1.3 + 2.\left( { - 3} \right) + 3.1}}{{\sqrt {\left( {1 + 4 + 9} \right)\left( {9 + 9 + 1} \right)} }}$ = 0

So, θ = $\frac{{\rm{\pi }}}{2}.$

 

6. Show that the plane 2x + 3y - 4z = 3 is parallel to the plane 10x + 15y - 20z = 12 and perpendicular to the plane 3x + 2y + 3z = 5

Solution:

The planes are:

2x + 3y – 4z = 3 ….(i)

10x + 15y – 20z = 12…(ii)

And 3x + 2y + 3z = 5 …(iii)

Now planes (i) and (ii) will be parallel if the direction ratios of the normal’s to the plane (i) and (ii) are proportional.

i.e. if $\frac{2}{{10}}$ = $\frac{3}{{15}}$ = $\frac{{ - 4}}{{ - 20}}$

i.e. $\frac{1}{5} = \frac{1}{5} = \frac{1}{5}.$

Hence, the plane (i) and (ii) are parallel.

Again, the plane (i) and (iii) will be perpendicular,

If a₁.a₂ + b₁b₂ + c₁c₂ = 0

i.e. if 2.3 + 3.2 + (-4).3 = 0

i.e. if 0 = 0.

So, a₁.a₂ + b₁.b₂ + c₁.c₂ = 0

So, the planes (i) and (iii) are perpendicular.

 

7. a) Find the equation of the plane through the point (1, 2, 3) and normal to the planes x - y - z = 5 and 2x - 5y - 3z = 7.

Solution:

Any plane through (1,2,3) is:

Or, a(x – 1) + b(y – 2) + c(z – 3) = 0 …(i)

Since the plane (i) is normal to the planes

x – y – z = 5 …(ii)

And 2x – 5y – 3z = 7 …(iii)

So, a.1 + b(-1) + c(-1) = 0     [from(i) and (ii)]

So, a – b – c = 0 ….(iv)

And 2a – 5b – 3c = 0 ….()    [from (i) and (iii)]

So, from (iv) and (v) we have, $\frac{{\rm{a}}}{{3 - 5}} = \frac{{\rm{b}}}{{ - 2 + 3}}$ = $\frac{{\rm{c}}}{{ - 5 + 2}}$.

So, $\frac{{\rm{a}}}{{ - 2}} = \frac{{\rm{b}}}{1} = \frac{{\rm{c}}}{{ - 3}}$ …(vi)

From (i) and (vi),

We have, - 2(x – 1) + 1(y – 2) – 3(z – 3) = 0

Or, - 2x + 2 + y – 2 – 3z + 9 = 0

So, 2x – y + 3z – 9 = 0. 

b) Find the equation of the plane through the point (2, 1, 4) and perpendicular to each of the planes 9x - 7y + 6z + 48 = 0 and x + y + z = 0

Solution:

Any plane through (2,1,4) is:

Or, a(x – 2) + b(y – 1) + c(z – 4) = 0….(i)

Since, the plane (i) is perpendicular to the plaes:

9x – 7y + 6z + 48 = 0 ….(ii)

And x + y + z = 0 ….(iii)

So, we have, 9a – 7b + 6c = 0

And a + b + c = 0.

So, $\frac{{\rm{a}}}{{ - 7 - 6}} = \frac{{\rm{b}}}{{6 - 9}} = \frac{{\rm{c}}}{{9 + 7}}$.

So, $\frac{{\rm{a}}}{{ - 13}} =  = \frac{{\rm{b}}}{{ - 3}} = \frac{{\rm{c}}}{{16}}$ …(iv)

Thus, from (i) and (iv) we have, - 13(x – 2) – 3(y – 1) + 16(z – 4) = 0.

Or, - 13x + 26 – 3y + 3 + 16z – 64 = 0.

So, 13x + 3y – 16z + 35 = 0.

 

8. a) Find the equation of the plane through P(a, b, c) and perpendicular to OP.

Solution:

Any plane through P(a,b,c) is:

Or, l(x – a) + m(y – b) + n(z – c) = 0 …(i)

Since the plane (i) is perpendicular to the line OP joining O(0,0,0) and P(a,b,c) so the line normal to the plane (i) is parallel to the line OP, so we have,

Or, $\frac{{\rm{l}}}{{{\rm{a}} - 0}} = \frac{{\rm{m}}}{{{\rm{b}} - 0}} = \frac{{\rm{n}}}{{{\rm{c}} - 0}}$.

So, $\frac{{\rm{l}}}{{\rm{a}}} = \frac{{\rm{m}}}{{\rm{b}}} = \frac{{\rm{n}}}{{\rm{c}}}$ ….(ii)

Thus, from (i) and (ii)

We have, a(x – a) + b(y – b) + c(z – c) = 0.

So, ax + by + cz = a² + b² + c², is the required plane.

 

b. Find the equation of the plane through the point (2, -3, 1) and perpendicular to the line joining the two points (3, 4, -1) and (2, -1, 5).

Solution:

Any plane through (2,-3,1) is

a (x – 2) + b(y + 3) + c(z – 1) = 0 ….(i)

Since the plane (i) is perpendicular to the line joining the points (3,4,-1) and B(2,-1,5) so the line normal to the plane (i) is parallel to the line AB, so we have,

Or, $\frac{{\rm{a}}}{{3 - 2}} = \frac{{\rm{b}}}{{4 + 1}} = \frac{{\rm{c}}}{{ - 1 - 5}}$     [3 – 4,4 + 1, - 1 – 5 are dr’s of AB]

So, $\frac{{\rm{a}}}{1} = \frac{{\rm{b}}}{5} = \frac{{\rm{c}}}{{ - 6}}$.

Thus, from (i) and (ii) we have,

1(x – 2) + 5(y + 3) – 6(z – 1) = 0

So, x + 5y – 6z + 19 = 0.

c) The foot of the perpendicular drawn from the origin to a plane is (3,-4, 2). Find the equation of the plane. 

Solution:

Since, the foot of the perpendicular to the plane is A(3,-4, 2). Therefore (3,-4, 2) is the point on the plane.

So, equation of the plane passing through the point (3,-4, 2) is:

a(x–3)+b(y+4)+c(z–2)=0.

Now, the direction ratios of the perpendicular line OA = 3-0,-4-0,2-0=3,-4, 2

Therefore, the required plane is:

3(x–3)-4(y+4)+2(z–2)=0

i.e. 3x-4y+2z-29=0

9. a) Find the equation of the plane through the point (2, 2, 1) and (9, 3, 6) and normal to the plane 2x + 6y + 6z = 9

Solution:

Any plane (2,2,1) is a(x – 2) + b(y – 2) + c(z – 1) = 0 ….(1)

The plane (1) passes through (9,3,6), so 7a + b + 5c = 0 ….(2)

The plane (1) is normal to the plane 2x + 6y + 6z + 9 , so

2a + 6b + 6c = 0 …(3)    [a₁.a₂ + b₁.b₂ + c₁.c₂ = 0]

From (2) and (3) we have, $\frac{{\rm{a}}}{{6 - 30}} = \frac{{\rm{b}}}{{10 - 42}} = \frac{{\rm{c}}}{{42 - 2}}$.

Or, $\frac{{\rm{a}}}{3} = \frac{{\rm{b}}}{4} = \frac{{\rm{c}}}{{ - 5}}$. …(4)

Eliminating a,b,c from (1) and (4), so,

So, 3(x – 2) + 4(y – 2) – 5(z – 1) = 0.

So, 3x + 4y – 5z – 9 = 0.

 

b. Find the equation of the plane through the points (-1, 1, -1) and (6, 2, 1) and perpendicular to the plane 2x + y + z = 5

Solution:

Any plane (-1,1,-1) is a(x + 1) + b(y – 1) + c(z + 1) = 0 ….(1)

The plane (1) passes through (6,2,1), so 7a + b + 2c = 0 ….(2)

The plane (1) is normal to the plane 2x + y + z = 5 , so

2a + b + c = 0 …(3)   

From (2) and (3) we have, $\frac{{\rm{a}}}{{1 - 2}} = \frac{{\rm{b}}}{{4 - 7}} = \frac{{\rm{c}}}{{7 - 2}}$ à $\frac{{\rm{a}}}{1} = \frac{{\rm{b}}}{3} = \frac{{\rm{c}}}{{ - 5}}$.

Eliminating a,b,c from (1) and (4), so,

So, 1(x + 1) + 3(y – 1) – 5(z + 1) = 0.

So, x + 3y – 5z = 7.

 

10. a) Find the equation of the plane through the intersection of the planes x - 2y + 3z + 10 = 0 and 3x + 5y - 2z - 14 = 0 and

(i) passing through the point (-1, 4, 3)

(ii) parallel to the x-axis.

Solution:

Solving the two given planes simultaneously, we get:

x - 2y + 3z = -10 ...(1)

3x + 5y - 2z = 14 ...(2)

Multiplying equation (1) by 3 and adding it to equation (2), we eliminate the variable x and get:

17y - 7z = 4 ...(3)

Multiplying equation (1) by 5 and subtracting it from equation (2), we eliminate the variable y and get:

13x + 17z = 64 ...(4)

Now, we have two equations with two variables (y and z). Solving them simultaneously, we get:

y = (4 + 7z)/17 ...(5)

x = (64 - 17z)/13 ...(6)

Substituting these values of x and y in equation (1), we get:

(64 - 17z)/13 - 2(4 + 7z)/13 + 3z = -10

Simplifying the above equation, we get:

3x + 4y - z = 9

Therefore, the equation of the plane passing through the intersection of the given planes is:

3x + 4y - z = 9

Now, we need to find the equation of the plane passing through the point (-1, 4, 3) and parallel to the x-axis. Since the plane is parallel to the x-axis, its normal vector is in the direction of the x-axis, which is (1, 0, 0).

Therefore, the equation of the plane can be written as:

(1)(x - (-1)) + (0)(y - 4) + (0)(z - 3) = 0

Simplifying the above equation, we get:

x + 1 = 0

Therefore, the equation of the plane passing through the point (-1, 4, 3) and parallel to the x-axis is:

x + 1 = 0

Alternatively, if you prefer the equation of the plane in the form of ax + by + cz = d, we can rewrite the above equation as:

x - 0y + 0z = -1

Therefore, the equation of the plane passing through the point (-1, 4, 3) and parallel to the x-axis is:

x - 0y + 0z = -1

which can also be written as:

x = -1

Finally, it is worth noting that there is a typo in part (ii) of the original question. The correct equation should be:

y - z = 4 

b. Find the equation of the plane through the intersection of the planes x + y + z = 6and 2x + 3y + 4z + 5 = 0 and perpendicular to the plane 4x + 5y - 3z = 8

Solution:

Any plane through the planes x + y + z – 6 = 0 and 2x + 3y + 4z + 5 = 0 is

X + y + z – 6 + k(2x + 3y + 4z + 5) = 0 ….(i)

Where, k is to be determined.

So, (1 + 2k)x + (1 + 3k)y + (1 + 4k)z + (-6 + 5k) = 0 ….(2)

The plane (2) is perpendicular to the plane 4x + 5y – 3z = 8, so,

Or, 4(1 + 2k) + 5(1 + 3k) – 3(1 + 4k) = 0

Or, 6 + 11k = 0

So, k = $ - \frac{6}{{11}}$.

Thus, from (1), the required plane is,

Or, 11(x + y + z – 6) – 6(2x + 3y + 4z + 5) = 0

So, x + 7y + 13z + 96 = 0.

 

11. Find the length of the perpendicular

(i) from the point (2, 5, 7) to the plane 6x + 6y + 3z = 11

Solution:

Given plane is 6x + 6y + 3z – 11 = 0 …(1)

So, the perpendicular distance from (2,5,7) to (1) is:

P =  $\frac{{\left| {6.2 + 6.5 + 3.7 - 11} \right|}}{{\sqrt {{6^2} + {6^2} + {3^2}} }}$ = $\frac{{52}}{9}$ units.

(ii) from the origin on the plane 3x - 2y + 6z = 17

Solution:

Required length, p = $\frac{{\left| {3.0 - 2.0 + 6.0 - 17} \right|}}{{\sqrt {9 + 4 + 36} }}$ = $\frac{{17}}{7}$ units.

 

12. Find the distance between the following planes

(i) x - y + 2z - 4 = 0 and 2x – 2y + 4z + 5 = 0

Solution:

The given planes are, x – y + 2z – 4 = 0 ….(1)

And, 2x – 2y + 4z + 5 = 0
So, x – y + 2z + $\frac{5}{2}$ = 0 ….(2)

Since, the planes (1) and (2) are parallel, so the distance between them ;s:

d = $\frac{{\left| {{{\rm{c}}_1} - {{\rm{c}}_2}} \right|}}{{\sqrt {{{\rm{a}}^2} + {{\rm{b}}^2} + {{\rm{c}}^2}} }}$

= $\frac{{\left| { - 4 - \frac{5}{2}} \right|}}{{\sqrt {{1^2} + {{\left( { - 1} \right)}^2} + {2^2}} }}$ = $\frac{{13}}{{2\sqrt 6 }}$ units.

 

(ii) 3x + 2y - 6z + 1 = 0 and 6x + 4y - 12z + 9 = 0
Solution:

Given planes are,6x + 4y – 12z + 9 = 0 ….(1)

And 3x + 2y – 6z + 1 = 0

So, 6x + 4y – 12z + 2 = 0 …(2)

Since, planes (1) and (2) are parallel so the distance between them is,

d = $\frac{{\left| {{{\rm{c}}_1} - {{\rm{c}}_2}} \right|}}{{\sqrt {{{\rm{a}}^2} + {{\rm{b}}^2} + {{\rm{c}}^2}} }}$ = $\frac{{9 - 2}}{{\sqrt {{6^2} + {4^2} + {{\left( { - 12} \right)}^2}} }}$ = $\frac{7}{{\sqrt {196} }}$ = $\frac{1}{2}$ units.

 

13. A variable plane is at a constant distance 3p from the origin and meets the axes in the points A, B, C. Prove that the locus of the centroid of the triangle ABC is$\frac{1}{{{{\rm{x}}^2}}} + \frac{1}{{{{\rm{y}}^2}}} + \frac{1}{{{{\rm{z}}^2}}} = \frac{1}{{{{\rm{p}}^2}}}$.

Solution:

Let the plane cut the axes at A(a,0,0), B(0,b,0) and C(0,0,C) so that the plane ABC is,

Or, $\frac{{\rm{x}}}{{\rm{a}}} + \frac{{\rm{y}}}{{\rm{b}}} + \frac{{\rm{z}}}{{\rm{c}}}$ = 1….(1)

Since, the length of the perpendicular from the origin on the plane (1) is 3p so,

Or, 3p = $\frac{{\left| {\frac{0}{{\rm{a}}} + \frac{0}{{\rm{b}}} + \frac{0}{{\rm{c}}} - 1} \right|}}{{\sqrt {\frac{1}{{{{\rm{a}}^2}}} + \frac{1}{{{{\rm{b}}^2}}} + \frac{1}{{{{\rm{c}}^2}}}} }}$ = $\frac{1}{{\sqrt {\frac{1}{{{{\rm{a}}^2}}} + \frac{1}{{{{\rm{b}}^2}}} + \frac{1}{{{{\rm{c}}^2}}}} }}$

So, $\frac{1}{{{{\rm{a}}^2}}} + \frac{1}{{{{\rm{b}}^2}}} + \frac{1}{{{{\rm{c}}^2}}}$ = $\frac{1}{{9{{\rm{p}}^2}}}$ …(2)

Let P (α,β,γ) be the centroid of $\Delta $ABC, so α = $\frac{{{\rm{a}} + 0 + 0}}{3}$ à a = 3α.

Simi;ar;y, b = 3β and c = 3γ.

Thus, from (2), we get, $\frac{1}{{9{{\rm{a}}^2}}} + \frac{1}{{9{\beta ^2}}} + \frac{1}{{9{\gamma ^2}}} = \frac{1}{{9{{\rm{p}}^2}}}$.

So, $\frac{1}{{{\alpha ^2}}} + \frac{1}{{{\beta ^2}}} + \frac{1}{{{\gamma ^2}}} = \frac{1}{{{{\rm{p}}^2}}}.$

So, the locus of (α,β,γ) is, $\frac{1}{{{{\rm{x}}^2}}} + \frac{1}{{{{\rm{y}}^2}}} + \frac{1}{{{{\rm{z}}^2}}} = \frac{1}{{{{\rm{p}}^2}}}$.