# RD Sharma Class 12 Ex 29.3 Solutions Chapter 29 The Plane

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## RD Sharma Class 12 Ex 29.3 Solutions Chapter 29 The Plane

### Question 1. Find the vector equation of a plane passing through a point having position vector  and perpendicular to the vector .

Solution:

As we know that the vector equation of a plane passing through a point and normal to is

Here,  and

So, the equation of the required equation is

⇒  = (4)(2) + (2)(-1) + (-3)(1)

⇒  = 8 – 2 – 3

⇒

Hence, the required equation is

### (i)

Solution:

Given vector equation of a plane,

Since denotes the position vector of an arbitrary point (x, y, z) on the plane.

Therefore, putting

⇒

⇒ (x)(12) + (y)(-3) + (z)(4) = -5

⇒ 12x – 3y + 4z + 5 = 0

So, this is the required cartesian equation of the plane.

### (ii)

Solution:

Given vector equation of a plane,

Since  denotes the position vector of an arbitrary point (x, y, z) on the plane.

Therefore, putting

⇒ (x)(-1) + (y)(1) + (z)(2) = 9

⇒ -x + y + 2z = 9

So, this is the required cartesian equation of the plane.

### Question 3. Find the vector equation of the coordinates planes.

Solution:

Vector equation of XY-Plane:

The XY- Plane passes through origin whose position vector is and

perpendicular to Z-axis whose position vector is

So the equation of the XY plane is

⇒

Vector equation of XZ-Plane:

The XZ- Plane passes through origin whose position vector is

and perpendicular to Y-axis whose position vector is

So the equation of the XZ plane is

⇒

Vector equation of YZ-Plane:

The YZ- Plane passes through origin whose position vector is

and perpendicular to X-axis whose position vector is

So the equation of the YZ plane is

⇒

Hence, the vector equation of the coordinates planes.

XY-Plane =

XZ-Plane =

YZ-Plane =

### (i) 2x – y + 2z =8

Solution:

Given equation of plane is,

2x – y + 2z = 8

So,

Therefore, the vector equation of the plane is

### (ii) x + y – z =5

Solution:

Given equation of plane is,

x + y – z = 5

So,

Therefore, Vector equation of the plane is

### (iii) x + y = 3

Solution:

Given equation of plane is,

x + y = 3

So,

Therefore, the vector equation of the plane is

### Question 5. Find the vector and cartesian equations of a plane passing through the point (1, -1, 1) and normal to the line joining the points (1, 2, 5) and (-1, 3, 1).

Solution:

As we know that the vector equation of a plane passing through a point and normal to is

…..(1)

So, according to the question it is given that the plane passes through the point (1, -1, 1) and

normal to the line joining the points A(1, 2, 5) and B(-1, 3, 1).

So,

= Position vector of  – Position vector of

Now, from eq (1), we get

⇒

⇒

⇒

⇒

⇒   …(2)

Now, Putting  in eq(2), we get

2x – y + 4z = 7

So, the vector equation is the plane is

and the cartesian equation of the plane is 2x – y + 4z = 7

### Question 6. If  is a vector of magnitude  √3 and is equally inclined with an acute angle with the coordinate axes. Find the vector and cartesian forms of equation of a plane which passes through (2, 1, -1) and is normal to

Solution:

Given that  = √3 and  makes equal angle with coordinate axes.

Let us considered  has direction cosine as u, v and w, and

it makes angle of α, β and γ with the coordinate axes.

So, α = β = γ

cos α = cos β = cos γ

Assuming u = v = w = p

We know that

u+ v2 + w2 = 1

p+ p+ p2 = 1

P2 = 1/3

P= ±1/√3

So,

u = ±1/√3

cos α = ±1/√3

Now, α = cos-1(-1/√3)

It gives, α is an obtuse angle so, neglect it.

Now, α = cos-1(1/√3)

It gives, α is an acute angle, so

cos α = ±1/√3

u = v = w = 1/√3

So,

= √3

Now, and

As we know that the vector equation of a plane passing through a point  and normal to  is

⇒

⇒

⇒   ….(1)

Now, Putting  in eq(1), we get

⇒

(x)(1) + (y)(1) + (z)(1) = 2

⇒ x + y + z = 2

Hence, the vector equation of the plane is

and the cartesian equation of the plane is x + y + z = 2

### Question 7. The coordinates of the foot of the perpendicular drawn from the origin to a plane are (12, -4, 3). Find the equation of the plane.

Solution:

According to the question it is given that the coordinates of the foot of the

perpendicular drawn from the origin O to a plane is P(12, -4, 3)

Thus, we can say that the required plane is passing through P(12, -4, 3) and perpendicular to OP.

As we know that the vector equation of a plane passing through a point  and normal to  is

…….(1)

Here,

On putting the values of  and  in equation (1)

⇒

⇒

⇒

⇒   ……..(2)

Now, on putting  in eq(2), we get

⇒

(x)(12) + (y)(-4) + (z)(3) = 169

⇒ 12x – 4y + 3z = 169

Hence, the vector equation of the plane is

and the cartesian equation of the plane is 12x- 4y+ 3z = 169

### Question 8. Find the equation of the plane passing through the point (2, 3, 1) given that the direction ratios of normal to the plane are proportional to (5, 3, 2).

Solution:

According to the question it is given that the plane is passing through P(2, 3, 1)

having (5, 3, 2) as the direction ratios of the normal to the plane.

As we know that the vector equation of a plane passing through a point  and normal to  is

…..(i)

So,

Now put all these values in equation (i),

………(2)

Now, putting,  in eq(2), we get

(x)(5) + (y)(3) + (z)(2) = 21

5x + 3y + 2z = 21

Hence, this is the required equation of plane

### Question 9. If the axes are perpendicular and P is the point (2, 3, -1), find the equation of the plane through P at right angles to OP.

Solution:

According to the question it is given that P is the point (2,3,-1) and

the plane is passing through P and OP is the vector normal to the plane.

So,

As we know that the vector equation of a plane passing through a point  and normal to  is

…..(i)

Here,

So,  = Position vector of P – Position vector of O

Now put, the value of  and  in equation (i),

……..(2)

Now putting, , in eq(2), we get

(x)(2) + (y)(3) + (z)(−1) = 14

2x + 3y − z = 14

So, this is the required equation of plane.

### Question 10. Find the intercepts made on the coordinate axes by the plane 2x + y -2z = 3 and find also the direction cosines of the normal to the plane.

Solution:

According to the question

The equation os plane = 2x + y -2z = 3

Now divide both sides of the equation by 3, we get

……(i)

Here, if a, b, c are the intercepts by a plane on the coordinates axes,

then the equation of the plane is:

……(ii)

On comparing the equation (i) and (ii), we get the value of a, b, and c

So, from the given equation of plane,

Hence, the vector normal to the plane is,

= √{(2)+ (1)+ (-2)2}

= √(4 + 1 + 4)

= √9

Hence, the unit vector perpendicular to

Hence, the direction cosine of normal to the plane = 2/3, 1/3, -2/3

### Question 11. A plane passes through the point (1, -2, 5) and is perpendicular to the line joining the origin to the point (). Find the vector and cartesian forms of the equation of the plane.

Solution:

As we know that the vector equation of a plane passing through a point and normal to  is

….(i)

Here,

= Position vector of P – Position of vector of O

Now, put, all these values in equation (i), we get,

….(2)

Now put  in eq(2), we get

(x)(3) + (y)(1) + (z)(−1) = -4

3x + y − z = -4

So, this is the required equation of plane.

### Question 12. Find the equation of the plane that bisects the line segment joining points (1, 2, 3) and (3, 4, 5) and is at right angle to it.

Solution:

As we know that the vector equation of a plane passing through a point  and normal to  is

…..(i)

Here, = mid-point of AB

So,

= Position vector of A + Position of vector of B/ 2

And,

= Position vector of B – Position of vector of A

Now put all these values in eq(1), we get

…..(2)

Now put  in eq(2), we get

(x)(2) + (y)(2) + (z)(2) = 18

2x + 2y + 2z = 18

or  we can write as

x + y + z = 9

So, this is the required equation of plane.

### (i) x – y + z – 2 = 0 and 3x + 2y – z + 4 = 0

Solution:

Given equations of planes are

x – y + z – 2 = 0 and 3x + 2y – z + 4 = 0

So first we solve, x – y + z – 2 = 0

…..(i)

Now we solve, 3x + 2y – z =- 4

…(ii)

So, from eq(i) and (ii), we conclude that

is normal to eq(i) and  is normal to eq(ii)

So,

=

= (1)(3) + (-1)(2) + (1)(-1)

=3 – 2 – 1

= 3 – 3 = 0

Hence, is perpendicular to

### (ii)  and

Solution:

Given equations of planes are

and

So first we solve,

…..(1)

Now we solve,

……(2)

So, from eq(i) and (ii), we conclude that

is normal to eq(i) and  is normal to eq(ii)

So,

=

= (2)(2) + (-1)(-2) + (3)(-2)

= 4 + 2 – 6

= 6 – 6

= 0

Hence, is perpendicular to

### Question 14. Show that the normal vector to the plane 2x + 2y + 2z = 3 is equally inclined with the coordinate axes.

Solution:

Equation of plane = 2x + 2y + 2z = 3

So,

So, the normal to the plane

and the direction ratio of

So, the direction cosine of  ….(1)

= √[(2)+ (2)+ (2)2]

= √[4 + 4 + 4]

= √12 = 2√3

Now put the value of |\vec{n}| in eq(1), we get

Direction cosine of |\vec{n}| =

So, u = 1/√3, v = 1/√3, W = 1/√3

Let us assume that the α, β, γ be the angle that normal \vec{n} makes with the coordinate axes.

So, u = cos α = 1/√3

α = cos-11/√3                   ….(2)

v = cos β = 1/√3

β = cos-11/√3                     ….(3)

w = cos γ = 1/√3

γ = cos-11/√3                     ….(4)

So, from equation (2), (3) and (4), we get

α = β = γ

Hence proved that the normal vector to the plane 2x + 2y + 2z = 3 is equally inclined with the coordinate axes.

### Question 15. Find a vector of magnitude 26 units normal to the plane 12x – 3y + 4y = 1.

Solution:

Given that,

The equation of plane is = 12x – 3y + 4y = 1

and the magnitude = 26 units

So,

The normal to the plane is

= √[(12)+ (-3)+ (4)2]

= √[144 + 9 + 16]

= √169 = 13

Hence, the unit vector  =

Now we find a vector normal to the plane with magnitude

So,

26 = 26

= 26

So, this is the required vector

### Question 16. If the line drawn from (4, -1, 2) meets a plane at right angles at the point (-10, 5, 4), find the equation of the plane.

Solution:

As we know that the vector equation of a plane passing through a point  and normal to  is

…..(i)

Here,  = position vector of B

So,

and

= Position vector of B – Position of vector of A

Now put all these values in eq(1), we get

…..(2)

Now put  in eq(2), we get

(x)(-14) + (y)(6) + (z)(2) = 178

-14x + 6y + 3z = 178

Or we can write as

7x – 2y – z = -89

So, this is the required equation of plane.

### Question 17. Find the equation of the plane which bisects the line segment joining the points (-1, 2, 3) and (3, -5, 6) at right angles.

Solution:

Let assume that point (-1, 2, 3) is A point and point (3, -5, 6) is B point and C be the line mid-point of line segment AB

As we know that the vector equation of a plane passing through a point  and normal to  is

…(i)

Here, = Position vector of C

So,      [Because c is the mid point of line AB]

Now,

= Position vector of B- Position vector of A

=

Now put all these values in eq(1), we get

= 28 ….(2)

Now put  in eq(2), we get

(x)(4) + (y)(-7) + (z)(3) = 28

4x – 7y + 3z = 28

So, this is the required equation of plane.

### Question 18. Find the vector and cartesian equations of the plane which passes through the point (5, 2, -4) and perpendicular to the line with direction ratios 2, 3, -1.

Solution:

According to the given question

As we know that the vector equation of a plane passing through a point  and normal to  is

So,

….(1)

For cartesian equation:

Put  in eq(1), we get

(x)(2) + (y)(3) + (z)(-1) = 20

2x + 3y -z = 20

So, this is the required equation of plane.

### Question 19.  If O be the origin and the coordinates of P be (1, 2, -3), then find the equation of the plane passing through P and perpendicular to OP.

Solution:

According to the question, a normal pass through point O(0, 0, 0) and P (1, 2, -3)

So,

and

As we know that the vector equation of a plane passing through a point  and normal to  is

So,

For cartesian equation:

Put  in eq(1), we get

(x)(1) + (y)(2) + (z)(-3) = 14

x + 2y – 3z = 14

So, this is the required equation of plane.

### Question 20. If O is the origin and the coordinates of A are (a, b, c). Find the direction cosines of OA and the equation of the plane through A at right angles to OA.

Solution:

According to the question it is given that, O is the origin and the coordinates of A are (a, b, c)

So,

Since, the direction ratios of OA are proportional to a, b, c

So, the direction cosines are:

So the equation of the line is,

ax + by + cz = a+ b+ c2

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