Equation of a Plane MCQ Quiz - Objective Question with Answer for Equation of a Plane - Download Free PDF

Last updated on May 19, 2025

Latest Equation of a Plane MCQ Objective Questions

Equation of a Plane Question 1:

If a, b and c are intercepts on X-axis, Y-axis and Z-axis respectively of the plane x + 2y + 3z = 1, then 2a + 4b + 3c = _________. 

  1. 19
  2. 5
  3. 6
  4. 17

Answer (Detailed Solution Below)

Option 2 : 5

Equation of a Plane Question 1 Detailed Solution

Calculation

Given: The equation of the plane is \(x + 2y + 3z = 1\).

set \(y = 0\) and \(z = 0\) in the equation of the plane:

\(x + 2(0) + 3(0) = 1\)

⇒ \(x = 1\)

So, \(a = 1\).

Set \(x = 0\) and \(z = 0\) in the equation of the plane:

\(0 + 2y + 3(0) = 1\)

⇒ \(2y = 1\)

⇒ \(y = \frac{1}{2}\)

\(b = \frac{1}{2}\).

Set \(x = 0\) and \(y = 0\) in the equation of the plane:

\(0 + 2(0) + 3z = 1\)

⇒ \(3z = 1\)

⇒ \(z = \frac{1}{3}\)

\(c = \frac{1}{3}\).

\(2a + 4b + 3c = 2(1) + 4\left(\frac{1}{2}\right) + 3\left(\frac{1}{3}\right)\)

⇒ \(2a + 4b + 3c = 2 + 2 + 1 = 5\)

Hence option 2 is correct.

Equation of a Plane Question 2:

The distance from a point (1, 1, 1) to a variable plane π is 12 units and the points of intersections of the plane л and X, Y, Z -axes are A, B, C respectively. If the point of intersection of the planes through the points A, B, C and parallel to the coordinate planes is P, then the equation of the locus of P is

  1. \(\rm \left(\frac{1}{xy}+\frac{1}{yz}+\frac{1}{zx}\right)=143\left(\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}\right)\)
  2. \(\rm \frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}=144\)
  3. \(\rm \left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}-1\right)^2=144\left(\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}\right)\)
  4. \(\rm \left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}-1\right)^2=144\left(\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}\right)^2\)

Answer (Detailed Solution Below)

Option 3 : \(\rm \left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}-1\right)^2=144\left(\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}\right)\)

Equation of a Plane Question 2 Detailed Solution

Concept:

Equation of a Plane:

  • The general equation of a plane passing through a point (a, b, c) and intercepting the X, Y, and Z axes at A, B, C respectively is given by:

x/A + y/B + z/C = 1

  • This form is called the Intercept Form of a plane.
  • The perpendicular distance from a point (x₀, y₀, z₀) to a plane ax + by + cz + d = 0 is:

Distance = |a·x₀ + b·y₀ + c·z₀ + d| / √(a² + b² + c²)

  • This formula helps in calculating the shortest distance from a point to a plane.
  • In this case, the point (1, 1, 1) lies at a distance of 12 units from the plane.

 

Calculation:

Given,

Point = (1, 1, 1)

Distance from the point to the plane = 12 units

Plane equation = x/A + y/B + z/C = 1

Distance = |1/A + 1/B + 1/C - 1| / √[(1/A)² + (1/B)² + (1/C)²]

⇒ |1/A + 1/B + 1/C - 1| / √[(1/A²) + (1/B²) + (1/C²)] = 12

⇒ (1/A + 1/B + 1/C - 1)² / [(1/A²) + (1/B²) + (1/C²)] = 144

⇒ (1/A + 1/B + 1/C)² = 144 × (1/A² + 1/B² + 1/C²)

Let x = A, y = B, z = C

⇒ (1/x + 1/y + 1/z)² = 144 × (1/x² + 1/y² + 1/z²)

∴ The required equation of the locus is (1/x + 1/y + 1/z)² = 144 (1/x² + 1/y² + 1/z²)

Equation of a Plane Question 3:

The equation of the plane which bisects the angle between the planes 3x − 6y + 2z + 5 = 0 and 4x − 12y + 3z − 3 = 0 which contains the origin is 

  1. 33x − 13y + 32z + 45 = 0
  2. x − 3y + z − 5 = 0
  3. 33x + 13y + 32z + 45 = 0
  4. None of these

Answer (Detailed Solution Below)

Option 4 : None of these

Equation of a Plane Question 3 Detailed Solution

Concept Used

The equation of the bisector plane is given by:

\(\frac{a_1x + b_1y + c_1z + d_1}{\sqrt{a_1^2 + b_1^2 + c_1^2}} = \pm \frac{a_2x + b_2y + c_2z + d_2}{\sqrt{a_2^2 + b_2^2 + c_2^2}}\)

Calculation

Given:

The equations of the planes are 3x - 6y + 2z + 5 = 0 and 4x - 12y + 3z - 3 = 0.

Here, a₁ = 3, b₁ = -6, c₁ = 2, d₁ = 5 and a₂ = 4, b₂ = -12, c₂ = 3, d₂ = -3.

\(\sqrt{3^2 + (-6)^2 + 2^2} = \sqrt{9 + 36 + 4} = \sqrt{49} = 7\)

\(\sqrt{4^2 + (-12)^2 + 3^2} = \sqrt{16 + 144 + 9} = \sqrt{169} = 13\)

The equations of the bisecting planes are:

\(\frac{3x - 6y + 2z + 5}{7} = \pm \frac{4x - 12y + 3z - 3}{13}\)

Case 1: (+)

13(3x - 6y + 2z + 5) = 7(4x - 12y + 3z - 3)

39x - 78y + 26z + 65 = 28x - 84y + 21z - 21

11x + 6y + 5z + 86 = 0

Case 2: (-)

13(3x - 6y + 2z + 5) = -7(4x - 12y + 3z - 3)

39x - 78y + 26z + 65 = -28x + 84y - 21z + 21

67x - 162y + 47z + 44 = 0

None of the option matches the equations

Hence, option 4 is correct.

Equation of a Plane Question 4:

The sum of the coordinates of a point lying in the yz- plane is 8. Its distance from xz -plane is thrice to its distance from xy - plane, then its coordinates are 

  1. (6, 2, 0)
  2. (0, 6, 2)
  3. (0, 2, 6)
  4. (2, 0, 6)

Answer (Detailed Solution Below)

Option 2 : (0, 6, 2)

Equation of a Plane Question 4 Detailed Solution

Explanation:

Point on yz-plane is (0, y, z)

point on xz-plane is (x, 0, z)

point on xy-plane is (x, y, 0)

Then by given condition, 

0 + y + z = 8 ⇒ y + z = 8....(i)

and

\(\sqrt{x^2+y^2}=3\sqrt{x^2+z^2}\)

⇒ x2 + y2 = 9(x2 + z2)....(ii)

The point (6, 2, 0) satisfying (i) but not (ii)

The point (0, 6, 2) satisfying both (i) and (ii)

Hence option (2) is true.

The point (0, 2, 6) and (2, 0, 6) also not satisfying both (i) and (ii).

Equation of a Plane Question 5:

Equation of a cylinder whose generators are parallel to y-axis and which passes through the curve of intersection of surfaces :

x2 + y2 + 2z2 = 12, x - y + z = 1, is-

  1. 11x2 + 11y2 + 10z2 + 24(-xy + xz - yz) = 0
  2. 2x2 + 3z2 + 2xz - 2x -2z - 11 = 0
  3. 3x2 + 2z2 + 2xz - 2x - 2z - 11 = 0
  4. 2x2 + 3y2 - 2xz + 2x + 2z - 11 = 0

Answer (Detailed Solution Below)

Option 2 : 2x2 + 3z2 + 2xz - 2x -2z - 11 = 0

Equation of a Plane Question 5 Detailed Solution

Explanation:

Surfaces is

x2 + y2 + 2z2 = 12...(i)

 x - y + z = 1...(ii)

x - y + z = 1 ⇒ y = x + z - 1

Putting in (i) we get

x2 + (x + z - 1)2 + 2z2 = 12

⇒ x2 + x2 + z2 + 1 + 2xz - 2x - 2z + 2z2 = 12

⇒ 2x2 + 3z2 + 2xz - 2x - 2z - 11 = 0

which is the required equation of a cylinder

Option (2) is true.

Top Equation of a Plane MCQ Objective Questions

The equation of the plane which contain the points (0, 6, 0) and (-2, -3, 4) and which is parallel to the ray with direction ratios (2, 3, -2) is: 

  1. 3x + 2y - 6z - 12 = 0 
  2. 3x + 2y + 6z - 12 = 0 
  3. 3x - 2y + 6z + 12 = 0 
  4. 3x - 2y - 6z + 12 = 0 

Answer (Detailed Solution Below)

Option 2 : 3x + 2y + 6z - 12 = 0 

Equation of a Plane Question 6 Detailed Solution

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Given:

The plane which contain the points (0, 6, 0) and (-2, -3, 4) 

and parallel to the ray with direction ratios (2, 3, -2)

Concept:

The equation of a plane which passed through a point \(\rm (x_1,y_1,z_1)\) with normal vector is \(\rm (a,b,c)\)

\(\rm a(x-x_1)+b(y-y_1)+c(z-z_1)=0\)

Calculation:

The equation of the plane passes through (0,6,0) is

\(\rm ax+b(y-6)+cz=0......(1)\)

This plane passes through (-2,-3,4) then

\(\rm a(-2)+b(-3-6)+c(4)=0\)

\(\rm \implies-2a-9b+4c=0.......(2)\)

Plain (1) is parallel to the ray with direction ratios (2, 3, -2)

\(\rm 2a+3b-2c=0......(3)\)

Now, solving the equations (1), (2) and (3)

\(\rm \begin{vmatrix} x & y-6& z\\ -2 & -9 & 4\\ 2 & 3 & -2 \end{vmatrix} =0\)

\(\rm \implies x(18-12)-(y-6)(4-8)+z(-6+18)=0\)

\(\rm \implies 6x+4y-24+12z=0\)

\(\rm \implies 3x+2y+6z-12=0\)

Hence the option (2) is correct.

 

Find the intercepts cut off by the plane 2x - y + z = 5 on the axes ?

  1. 5/2, -5, 5
  2. 5/2, 5, 5
  3. -5/2, 5, 5
  4. None of these

Answer (Detailed Solution Below)

Option 1 : 5/2, -5, 5

Equation of a Plane Question 7 Detailed Solution

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CONCEPT:

Let the plane make intercepts a, b, c on x, y and z axes, respectively.

D9

The equation of the plane in the intercept form is given by: \(\frac{x}{a}\; + \;\frac{y}{b}\; + \;\frac{z}{c}\; = \;1\)

CALCULATION:

Given: Equation of plane is 2x - y + z = 5

The given equation can be re-written in the intercept form as: \(\frac{x}{\frac{5}{2}}\; + \;\frac{y}{-5}\; + \;\frac{z}{5}\; = \;1\)

So, by comparing the equation \(\frac{x}{\frac{5}{2}}\; + \;\frac{y}{-5}\; + \;\frac{z}{5}\; = \;1\) with \(\frac{x}{a}\; + \;\frac{y}{b}\; + \;\frac{z}{c}\; = \;1\) we get,

⇒ a = 5/2, b = - 5 and c = 5

So, the intercepts made by the given plane on the axes are 5/2, - 5 and 5

Hence, option A is the correct answer.

The equation of the plane that passes through (1, -1, 2) and has direction ratios (1, 2, 3) is:

  1. x + 2y + 3z = 5
  2. x + 3y + 2z = 5
  3. 2x + y + 3z = 5
  4. 3x + 2y + 2z = 5

Answer (Detailed Solution Below)

Option 1 : x + 2y + 3z = 5

Equation of a Plane Question 8 Detailed Solution

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Explanation:

The Plane passes through the point having a position vector \(\overrightarrow{a}=i-j+2k\) and is the normal vector \(\overrightarrow{n}=i+2j+3k\)

So, the vector equation of the plane is

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

⇒ \(\overrightarrow{r}.\overrightarrow{n}=\overrightarrow{a}.\overrightarrow{n}\)

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

Let \(\overrightarrow{r}=xi+yj+zk\)

⇒ \((xi+yj+zk).(i+2j+3k)=(i-j+2k).(i+2j+3k)\)

⇒ x + 2y + 3z = 1 - 2 + 6

⇒ x + 2y + 3z = 5

Hence, the cartesian equation of the plane is x + 2y + 3z = 5.

Find the intercepts cut off by the plane 2x + 3y - z = 6 on the axes ?

  1. 3, -2, -6
  2. 3, 2, 6
  3. 3, 2, -6
  4. None of these

Answer (Detailed Solution Below)

Option 3 : 3, 2, -6

Equation of a Plane Question 9 Detailed Solution

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CONCEPT:

Let the plane make intercepts a, b, c on x, y and z axes, respectively.

D9

The equation of the plane in the intercept form is given by: \(\frac{x}{a}\; + \;\frac{y}{b}\; + \;\frac{z}{c}\; = \;1\)

CALCULATION:

Given: Equation of plane is 2x + 3y - z = 6

The given equation can be re-written in the intercept form as: \(\frac{x}{3}\; + \;\frac{y}{2}\; + \;\frac{z}{-6}\; = \;1\)

So, by comparing the equation \(\frac{x}{3}\; + \;\frac{y}{2}\; + \;\frac{z}{-6}\; = \;1\) with \(\frac{x}{a}\; + \;\frac{y}{b}\; + \;\frac{z}{c}\; = \;1\) we get,

⇒ a = 3, b = 2 and c = - 6

So, the intercepts made by the given plane on the axes are 3, 2 and - 6

Hence, option C is the correct answer.

Find the equation of the plane which is at a distance of 1/3 unit from the origin and \(\hat{i}+2\hat{j}+2\hat{k}\) is the normal vector from the origin to the plane ?

  1. x + 2y + 2z = 1
  2. x + 2y + 2z = 3
  3. x + 2y + 2z = 1/3
  4. x + 2y + 2z = 0

Answer (Detailed Solution Below)

Option 1 : x + 2y + 2z = 1

Equation of a Plane Question 10 Detailed Solution

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Concept:

Vector equation of a plane whose perpendicular distance from the origin is d and \(\hat n\) is the unit normal vector to the plane through origin is given by \(\vec{r}.\hat{n}=d\)

Calculation:

Let \(\vec{n}=\hat{i}+2\hat{j}+2\hat{k}\) is the normal to the required plane from the origin and it is at a distance of 1/3 units from the origin. 

⇒ \(|\vec n|=|\hat{i}+2\hat{j}+2\hat{k}| =\sqrt{1^2+2^2+2^2}=3\).

So, the unit normal vector \(\hat{n}=\frac{1}{3}\hat{i}+\frac{2}{3}\hat{j}+\frac{2}{3}\hat{k}\)

As we know that, the equation of a plane whose perpendicular distance from the origin is d and \(\hat n\) is the unit normal vector to the plane through origin is given by \(\vec{r}.\hat{n}=d\)

Here, d = 1/3, \(\hat{n}=\frac{1}{3}\hat{i}+\frac{2}{3}\hat{j}+\frac{2}{3}\hat{k}\) and let \(\vec r = x\hat i + y\hat j + z \hat k\)

⇒ \(\frac{1}{3}x+\frac{2}{3}y+\frac{2}{3}z=\frac{1}{3}\)

⇒  x + 2y + 2z = 1.

So, the equation of the required plane is x + 2y + 2z = 1

Hence, option 1 is correct.

Equation of a plane parallel to xz plane at unit distance is 

  1. x + z = 1
  2. y = 1
  3. x = z = 1
  4. x = z = 0

Answer (Detailed Solution Below)

Option 2 : y = 1

Equation of a Plane Question 11 Detailed Solution

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Concept:

Standard equations for some special plane:

  • A plane parallel to the x-y-plane must have a standard equation z = d, where d = distance of a plane from xy plane
  • A plane parallel to the y-z-plane has equation x = d, where d = distance of a plane from yz plane
  • A plane parallel to the x-z-plane has equation y = d, where d = distance of a plane from xz plane

 

 

Calculation:

Equation of xz plane is y = 0.

We know, a plane parallel to the x-z-plane has equation y = d

Here, d = 1

∴The equation of plane parallel to xz plane at unit distance will be, y = 1

Hence, option (2) is correct.

Determine the vector equation of the plane passing through the intersection of the planes \(\vec{r} \cdot(\hat{\imath}+\hat{\jmath}+\hat{k})=6\) and \(\vec{r}. (2 \hat{\imath}+3 \hat{\jmath}+4 \hat{k})=-5\), and the point (1, 1, 1)?

  1. \(\vec{r} \cdot(10 \hat{\imath}+13 \hat{\jmath}+23 \hat{k})=69\)
  2. \(\vec{r} \cdot(20 \hat{\imath}+23 \hat{\jmath}+26 \hat{k})=69\)
  3. \(\vec{r} \cdot(30 \hat{\imath}+23 \hat{\jmath}+13 \hat{k})=69\)
  4. \(\vec{r} \cdot(10 \hat{\imath}+23 \hat{\jmath}+13 \hat{k})=69\)

Answer (Detailed Solution Below)

Option 2 : \(\vec{r} \cdot(20 \hat{\imath}+23 \hat{\jmath}+26 \hat{k})=69\)

Equation of a Plane Question 12 Detailed Solution

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Concept:

As we know that, if a1x + b1y + c1z + d1 = 0 and a2x + b2y + c2z + d2 = 0 represents two different planes, then equation of plane passing through the intersection of these planes is given by:

(a1x + b1y + c1z + d1) + λ × (a2x + b2y + c2z + d2) = 0.

Calculation:

Given:

\(\vec{r} \cdot(\hat{\imath}+\hat{\jmath}+\hat{k})=6\) and \(\vec{r}. (2 \hat{\imath}+3 \hat{\jmath}+4 \hat{k})=-5\)

Convert these planes in the cartesian form.

Put \(\rm \vec r=x\hat i+y\hat j+z\hat k\)

So \(\rm \vec r.(\hat i+\hat j+\hat k)=6\)

\(\rm⇒ (x\hat i+j\hat j+z\hat k).(\hat i+\hat j+\hat k)=6\)

⇒ x + y + z = 6      .........(1)

Similarly,

\(\rm \vec r.(2\hat i+3\hat j+4 \hat k)=-5\)

⇒ 2x + 3y + 4z = -5      .........(2)

So, the plane passing through the intersection of two given planes is:

⇒ (x + y + z – 6) + λ × (2x +3y + 4z + 5) = 0 

∵ it is given that the plane passing through the intersection of two given planes also passes through the point (1, 1, 1)

⇒ The point (3, 2, 1) will satisfy equation (1)

⇒ (1 + 1 + 1 - 6) + λ × (2 + 3 + 4 + 5) = 0 ⇒ λ = 3/14.

So, by substituting the value of λ in equation (1), we get

⇒ (x + y + z – 6) + (3/14) × (2x +3y + 4z + 5) = 0 

20x +23y + 26z = 69

\(\vec{r} \cdot(20 \hat{\imath}+23 \hat{\jmath}+26 \hat{k})=69\)

What is the equation of the plane which cuts an intercept 5 units on the z-axis and is parallel to xy-plane?

  1. x + y = 5
  2. z = 5
  3. z = 0
  4. x + y + z = 5

Answer (Detailed Solution Below)

Option 2 : z = 5

Equation of a Plane Question 13 Detailed Solution

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Concept:

Plane parallel to xy- plane so, x-intercept = y-intercept = 0

 

Calculation:

Here, z-intercept = 5, and parallel to xy-plane

∴Equation of plane: z = 5

Hence, option (2) is correct.

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?

  1. x + 2y - 3z = 18
  2. 2x + 3y - z = 20
  3. 2x + 3y - 5y = 24
  4. 2x + 5y - 7z = 15

Answer (Detailed Solution Below)

Option 2 : 2x + 3y - z = 20

Equation of a Plane Question 14 Detailed Solution

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Given:

The plane which passes through the point (5, 2, -4)

Plane is perpendicular to the line with direction ratios 2, 3, -1

Concept:

Cartesian equation of the plane passing through (x1 , y1 , z1) and perpendicular to line having drs a, b, c is given by :

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

Solution:

Equation of plane :

2(x - 5) + 3(y - 2) -(z - (-4)) = 0

⇒ 2x + 3y - z = 20

Find the equation of the plane passing through the point (1, 0, 1) and perpendicular to the planes 2x + 3y - z = 2 and x - y + 2z = 1

  1. x + y - z = 0
  2. x - y + z = 2
  3. x + y + z = 2
  4. x - y - z = 0

Answer (Detailed Solution Below)

Option 4 : x - y - z = 0

Equation of a Plane Question 15 Detailed Solution

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Concept:

Let two lines having direction ratios a1, b1, c1, and a2, b2, c2 respectively.

Condition for perpendicular lines: a1a2 + b1b2 + c1c2 = 0

Condition for parallel lines: \(\rm \frac {a_1}{a_2} = \frac {b_1}{b_2} = \frac {c_1}{c_2}\)

 

Calculation:

The equation of the plane passing through the given point is

a(x - 1) + b(y - 0) + c(z - 1) = 0

Given perpendicular planes are 2x + 3y - z = 2 and x - y + 2z = 1

∴ 2a + 3b - c = 0                ....(i)

Also,

a - b + 2c = 0                   ....(ii)

On substracting 2 × (ii) from (i),

5b - 5c = 0

b = c

Putting it in equation (ii)

a - c + 2c = 0

a = -c

Now putting the values of a and b in the equation of the plane

-c(x - 1) + c(y - 0) + c(z - 1) = 0

-x + 1 + y + z - 1 = 0

x - y - z = 0

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