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Class 12 RD Sharma Solutions – Chapter 29 The Plane – Exercise 29.3 | Set 1

  • Last Updated : 19 Apr, 2021

Question 1. Find the vector equation of a plane passing through a point having position vector 2\hat{i}-\hat{j}+\hat{k}    and perpendicular to the vector 4\hat{i}+2\hat{j}-3\hat{k}   .

Solution:

As we know that the vector equation of a plane passing through a point \vec{a}   and normal to \vec{n}   is

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(\vec{r}-\vec{a}).\vec{n} =0



\vec{r}.\vec{n}=\vec{a}.\vec{n}

Here, \vec{a}=2\hat{i}-\hat{j}+\hat{k}    and \vec{n}=4\hat{i}+2\hat{j}-3\hat{k}

So, the equation of the required equation is 

\vec{r}.(4\hat{i}+2\hat{j}-3\hat{k})=(2\hat{i}-\hat{j}+\hat{k})(4\hat{i}+2\hat{j}-3\hat{k})

⇒ \vec{r}.(4\hat{i}+2\hat{j}-3\hat{k})  = (4)(2) + (2)(-1) + (-3)(1)

⇒ \vec{r}.(4\hat{i}+2\hat{j}-3\hat{k})  = 8 – 2 – 3

⇒ \vec{r}.(4\hat{i}+2\hat{j}-3\hat{k})=3

Hence, the required equation is 



\vec{r}.(4\hat{i}+2\hat{j}-3\hat{k})-3=0

Question 2. Find the cartesian form of equation of a plane whose vector equation is

(i) \vec{r}.(12\hat{i}-3\hat{j}+4\hat{k})+5=0                      

Solution:

Given vector equation of a plane,

\vec{r}.(12\hat{i}-3\hat{j}+4\hat{k})+5=0

Since \hat{i}, \hat{j}, \hat{k}   denotes the position vector of an arbitrary point (x, y, z) on the plane. 

Therefore, putting \vec{r}=x\hat{i}+y\hat{j}+j\hat{k}

⇒ (x\hat{i}+y\hat{j}+j\hat{k}).(12\hat{i}-3\hat{j}+4\hat{k})+5=0

(x)(12) + (y)(-3) + (z)(4) = -5

⇒ 12x – 3y + 4z + 5 = 0

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



(ii) \vec{r}.(-\hat{i}+\hat{j}+2\hat{k})=9

Solution:

Given vector equation of a plane,

\vec{r}.(-\hat{i}+\hat{j}+2\hat{k})=9

Since \hat{i}, \hat{j}, \hat{k}    denotes the position vector of an arbitrary point (x, y, z) on the plane. 

Therefore, putting \vec{r}=x\hat{i}+y\hat{j}+j\hat{k}                

 (x\hat{i}+y\hat{j}+j\hat{k}).(-\hat{i}+\hat{j}+2\hat{k})=9

(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 \vec{a}=\vec{0}   and 

perpendicular to Z-axis whose position vector is \vec{k}

So the equation of the XY plane is \vec{r}.\vec{n}=\vec{a}.\vec{n}

⇒ \vec{r}.\vec{k}=\vec{0}.\vec{k}

\vec{r}.\vec{k}=0

Vector equation of XZ-Plane:

The XZ- Plane passes through origin whose position vector is \vec{a}=\vec{0}

and perpendicular to Y-axis whose position vector is \vec{j}

So the equation of the XZ plane is \vec{r}.\vec{n}=\vec{a}.\vec{n}



⇒ \vec{r}.\vec{j}=\vec{0}.\vec{j}

\vec{r}.\vec{j} =0

Vector equation of YZ-Plane:

The YZ- Plane passes through origin whose position vector is \vec{a}=\vec{0}       

and perpendicular to X-axis whose position vector is \vec{i}

So the equation of the YZ plane is \vec{r}.\vec{n}=\vec{a}.\vec{n}

⇒ \vec{r}.\vec{i}=\vec{0}.\vec{i}

\vec{r}.\vec{i}=0

Hence, the vector equation of the coordinates planes.

XY-Plane = \vec{r}.\vec{k}=0



XZ-Plane = \vec{r}.\vec{j}=0

YZ-Plane = \vec{r}.\vec{i}=0

Question 4. Find the vector equation of each one of following planes:

(i) 2x – y + 2z =8

Solution:

Given equation of plane is,

2x – y + 2z = 8

So, 

(x\hat{i}+y\hat{j}+z\hat{k})(2\hat{i}-\hat{j}+2\hat{k})=8

\vec{r}.(2\hat{i}-\hat{j}+2\hat{k})=8

Therefore, the vector equation of the plane is \vec{r}.(2\hat{i}-\hat{j}+2\hat{k})=8

(ii) x + y – z =5

Solution:

Given equation of plane is,

x + y – z = 5

So, 

(x\hat{i}+y\hat{j}+z\hat{k})(\hat{i}+\hat{j}-\hat{k})=5

\vec{r}.(\hat{i}+\hat{j}-\hat{k})=5

Therefore, Vector equation of the plane is \vec{r}.(\hat{i}+\hat{j}-\hat{k})=5

(iii) x + y = 3 

Solution:

Given equation of plane is,

x + y = 3

So, 



(x\hat{i}+y\hat{j}+z\hat{k})(\hat{i}+\hat{j})=3

\vec{r}.(\hat{i}+\hat{j})=3

Therefore, the vector equation of the plane is \vec{r}.(\hat{i}+\hat{j})=3

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 \vec{a}   and normal to \vec{n}   is

(\vec{r}-\vec{a}).\vec{n} = 0      …..(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, 

\vec{a}=\hat{i}-\hat{j}+\hat{k}



\vec{n}=\overrightarrow{AB}

\overrightarrow{AB} = Position vector of \vec{B}    – Position vector of \vec{A}

(-\hat{i}+3\hat{j}+\hat{k})-(\hat{i}+2\hat{j}+5\hat{k})

-2\hat{i}+\hat{j}-4\hat{k}

Now, from eq (1), we get

[\vec{r}-(\hat{i}-\hat{j}+\hat{k})].(-2\hat{i}+\hat{j}-4\hat{k})=0

⇒ \vec{r}.(-2\hat{i}+\hat{j}-4\hat{k})-(\hat{i}-\hat{j}+\hat{k}).(-2\hat{i}+\hat{j}-4\hat{k})=0

⇒ \vec{r}.(-2\hat{i}+\hat{j}-4\hat{k})-[-2-1-4]=0

⇒ \vec{r}.(-2\hat{i}+\hat{j}-4\hat{k})+7=0

⇒ \vec{r}.(-2\hat{i}+\hat{j}-4\hat{k})=-7



⇒ \vec{r}.(2\hat{i}-\hat{j}+4\hat{k})=7   …(2)

Now, Putting \vec{r}=x\hat{i}+y\hat{j}+j\hat{k}  in eq(2), we get

 (x\hat{i}+y\hat{j}+j\hat{k}).(2\hat{i}-\hat{j}+4\hat{k})=7

2x – y + 4z = 7

So, the vector equation is the plane is \vec{r}.(2\hat{i}-\hat{j}+4\hat{k})=7

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

Question 6. If \vec{n}    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 \vec{n}               

Solution:

Given that \vec{n}    = √3 and \vec{n}    makes equal angle with coordinate axes.

Let us considered \vec{n}    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

u2 + v2 + w2 = 1

p2 + p2 + 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, 

\vec{n} = |\vec{n}|(l\vec{i}+m\vec{j}+n\vec{k})

= √3(\frac{1}{√3}\vec{i}+\frac{1}{√3}\vec{j}+\frac{1}{√3}\vec{k})

Now, \vec{n}=\hat{i}+\hat{j}+\hat{k}         and \vec{a}=2\hat{i}+\hat{j}-\hat{k}

As we know that the vector equation of a plane passing through a point \vec{a}    and normal to \vec{n}    is

(\vec{r}-\vec{a}).\vec{n}=0

[\vec{r}-(2\hat{i}+\hat{j}-\hat{k})].(\hat{i}+\hat{j}+\hat{k})=0

⇒ \vec{r}(\hat{i}+\hat{j}+\hat{k})-(2\hat{i}+\hat{j}-\hat{k})(\hat{i}+\hat{j}+\hat{k})=0

⇒ \vec{r}(\hat{i}+\hat{j}+\hat{k})-(2+1-1)=0

⇒ \vec{r}(\hat{i}+\hat{j}+\hat{k})=2   ….(1)

Now, Putting \vec{r}=x\hat{i}+y\hat{j}+j\hat{k}  in eq(1), we get

⇒ (x\hat{i}+y\hat{j}+j\hat{k})(\hat{i}+\hat{j}+\hat{k})=2

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



⇒ x + y + z = 2

Hence, the vector equation of the plane is \vec{r}(\hat{i}+\hat{j}+\hat{k})=2

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 \vec{a}    and normal to \vec{n}    is

(\vec{r}-\vec{a}).\vec{n}=0       …….(1)

Here, \vec{a}=12\hat{i}-4\hat{j}+3\hat{k}

\vec{n}=(12\hat{i}-4\hat{j}+3\hat{k})-(0\hat{i}+0\hat{j}+0\hat{k})

\vec{n}=12\hat{i}-4\hat{j}+3\hat{k}

On putting the values of \vec{a}    and \vec{n}    in equation (1)

[\vec{r}-(12\hat{i}-4\hat{j}+3\hat{k})].(12\hat{i}-4\hat{j}+3\hat{k})=0

⇒ \vec{r}(12\hat{i}-4\hat{j}+3\hat{k})-(12\hat{i}-4\hat{j}+3\hat{k})(12\hat{i}-4\hat{j}+3\hat{k})=0

⇒ \vec{r}(12\hat{i}-4\hat{j}+3\hat{k})-[(12)(12)+(-4)(-4)+(3)(3)]=0

⇒ \vec{r}(12\hat{i}-4\hat{j}+3\hat{k})-(144+16+9)=0

⇒ \vec{r}(12\hat{i}-4\hat{j}+3\hat{k})=169   ……..(2)

Now, on putting \vec{r}=x\hat{i}+y\hat{j}+z\hat{k}  in eq(2), we get

⇒ (x\hat{i}+y\hat{j}+z\hat{k})(12\hat{i}-4\hat{j}+3\hat{k})-169=0



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

⇒ 12x – 4y + 3z = 169

Hence, the vector equation of the plane is \vec{r}(12\hat{i}-4\hat{j}+3\hat{k})=169

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 \vec{a}    and normal to \vec{n}    is

(\vec{r}-\vec{a}).\vec{n}=0                  …..(i)

So, \vec{a}=2\hat{i}+3\hat{j}+\hat{k}

\vec{n}=5\hat{i}+3\hat{j}+2\hat{k}

Now put all these values in equation (i),

[\vec{r}-(2\hat{i}+3\hat{j}+\hat{k})].(5\hat{i}+3\hat{j}+2\hat{k})=0

\vec{r}(5\hat{i}+3\hat{j}+2\hat{k})-(2\hat{i}+3\hat{j}+\hat{k})(5\hat{i}+3\hat{j}+2\hat{k})=0

\vec{r}(5\hat{i}+3\hat{j}+2\hat{k})-[(2)(5)+(3)(3)+(1)(2)]=0

\vec{r}(5\hat{i}+3\hat{j}+2\hat{k})-[10+9+2]=0

\vec{r}(5\hat{i}+3\hat{j}+2\hat{k})-21=0   ………(2)

Now, putting, \vec{r}=x\hat{i}+y\hat{j}+z\hat{k}    in eq(2), we get

(x\hat{i}+y\hat{j}+z\hat{k})(5\hat{i}+3\hat{j}+2\hat{k})-21=0

(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 \vec{a}    and normal to \vec{n}    is

(\vec{r}-\vec{a}).\vec{n}=0                   …..(i)

Here, \vec{a}=2\hat{i}+3\hat{j}-\hat{k}

\vec{n}=\overrightarrow{OP}

So, \overrightarrow{OP}    = Position vector of P – Position vector of O

(2\hat{i}+3\hat{j}-\hat{k})-(0\hat{i}+0\hat{j}+0\hat{k})

\vec{n}=(2\hat{i}+3\hat{j}-\hat{k})

Now put, the value of \vec{a}    and \vec{n}    in equation (i),

[\vec{r}-(2\hat{i}+3\hat{j}-\hat{k})].(2\hat{i}+3\hat{j}-\hat{k})=0

\vec{r}(2\hat{i}+3\hat{j}-\hat{k})-[2\hat{i}+3\hat{j}-\hat{k})(2\hat{i}+3\hat{j}-\hat{k})]=0

\vec{r}(2\hat{i}+3\hat{j}-\hat{k})-[(2)(2)+(3)(3)+(-1)(-1)]=0

\vec{r}(2\hat{i}+3\hat{j}-\hat{k})-[4+9+1]=0

\vec{r}(2\hat{i}+3\hat{j}-\hat{k})=14    ……..(2)

Now putting, \vec{r}=x\hat{i}+y\hat{j}+z\hat{k} , in eq(2), we get



(x\hat{i}+y\hat{j}+z\hat{k})(2\hat{i}+3\hat{j}-\hat{k})=14

(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

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

\frac{x}{3/2}{}+\frac{y}{3}+\frac{z}{-3/2}=1             ……(i)

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

then the equation of the plane is:

\frac{x}{a}+\frac{y}{b}+\frac{z}{c}=1                          ……(ii)

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

a=\frac{3}{2}, b=\frac{3}{1}, c=-\frac{3}{2}

So, from the given equation of plane,

(x\hat{i}+y\hat{j}+z\hat{k})(2\hat{i}+\hat{j}-2\hat{k})=3

\vec{r}(2\hat{i}+\hat{j}-2\hat{k})=3

Hence, the vector normal to the plane is,

\vec{n}=2\hat{i}+\hat{j}-2\hat{k}

|\vec{n}| = √{(2)2 + (1)2 + (-2)2}

= √(4 + 1 + 4)

= √9

|\vec{n}|=3

Hence, the unit vector perpendicular to \vec{n}= \frac{\vec{n}}{|\vec{n}|}= \frac{2\vec{i} + \vec{j}-2\vec{k}}{3}

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




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