RD Sharma Class 12 Ex 29.3 Solutions Chapter 29 The Plane

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TextbookNCERT
ClassClass 12th
SubjectMaths
Chapter29
Exercise29.3
CategoryRD Sharma Solutions

Table of Contents

RD Sharma Class 12 Ex 29.3 Solutions Chapter 29 The Plane

Question 1: Ten cards numbered 1 through 10 are placed in a box, mixed

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

(\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

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, 

\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)+ (1)+ (-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 

Question 11. A plane passes through the point (1, -2, 5) and is perpendicular to the line joining the origin to the point (3\hat{i}+\hat{j}-\hat{k}  ). 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 \vec{a}  and normal to \vec{n}   is

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

Here, \vec{a}=\hat{i}-2\hat{j}+5\hat{k}

\vec{n}=\overrightarrow{OP}

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

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

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

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

[\vec{r}-(\hat{i}-2\hat{j}+5\hat{k})](3\hat{i}+\hat{j}-\hat{k})=0
\vec{r}.(3\hat{i}+\hat{j}-\hat{k})-(\hat{i}-2\hat{j}+5\hat{k})(3\hat{i}+\hat{j}-\hat{k})=0
\vec{r}.(3\hat{i}+\hat{j}-\hat{k})-[(1)(3)+(-2)(1)+(5)(-1)]=0
\vec{r}.(3\hat{i}+\hat{j}-\hat{k})-[3-2-5]=0
\vec{r}.(3\hat{i}+\hat{j}-\hat{k})-[-4]=0

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

Now put \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}).(3\hat{i}+\hat{j}-\hat{k})=-4

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

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

Here, \vec{a}   = mid-point of AB

So, \vec{n}=\frac{\overrightarrow{AB}}{2}      

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

\frac{\hat{i}+2\hat{j}+3\hat{k}+3\hat{i}+4\hat{j}+5\hat{k}}{2}

\vec{a}=\frac{4\hat{i}+6\hat{j}+8\hat{k}}{2}

\vec{a}=2\hat{i}+3\hat{j}+4\hat{k}

And, \vec{n}=\overrightarrow{AB}

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

(3\hat{i}+4\hat{j}+5\hat{k})-(\hat{i}+2\hat{j}+3\hat{k})

3\hat{i}+4\hat{j}+5\hat{k}-\hat{i}-2\hat{j}-3\hat{k}

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

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

\vec{r}-(2\hat{i}+3\hat{j}+4\hat{k})(2\hat{i}+2\hat{j}+2\hat{k})=0
\vec{r}(2\hat{i}+2\hat{j}+2\hat{k})-[(2\hat{i}+3\hat{j}+4\hat{k})(2\hat{i}+2\hat{j}+2\hat{k})]=0
\vec{r}(2\hat{i}+2\hat{j}+2\hat{k})-[(2)(2)+(3)(2)+(4)(2)]=0
\vec{r}(2\hat{i}+2\hat{j}+2\hat{k})-[4+6+8]=0

\vec{r}(2\hat{i}+2\hat{j}+2\hat{k})=18    …..(2)

Now put \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}+2\hat{j}+2\hat{k})=18

(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.

Question 13. Show that the normal to the following pairs of planes are perpendicular to each other:

(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

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

\vec{r}.\vec{n_1} =2                 …..(i)

Now we solve, 3x + 2y – z =- 4

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

\vec{r}.\vec{n_2} =-4               …(ii)

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

\vec{n_1} is normal to eq(i) and \vec{n_2}  is normal to eq(ii)

So, 

\vec{n_1}\vec{n_2} =(\hat{i}-\hat{j}+\hat{k}).(3\hat{i}+2\hat{j}-\hat{k})

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

=3 – 2 – 1

= 3 – 3 = 0

Hence,\vec{n_1}  is perpendicular to \vec{n_2}.

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

Solution: 

Given equations of planes are

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

So first we solve, \vec{r}.(2\hat{i}-\hat{j}+3\hat{k})=5

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

\vec{r}.\vec{n_1} =5    …..(1)

Now we solve, \vec{r}.(2\hat{i}-2\hat{j}-2\hat{k})=5

\vec{r}.(2\hat{i}-2\hat{j}-2\hat{k})=5

\vec{r}.\vec{n_2} =5      ……(2)

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

\vec{n_1} is normal to eq(i) and \vec{n_2}  is normal to eq(ii)

So, 

\vec{n_1}.\vec{n_2}  = (2\hat{i}-\hat{j}+3\hat{k})(2\hat{i}-2\hat{j}-2\hat{k})

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

= 4 + 2 – 6

= 6 – 6

= 0

Hence,\vec{n_1}  is perpendicular to \vec{n_2}.

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,

(x\hat{i}+y\hat{j}+z\hat{k})(2\hat{i}+2\hat{j}+2\hat{k})=3
\vec{r}.(2\hat{i}+2\hat{j}+2\hat{k})=3
\vec{r}.\vec{n} =d

So, the normal to the plane \vec{n}=(2\hat{i}+2\hat{j}+2\hat{k})

and the direction ratio of \vec{n} = 2,2,2

So, the direction cosine of \vec{n} = \frac{2}{|\vec{n}|} ,\frac{2}{|\vec{n}|},\frac{2}{|\vec{n}|}  ….(1)

|\vec{n}| = √[(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}| = \frac{2}{2√3},\frac{2}{2√3},\frac{2}{2√3}

\frac{1}{√3}, \frac{1}{√3}, \frac{1}{√3}

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, 

(x\hat{i}+y\hat{j}+z\hat{k})(12\hat{i}-3\hat{j}+4\hat{k})=1
\vec{r}.\vec{n} =1

The normal to the plane is

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

|\vec{n}|  = √[(12)+ (-3)+ (4)2]

= √[144 + 9 + 16]

= √169 = 13

Hence, the unit vector \hat{n}  = (12\hat{i}-3\hat{j}+4\hat{k})/13

=\frac{12}{13}\hat{i}-\frac{3}{13}\hat{j}+\frac{4}{13}\hat{k}

Now we find a vector normal to the plane with magnitude

So, 

26 = 26 \vec{n}

= 26 (\frac{12}{13}\hat{i}-\frac{3}{13}\hat{j}+\frac{4}{13}\hat{k})

24\hat{i}-6\hat{j}+8\hat{k}

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 \vec{a}   and normal to \vec{n}   is

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

Here, \vec{a}  = position vector of B

So, \vec{a}=(-10\hat{i}+5\hat{j}+4\hat{k})

and \vec{n}=\overrightarrow{AB}

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

(-10\hat{i}+5\hat{j}+4\hat{k})-(4\hat{i}-\hat{j}+2\hat{k})

-10\hat{i}+5\hat{j}+4\hat{k}-4\hat{i}+\hat{j}-2\hat{k}

\vec{n} = -14\hat{i}+6\hat{j}+2\hat{k}

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

[\vec{r}-(-10\hat{i}+5\hat{j}+4\hat{k})].(-14\hat{i}+6\hat{j}+2\hat{k})=0
\vec{r}(-14\hat{i}+6\hat{j}+2\hat{k})-(-10\hat{i}+5\hat{j}+4\hat{k})(-14\hat{i}+6\hat{j}+2\hat{k})=0
\vec{r}(-14\hat{i}+6\hat{j}+2\hat{k})-[(-10)(-14)+(5)(6)+(4)(2)]=0
\vec{r}(-14\hat{i}+6\hat{j}+2\hat{k})-[140+30+8]=0

\vec{r}(-14\hat{i}+6\hat{j}+2\hat{k})=178   …..(2)

Now put \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}).(-14\hat{i}+6\hat{j}+2\hat{k})=178

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

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

Here, \vec{a}   = Position vector of C

So, \vec{a}=\frac{\overrightarrow{AB}}{2}       [Because c is the mid point of line AB]             

\vec{a}=\frac{-\hat{i}+2\hat{j}+3\hat{k}+3\hat{i}-5\hat{k}+6\hat{k}}{2}
=\frac{2\hat{i}}{2}-\frac{3\hat{j}}{2}+\frac{9\hat{k}}{2}
\vec{a}=\hat{i}-\frac{3}{2}\hat{j}+\frac{9}{2}\hat{k}

Now, \vec{n}=\overrightarrow{AB}

\overrightarrow{AB}  = Position vector of B- Position vector of A

=(3\hat{i}-5\hat{j}+6\hat{k})-(-\hat{i}+2\hat{j}+3\hat{k})

3\hat{i}-5\hat{j}+6\hat{k}+\hat{i}-2\hat{j}-3\hat{k}

4\hat{i}-7\hat{j}+3\hat{k}

\vec{n}=4\hat{i}-7\hat{j}+3\hat{k}

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

[\vec{r}-(\hat{i}-\frac{3}{2}\hat{j}+\frac{9}{2}\hat{k})](4\hat{i}-7\hat{j}+3\hat{k})=0
\vec{r}(4\hat{i}-7\hat{j}+3\hat{k})-(\hat{i}-\frac{3}{2}\hat{j}+\frac{9}{2}\hat{k})(4\hat{i}-7\hat{j}+3\hat{k})=0

\vec{r}(4\hat{i}-7\hat{j}+3\hat{k})  = 28 ….(2)

Now put \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}).(4\hat{i}-7\hat{j}+3\hat{k})= 28

(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

\vec{a}=5\hat{i}+2\hat{j}-4\hat{k}
\vec{n}=2\hat{i}+3\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

So, 

[\vec{r}-(5\hat{i}+2\hat{j}-4\hat{k})].(2\hat{i}+3\hat{j}-\hat{k})=0
\vec{r}(2\hat{i}+3\hat{j}-\hat{k})-(5\hat{i}+2\hat{j}-4\hat{k}).(2\hat{i}+3\hat{j}-\hat{k})=0
\vec{r}(2\hat{i}+3\hat{j}-\hat{k})-10-6-4=0

\vec{r}(2\hat{i}+3\hat{j}-\hat{k})=20   ….(1)

For cartesian equation:

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

(x\hat{i}+y\hat{j}+z\hat{k}).(2\hat{i}+3\hat{j}-\hat{k})=20

(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, \vec{n}=\hat{i}+2\hat{j}-3\hat{k}

and \vec{a}=\hat{i}+2\hat{j}-3\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

So,

[\vec{r}-(\hat{i}+2\hat{j}-3\hat{k})].(\hat{i}+2\hat{j}-3\hat{k})=0
\vec{r}(\hat{i}+2\hat{j}-3\hat{k})-(\hat{i}+2\hat{j}-3\hat{k})(\hat{i}+2\hat{j}-3\hat{k})=0
\vec{r}(\hat{i}+2\hat{j}-3\hat{k})-1-4-9=0
\vec{r}(\hat{i}+2\hat{j}-3\hat{k})=14

For cartesian equation:

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

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

(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, \overrightarrow{OA}=a\hat{i}+b\hat{j}+c\hat{k}

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

So, the direction cosines are:

\frac{a}{\sqrt{a^2+b^2+c^2}},\frac{b}{\sqrt{a^2+b^2+c^2}},\frac{c}{\sqrt{a^2+b^2+c^2}}

So the equation of the line is,

[(x\hat{i}+y\hat{j}+z\hat{k})-(a\hat{i}+b\hat{j}+c\hat{k})].(a\hat{i}+b\hat{j}+c\hat{k})=0

ax + by + cz = a+ b+ c2

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