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All questions of Equations of Lines and Planes for Grade 12 Exam

Find the equation of the set of points which are equidistant from the points (1, 2 , 3) and (3, 2, -1)​
a) x + 2z = 0
b) y + 2z = 0
c) x – 2z = 0
d) x – 2y = 0 
Correct answer is option 'C'. Can you explain this answer?

Aryan Khanna answered
Pt. A(1, 2 , 3)
Pt. B(3, 2, -1)
Let P(x,y,z)
So, AP = BP
((x-1)2 + (y-2)2 + (z-3)2)1/2 = ((x-3)2 + (y-2)2 + (z+1)2)1/2
(x-1)2 + (y-2)2 + (z-3)2) = (x-3)2 + (y-2)2 + (z+1)2
x2 +1 -2x + y2 + 4 - 4y + z2 + 9 – 6z = x2 +9 -6x + y2 + 4 - 4y + z2 + 1 + 2z
4x – 8z = 0
x – 2z = 0
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The direction cosines of the line joining the points (2, -1, 8) and (-4, -3, 5) are:
  • a)
  • b)
  • c)
  • d)
Correct answer is option 'B'. Can you explain this answer?

Geetika Shah answered
Pt. A(2, -1, 8)
Pt. B(-4, -3, 5)
Direction Ratio DR of AB : ( -4-2 , -3+1 , 5-8 )
: (-6,-2,-3)
Direction cosine of AB : ( -6/(62+22+32)1/2 , -2/(62+22+32)1/2 , -3/(62+22+32)1/2)
: ( -6/7, -2/7, -3/7)
 

The equation of the plane passing through the line of intersection of the planes x-2y+3z+8=0 and 2x-7y+4z-3=0 and the point (3, 1, -2) is:​
  • a)
    6x-15y+12z+29=0
  • b)
    6x-15y+16z+29=0
  • c)
    6x-15y+12z+32=0
  • d)
    2x-5y+4z+9=0
Correct answer is option 'B'. Can you explain this answer?

Tejas Verma answered
(x - 2y + 3z + 8) + μ(2x - 7y + 4z - 3) = 0
i.e, (1 + 2μ)x - (2 + 7μ)y + (3 + 4μ)z + (8 - 3μ) = 0......(1)
 the required plane is passing through (3, 1, -2)
so, 3(1 + 2μ) - (1)(2 + 7μ) + (-2)(3 + 4μ)+ (8 - 3μ) = 0
3 + 6μ - 2 - 7μ -6 -8μ + 8 - 3μ = 0
by solving, μ = 1/4
putting μ in equation (1)
we get the required equation of plane as :- 6x - 15y + 16z + 29 = 0

If plane cuts off intercepts OA = a, OB = b, OC = c from the coordinate axes, then the area of the triangle ABC equal to
  • a)
  • b)
  • c)
  • d)
     
Correct answer is option 'A'. Can you explain this answer?

Neha Joshi answered
 =AC = −ai^ + ck^
AB = −ai^ + bj^
 Area of △ABC= ½|AB × AC∣ 
|AB × AC∣ =

−(bc)i^− (ac)j^ − (ab)k^
∣AB × AC∣ = (b2c2 + a2c2 + a2b2)1/2
Area = 1/2(a2b2 + b2c2 + c2a2)1/2

If the sum of the squares of the distances of a point from the three coordinate axes be 36, then its distance from the origin is
  • a)
    6
  • b)
    3√2
  • c)
    2√3
  • d)
    6√2
Correct answer is option 'B'. Can you explain this answer?

Yash Patel answered
Let (x,y,z) be the point.
Given sum of the squares of distance from point to the axes is 36. 
⇒(x2+y2)+(y2+z2)+(z2+x2)=36
⇒2(x2+y2+z2)=36⇒x2+y2+z2=18
So the distance of the point from the origin is =3(2)1/2

If a line has the direction ratios -4, 18, -12 then what are its direction cosines?​
  • a)
    -2, 9, -6
  • b)
    -4, 18, -12
  • c)
    2/11, 9/11, 6/11
  • d)
    -2/11, 9/11, -6/11
Correct answer is option 'D'. Can you explain this answer?

Vikas Kapoor answered
DR of the line :  (-4, 18 -12)
DC of the line : (-4/k, 18/k, -12/k)
where k = ((42) + (182) + (12)2)1/2
= (16 + 324 + 144)1/2
= (484)1/2
= 22
So, DC : (-4/22, 18/22, -12/22)
: (-2/11 , 9/11 , -6/11)

The length of the perpendicular from the origin to the plane 3x + 2y – 6z = 21 is:​
  • a)
    3
  • b)
    14
  • c)
    21
  • d)
    7
Correct answer is option 'A'. Can you explain this answer?

Leelu Bhai answered
Given equation of plane is : 3x + 2y - 6z - 21= 0
the length of perpendicular from a given point
(x' , y', z') on a plane ax + by + cz + d = 0 is given as :-

d = modulus of [{ax' + by' + cz' + d}/{√(a² + b² + c)²}]

so, d = modulus of [{(3*0) + (2*0) + (-6*0) + (-21)}/{√(3² + 2² + (-6)²)}]

d= modulus of (-21/√49) = (-21/7) = 3 units
hence option A is correct....

The distance of the point (2, 3, – 5) from the plane x + 2y – 2z = 9 is:​
  • a)
    2 units
  • b)
    3/2 units
  • c)
    3 units
  • d)
    10/3 units
Correct answer is option 'C'. Can you explain this answer?

Nikita Singh answered
 Length of perpendicular from (2,3,-5) to the plane x + 2y − 2z − 9 = 0.
= |(2 + 2×3 −2×(−5) − 9)|√12 + 22 + (−2)2
= |2 + 6 + 10 − 9|/√9
= 9/3
= 3 units.

The equation of the plane passing through the intersection of the planes  and and the point (1, 2, 1) is:​
  • a)
    18x+6y+14z-23=0
  • b)
    18x+7y+14z-46=0
  • c)
    9x+3y+7z-23=0
  • d)
    18x+7y+14z-38=0
Correct answer is option 'B'. Can you explain this answer?

Aryan Khanna answered
n1 = 2i + j + k
n2 = 2i + 3j - 4k
p1 = 4,   p2 = -6
r.(n1 + λn2) = p1 + λp2
=> r . [2i + j + k + λ(2i + 3j - 4k)] = 4 - 6λ
=> r . [ i(2 + 2λ) + j(1 + 3λ) + k(1 - 4k)] = 4 - 6λ
Taking r = xi + yj + zk
(2 + 2λ)x + (1 + 3λ)y + (1 - 4k)z = 4 - 6λ
(2x + y + - z - 4) + λ(2x + 3y - 4k + 6) = 0
Given points are (1,2,1) 
(2 + 2 - 1 - 4) + λ(2 + 6 - 4 + 6) = 0
-1 + λ(10) = 0
 λ = 1/10
Substitute  λ = 1/10, we get
18x + 7y + 14z - 46=0

Find the direction cosines of the x axis.​
  • a)
    1, 0, 0
  • b)
    0, 0, 0
  • c)
    0, 1, 0
  • d)
    0, 0, 1
Correct answer is option 'A'. Can you explain this answer?

Infinity Academy answered
To find Direction Cosines of X-axis.
Take any two points on X-axis : A(a,0,0) & B(b,0,0)
DR of AB : (b-a,0,0)
DC of AB : ((b-a)/(((b-a)2 + 0 + 0)1/2), 0, 0)
: ((b-a)/(b-a) , 0 , 0)
: (1,0,0)

The equation of the plane passing through the point (3, – 3, 1) and perpendicular to the line joining the points (3, 4, – 1) and (2, – 1, 5) is:​
  • a)
    – x – 5y + 6z + 18 = 0
  • b)
    x – 5y + 6z + 18 = 0
  • c)
    x + 5y – 6z + 18 = 0
  • d)
    – x – 5y – 6z + 18 = 0
Correct answer is option 'C'. Can you explain this answer?

Preeti Iyer answered
The equation of the plane passing through the point (3, – 3, 1) is:
a(x – 3) + b(y + 3) + c(z – 1) = 0 and the direction ratios of the line joining the points
(3, 4, – 1) and (2, – 1, 5) is 2 – 3, – 1 – 4, 5 + 1, i.e., – 1, – 5, 6.
Since the plane is perpendicular to the line whose direction ratios are – 1, – 5, 6, therefore, direction ratios of the normal to the plane is – 1, – 5, 6.
So, required equation of plane is: – 1(x – 3) – 5(y + 3) + 6(z – 1) = 0
i.e., x +  5y – 6z + 18 = 0.

A point moves so that the sum of the squares of its distances from the six faces of a cube given by x = ± 1, y = ± 1, z = ± 1 is 10 units. The locus of the point is
  • a)
    x2 + y2 + z2 = 1
  • b)
    x2 + y2 + z2 = 2
  • c)
    x + y + z = 1
  • d)
    x + y + z = 2
Correct answer is option 'B'. Can you explain this answer?

Ram Mohith answered
Let the point be (x,y,z)
Distance of this point from x = 1 is |x -1| and from x = +1 is |x + 1|. Similarly you can find the distance from the other faces. The sum of squares of distances will be,
(x - 1)^2 + (x + 1)^2 + (y - 1)^2 + (y + 1)^2 + (z - 1)^2 + (z + 1)^2 = 10

2(x^2 + y^2 + z^2 ) + 6 = 10

x^2 + y^2 + z^2 = 2

The equation of the plane, which is at a distance of 5 unit from the origin and has  as a normal vector, is:
  • a)
  • b)
  • c)
  • d)
Correct answer is option 'D'. Can you explain this answer?

Preeti Iyer answered
x = 3i - 2j - 6k
|x| = ((3)2 + (2)2 + (6)2)
|x| = (49)½ 
|x| = 7
x = x/|x|
= (3i - 2j - 6k)/7
The required equation of plane is r.x = d
⇒  r.(3i - 2j - 6k)/7 = 5
⇒  r.(3i - 2j - 6k) = 35

Direction: Read the following text and answer the following questions on the basis of the same:
A mobile tower stands at the top of a hill. Consider the surface on which the tower stands as a plane having points A(1, 0, 2), B(3, –1, 1) and C(1, 2, 1) on it. The mobile tower is tied with 3 cables from the points A, B and C such that it stands vertically on the ground. The top of the tower is at the point (2, 3, 1) as shown in the figure.
Q. The equation of the plane passing through the points A, B and C is
  • a)
    3x – 2y + 4z = –11
  • b)
    3x + 2y + 4z = 11
  • c)
    3x – 2y – 4z = 11
  • d)
    –3x + 2y + 4z = –11
Correct answer is option 'B'. Can you explain this answer?

Krishna Iyer answered
A(1, 0, 2), B(3, –1, 1) and C(1, 2, 1)
(x - 1)(1 + 2) -y (-2, 0) + (z - 2) (4 - 0)
3x - 3 + 2y + 4z - 8 = 0
3x + 2y + 4z = 11

The equation of plane through the intersection of planes (x+y+z =1) and (2x +3y – z+4) =0 is​
  • a)
    x(1 + 2k) + y(1 + 3k) + z(1 – k) + (-1 + 4k) = 0
  • b)
    x(1+2k)+y(1-3k)+z(1-k)+(-1+4k) = 0
  • c)
    x(1+2k) +y(1+3k)+z(1-k) +(-1 – 4k) = 0
  • d)
    x (1-2k) + y(1+3k) +z(1-k) +(-1+4k) = 0
Correct answer is option 'A'. Can you explain this answer?

Kritika Sarkar answered
To find the equation of the plane through the intersection of two given planes, we need to first determine the line of intersection and then find a normal vector for the plane.

1. Finding the line of intersection:
The two given planes are:
Plane 1: x + y + z = 1
Plane 2: 2x + 3y + z + 4 = 0

To find the line of intersection, we can set the equations of the two planes equal to each other:
x + y + z = 1
2x + 3y + z + 4 = 0

By subtracting the second equation from the first equation, we can eliminate z:
x + y + z - (2x + 3y + z + 4) = 1 - 0
-x - 2y - 4 = -1

Simplifying the equation, we get:
x + 2y = 3

This equation represents the line of intersection of the two planes.

2. Finding a normal vector for the plane:
Since the line of intersection lies on both planes, the normal vector of the required plane should be perpendicular to this line. Therefore, we can choose the direction ratios of the line, -1 and 2, as coefficients of the normal vector.

The normal vector of the required plane is given by the cross product of the direction ratios of the line of intersection:
n = (1, 2, 0) x (-1, 2, 0) = (0, 0, -4)

3. Writing the equation of the plane:
Now we have a point on the plane, (1, 0, 0), and a normal vector, (0, 0, -4). We can use the point-normal form of the equation of a plane to write the equation of the required plane.

The equation of the plane is given by:
0(x - 1) + 0(y - 0) + (-4)(z - 0) = 0

Simplifying the equation, we get:
-4z + 4 = 0
z = 1

Therefore, the equation of the plane through the intersection of the given planes is:
x + 2y - 4z + 4 = 0

Comparing this equation with the options provided, we see that option A is the correct answer:
x(1 2k) + y(1 3k) + z(1 k) + (-1 4k) = 0

For which value of a lines  and  are perpendicular?
  • a)
    11/70
  • b)
    5
  • c)
    1
  • d)
    70/11
Correct answer is option 'B'. Can you explain this answer?

Tejas Verma answered
(x-1)/(-3) = (y-2)/(2p/7) = (z-3)/2 
(x-1)(-3p/7) = (y - 5)/1 = (z - 6)/(-5)
The direction ratio of the line are -3, 2p/7, -2 and (-3p)/7, 1, -5
Two lines with direction ratios, a1, b1, c1 and a2, b2, c2 are perpendicular to each other if a1a2 + b1b2 + c1c2 = 0
Therefore, (-3)(-3p/7) + (2p/7)(1) + 2(-5) = 0
(9p/7) + (2p/7) = 10
11p = 70
p = 70/1

Determine the direction cosines of the normal to the plane and the distance from the origin. Plane z = 2
  • a)
    0, 1, 0; 2
  • b)
    1, 0, 0; 3
  • c)
    1, 0, 1; 3
  • d)
    0, 0, 1; 2
Correct answer is option 'D'. Can you explain this answer?

Priya Basu answered
Understanding the Plane Equation
The equation of the plane given is z = 2. This indicates that the plane is parallel to the x-y plane and is located at a height of 2 units above it.
Normal Vector to the Plane
- The normal vector to a plane defined by the equation Ax + By + Cz + D = 0 can be derived from the coefficients A, B, and C.
- For the plane z = 2, we can rewrite it as 0x + 0y + 1z - 2 = 0.
- Here, A = 0, B = 0, C = 1.
Direction Cosines
- The direction cosines of a normal vector (n_x, n_y, n_z) are given by the ratios of the components of the normal vector to its magnitude.
- For our normal vector (0, 0, 1), the magnitude is √(0² + 0² + 1²) = 1.
- Therefore, the direction cosines are:
- n_x = 0/1 = 0
- n_y = 0/1 = 0
- n_z = 1/1 = 1
Distance from the Origin
- The distance d from the origin (0, 0, 0) to the plane z = 2 can be calculated as the absolute value of the constant term divided by the magnitude of the normal vector.
- Here, d = |D| / √(A² + B² + C²) = |2| / 1 = 2.
Conclusion
- The direction cosines of the normal to the plane z = 2 are (0, 0, 1), and the distance from the origin to the plane is 2.
- Hence, the correct answer is option 'D': (0, 0, 1).

The non zero value of ‘a’ for which the lines 2x – y + 3z + 4 = 0 = ax + y – z + 2 and x – 3y + z = 0 = x + 2y + z + 1 are co-planar is
  • a)
    –2
  • b)
    4
  • c)
    6
  • d)
    0
Correct answer is option 'A'. Can you explain this answer?

Parth Yadav answered
2x - y +3z + 4 = 0
x - 3y + z = 0
x + 2y + z + 1 = 0
x = 12/5
y = -1/5
z= -3
this point also satisfied by
ax - y + z + 2 = 0
a(12 / 5) - (-1/5) + (-3) = 0
⇒ 12a/5 + 1/5 -3 = 0
a= -2

The equation of the plane passing through the point (1, – 3, –2) and perpendicular to planes x + 2y + 2z = 5 and 3x + 3y + 2z = 8, is
  • a)
    2x – 4y + 3z – 8 = 0
  • b)
    2x – 4y – 3z + 8 = 0
  • c)
    2x – 4y + 3z + 8 = 0
  • d)
    None of these
Correct answer is option 'A'. Can you explain this answer?

Nidhi Sen answered
-2, 3) and perpendicular to the vector (2, 1, -4) can be found using the point-normal form of the equation of a plane:

(x - x1)(a) + (y - y1)(b) + (z - z1)(c) = 0

where (x1, y1, z1) is the given point and (a, b, c) is the normal vector to the plane.

Substituting the given values, we get:

(x - 1)(2) + (y + 2)(1) + (z - 3)(-4) = 0

Expanding and simplifying, we get:

2x + y - 4z - 15 = 0

Therefore, the equation of the plane is:

2x + y - 4z - 15 = 0

Equation of a plane which is at a distance d from the origin and the direction cosines of the normal to the plane are l, m, n is.
  • a)
    lx – my + nz = d
  • b)
    – lx + my + nz = d
  • c)
    lx + my + nz = d
  • d)
    lx + my + nz = – d
Correct answer is option 'C'. Can you explain this answer?

Bhargavi Sen answered
Equation of a Plane with Given Normal Vector
To find the equation of a plane in 3D space, we need a point on the plane and a normal vector to the plane. In this case, we are given the direction cosines of the normal vector and the distance of the plane from the origin.

Given Information:
- Direction cosines of the normal to the plane: l, m, n
- Distance of the plane from the origin: d

Equation of the Plane:
The general form of the equation of a plane in 3D space is:
lx + my + nz = d

Explanation:
- The direction cosines of the normal vector to the plane are l, m, n. This means that the coefficients of x, y, and z in the equation of the plane will be l, m, and n respectively.
- The distance of the plane from the origin is represented by the constant term d in the equation.

Correct Answer:
The correct equation of the plane in this case is:
lx + my + nz = d

Conclusion:
When given the direction cosines of the normal vector and the distance of the plane from the origin, the equation of the plane can be determined by using the general form lx + my + nz = d.

Chapter doubts & questions for Equations of Lines and Planes - Mathematics for Grade 12 2025 is part of Grade 12 exam preparation. The chapters have been prepared according to the Grade 12 exam syllabus. The Chapter doubts & questions, notes, tests & MCQs are made for Grade 12 2025 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests here.

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