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

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.

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)

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.

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).

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.

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.

CONCEPT:

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

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.

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.

CONCEPT:

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

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.

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.

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.

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

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

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.

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

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