Ans. (a) Given: Equation of the plane is Therefore, the direction ratios of the normal to the plane are 0, 0, 1.

  a=0,b=0,c=1a=0,b=0,c=1

  a2+b2+c2−−−−−−−−−√=(0)2+(0)2+(1)2−−−−−−−−−−−−−−√=1a2+b2+c2=(0)2+(0)2+(1)2=1

Therefore, 0x1+0y1+z1=210x1+0y1+z1=21

Comparing with lx+my+nz=p,lx+my+nz=p, we get p=21p=21

Therefore, direction cosines of normal to the plane are coefficients of  in , i.e., 0, 0, 1 and length of perpendicular from the origin to the plane is .

(b) Given: Equation of the plane is 

 a=1,b=1,c=1a=1,b=1,c=1

a2+b2+c2−−−−−−−−−√a2+b2+c2 = 

Therefore, x3√+y3√+z3√=13√x3+y3+z3=13

Comparing with lx+my+nz=p,lx+my+nz=p, we get p=13√p=13

Therefore direction cosines of the normal to the plane are the coefficients of  in , i.e., and length of perpendicular from the origin to the plane is .

(c) Equation of the plane is 

 a=2,b=3,c=−1a=2,b=3,c=−1

a2+b2+c2−−−−−−−−−√a2+b2+c2 = 

Therefore, 2x14√+3y14√−z14√=514√2×14+3y14−z14=514

Comparing with lx+my+nz=p,lx+my+nz=p, we get

Therefore direction cosines of the normal to the plane are the coefficients of  in , i.e., and length of perpendicular from the origin to the plane is .

(d) Given: Equation of the plane is    

 a=0,b=−5,c=0a=0,b=−5,c=0

a2+b2+c2−−−−−−−−−√a2+b2+c2 = 

Therefore, 0x5+−5y5+0z5=850x5+−5y5+0z5=85

Comparing with lx+my+cz=p,lx+my+cz=p, we get

Therefore direction cosines of the normal to the plane are the coefficients of  in , i.e., and length of perpendicular from the origin to the plane is .

Ans. Here   The unit vector perpendicular to the plane is

Also (given)

Therefore the equation of the required plane is 

 

 

Ans. (a) Vector equation of the plane is  ……….(i)Putting  in eq. (i) as in 3-D, Cartesian equation of the plane is 

    

(b) Since,  is the position vector of any arbitrary point P on the plane.

    

which is the required Cartesian equation.

(c) Vector equation of the plane is 

Since, Since,  is the position vector of any arbitrary point P on the plane.

  

  which is the required Cartesian equation.

Ans. (a) Given: Equation of the plane is …….(i) and point is O (0, 0, 0)Let M be the foot of the perpendicular drawn from the origin (0, 0, 0) to the given plane.

Since, direction ratios of perpendicular OM to plane are coefficients of  in , i.e., 2, 3, 4 =  (say)

  Equation of the perpendicular OM is  (say)

    

 

Therefore, point M on this line OM is M ………..(ii)

But point M lies on plane (i)

 Putting  in eq. (i), we have

2(2λ)+3(3λ)+4(4λ)−12=02(2λ)+3(3λ)+4(4λ)−12=0

    

Hence, putting  in equation (ii), the coordinates of foot of the perpendicular is 

(b) Given: Equation of the plane is …….(i) and point is O (0, 0, 0)

Let M be the foot of the perpendicular drawn from the origin (0, 0, 0) to the given plane.

Since, direction ratios of perpendicular OM to plane are coefficients of  in , i.e., 0, 3, 4 =  (say)

  Equation of the perpendicular OM is  (say)

    

 

Therefore, point M on this line OM is M ………..(ii)

But point M lies on plane (i)

 Putting  in eq. (i), we have

3(3λ)+4(4λ)−6=03(3λ)+4(4λ)−6=0

    

Hence, putting  in equation (ii), the coordinates of foot of the perpendicular is 

(c) Given: Equation of the plane is …….(i) and point is O (0, 0, 0)

Let M be the foot of the perpendicular drawn from the origin (0, 0, 0) to the given plane.

Since, direction ratios of perpendicular OM to plane are coefficients of  in , i.e., 1, 1, 1 =  (say)

  Equation of the perpendicular OM is  (say)

    

 

Therefore, point M on this line OM is M ………..(ii)

But point M lies on plane (i)

 Putting  in eq. (i), we have

   

Hence, putting  in equation (ii), the coordinates of foot of the perpendicular is 

(d) Given: Equation of the plane is …….(i) and point is O (0, 0, 0)

Let M be the foot of the perpendicular drawn from the origin (0, 0, 0) to the given plane.

Since, direction ratios of perpendicular OM to plane are coefficients of  in , i.e., 0, 5, 0 =  (say)

  Equation of the perpendicular OM is  (say)

    

 

Therefore, point M on this line OM is M ………..(ii)

But point M lies on plane (i)

 Putting  in eq. (i), we have

   

Hence, putting  in equation (ii), the coordinates of foot of the perpendicular is .

Ans. (a) Vector form: The given point on the plane is   The position vector of the given point is  = 

Also Normal vector to the plane is 

 Vector equation of the required line is 

  

Putting the values of  and ,

  

   

Cartesian form: The plane passes through the point  = 

Normal vector to the plane is 

 Direction ratios of normal to the plane are coefficients of  in  are 

 Cartesian form of equation of plane is 

    

 

(b) Vector form: The given point on the plane is 

  The position vector of the given point is 

Also Normal vector to the plane is 

 Vector equation of the required line is 

  

Putting the values of  and ,

  

   

Cartesian form: The plane passes through the point  = 

Normal vector to the plane is 

 Direction ratios of normal to the plane are coefficients of  in  are 

 Cartesian form of equation of plane is 

    

 

Ans. We know that through three collinear points A, B, C i.e., through a straight line, we can pass an infinite number of planes.(a) The three given points are A B and C

Now direction ratios of line AB are  

=  (say)

Again direction ratios of line BC are 

=  (say)

Now   

Since, 

Therefore, line AB and BC are parallel and B is their common point.

  Points A, B and C are collinear and hence an infinite number of planes can be drawn through the three given collinear points i.e. no unique plane can be drawn.

(b) The three given points are A B and C

Now direction ratios of line AB are  

=   (say)

Again direction ratios of line BC are 

=  (say)

Now    a1a2=0,b1b2=∞,=c1c2=−12a1a2=0,b1b2=∞,=c1c2=−12

Since, 

  Points A, B and C are not collinear and hence the unique plane can be drawn through the three given collinear points, i.e.,

  

  

Expanding along first row,

 

 

 

 

 

Hence the equation of required plane is .

Ans. Equation of the plane is   

 

Comparing with intercept form , we have a=52,b=5,c=−5a=52,b=5,c=−5  which are intercepts cut off by the plane on axis, axis and axis respectively.

Ans. Since equation of ZOX plane is   Equation of any plane parallel to ZOX plane is  ……….(i)

[ Equation of any plane parallel to the plane  is  i.e., change only the constant term]

Now, Plane (i) makes an intercept 3 on the axis ( and ) i.e., plane (i) passes through (0, 3, 0).

Putting  and  in eq. (i),  

Putting  in eq. (i), equation of required plane is 

Ans. Equations of given planes are  and Since, equation of any plane through the intersection of these two planes is

L.H.S. of plane I +  (L.H.S. of plane II) = 0

   ……….(i)

Now, required plane (i) passes through the point (2, 2, 1).

Putting  in eq. (i),

 

 

 

Now putting  in eq. (i) of required plane is

 

 

Ans. Equation of first plane is   ……….(i)

Again equation of the second plane is 

  ……….(ii)

Since, equation of any plane passing through the line of intersection of two planes is

L.H.S. of plane I +  (L.H.S. of plane II) = 0

   ……….(iii)

Now, the plane (iii) passes through the point (2, 1, 3) = 

  

Putting this value of  in eq. (iii),

 

   

Putting  in eq. (iii) of required plane is

 

 

 

Ans. Equations of the given planes are  and   and 

Since, equation of any plane passing through the line of intersection of two planes is

L.H.S. of plane I +  (L.H.S. of plane II) = 0

  ……….(i)

 

 

According to the question, this plane is perpendicular to the plane 

  

 

 

   

Putting  in eq. (i) of required plane is

 3x+3y+3z−3−2x−3y−4z+5=03x+3y+3z−3−2x−3y−4z+5=0

 

Ans. Equation of one plane is ……….(i)Comparing this equation with , we have

Normal vector to plane (i) is 

Again, equation of second plane is  ……….(ii)

Comparing this equation with , we have

Normal vector to plane (i) is 

Let  be the acute angle between plane (i) and (ii).

  angle between normals  and  to planes (i) and (ii) is also 

 

= 

= 

 

Ans. (a) Equations of the given planes are   and 

Here, 

Since 

Therefore, the given two planes are not parallel.

Again  = 21 – 5 – 60 = 21 – 65 = –44

Since 

Therefore, the given two planes are not perpendicular.

Now let  be the angle between the two planes.

  

= 

= 

=  =  = 

  

(b) equations of the given planes are   and  i.e., x−2y+0z+5=0x−2y+0z+5=0 

Here, 

Since 

Therefore, the given two planes are not parallel.

Again  =  = 2 – 2 + 0 = 0

Since 

Therefore, the given two planes are perpendicular.

(c) equations of the given planes are   and  

Here, 

Since 

Therefore, the given two planes are parallel.

(d) equations of the given planes are   and  

Here, 

Since 

Therefore, the given two planes are parallel.

(e) equations of the given planes are   and y+z−4=0y+z−4=0 i.e., 

Here, 

Since 

Therefore, the given two planes are not parallel.

Again  = 4 x 0 + 8 x 1 + 1 x 1 = 0 + 8 + 1 = 9

Since 

Therefore, the given two planes are not perpendicular.

Now let  be the angle between the two planes.

  

= 

=  =  =  = 

  

Ans. (a) Distance (of course perpendicular) of the point (0, 0, 0) from the plane 3x−4y+12z=33x−4y+12z=3      is

= 

=  = 

(b) Length of perpendicular from the point  on the plane  is

= 
=  = 

(c) Length of perpendicular from the point  on the plane    is

= 
=  = 

(d) Length of perpendicular from the point  on the plane  is

= 
=  =