Distance of a point from a plane MCQ Quiz - Objective Question with Answer for Distance of a point from a plane - Download Free PDF

Calculation

Let the equation of the plane be \( ax+by+cz=d \)

Then point \( (1,2,2) \) lies on the plane

So, \( a+2b+2c=d \) .......(1)

Since the plane is perpendicular to the given two planes, so dot product of the direction ratios of the perpendicular of the plane with the direction ratio of the perpendicular of the given two plane must be zero.

So,

⇒ \( a-b+2c=0 \)  ....... (2)

And \( 2a-2b+c=0 \) ........(3)

From (1) and (2), we get

⇒ \( 3b=d \),

Similarly,

On solving above 3 equations we get

⇒ \( c=0 \) and \( a=b \) and \( d=3a \)

Put the value of \( b,c \) and \( d \) in terms of \( a \) and get the equation of the plane as

⇒ \( x+y=3 \)

Hence \( d= \dfrac {|1-2-3|}{\sqrt2}=2\sqrt2 \)

Hence option 2 is correct

Calculation

\( \overline{n} = \begin{bmatrix} i & j & k \\ 1 & -2 & 3 \\ 2 & -1 & -1 \end{bmatrix} \)

\(\Rightarrow 5i + 7j + 3k \)

Equation of plane is \( \overline{r} \cdot \overline{n} = \overline{n} \cdot \overline{a} \)

⇒ \( \overline{r} \cdot (5i + 7j + 3k) = (5i + 7j + 3k) \cdot (i - j - k) \)

⇒ \( \overline{r} \cdot (5i + 7j + 3k) = -5 \)

Distance = \( \dfrac{5 + 21 - 21 + 5}{\sqrt{5^2 + 7^2 + 3^2}} = \dfrac{10}{\sqrt{83}} \)

Hence option 2 is correct

Normal to plane \(x-y+z+1=0\) is \(\vec{n_2}=\hat{i}-\hat{j}-\hat{k}\)

Normal to plane \(x+2y-z-3=0\) is \(\vec{n_3}=\hat{i}+2\hat{j}-\hat{k}\)

Normal to plane \(3x-y+2z-1=0\) is \(\vec{n_4}=3\hat{i}-\hat{j}+2\hat{k}\)

So line of intersection of planes 1 and 2 is along \(\vec{n_1}\times\vec{n_2}=\hat{i}+\hat{j}:\vec{L_1}\)

similarly the line of intersection of planes 3 and 4 is along \(\vec{n_3}\times \vec{n_4}=3\hat{i}-5\hat{j}-7\hat{k}:\vec{L_2}\)

The normal to the plane containing \(\vec{L_1}\) and \(\vec{L_2}\) is along \((\vec{i}+\vec{j})\times (3\hat{i}-5\hat{j}-7\hat{k})={-7\hat{i}}+ {7\hat{j}}-{3\hat{k}}\)

By inspection \(\vec{L_1}\) passes through \((0,5,4)\)

so the equation of the required plane is \(-7x+7y-8z-3=0\)

and its distance from origin is \(\dfrac{3}{(7^2+7^2+8^2)^\frac{1}{2}}=\dfrac{3}{\sqrt{162}}=\dfrac{1}{3\sqrt2}\)

Calculation

Using the concept of family of planes, the equation of \(P_3\) can be written as \((x+z-1) +\lambda y = 0\).

The distance of \((0,1,0)\) from \(P_3\) is \(1\).

Hence,

\(\dfrac{-1 +\lambda} {\sqrt{\lambda^2 +2 }} = 1 \)

On squaring,

\(\lambda^2 -2 \lambda +1 = \lambda^2 +2 \)

\(\lambda = -\dfrac{1}{2}\)

Hence, \(P_3 : 2x-y+ 2z -2 =0 \)

The distance of \((\alpha, \beta, \gamma) \) from \(P_3\) is \(2\).

Hence,

\(\left | \dfrac{ 2\alpha - \beta +2\gamma -2 }{3} \right| = 2 \)

\(\Rightarrow 2\alpha - \beta +2\gamma -8 = 0 \)

or

\( 2\alpha - \beta +2\gamma +4 =0\)

Hence, options 2 and 4 are correct.

Concept Used:

The perpendicular distance (d) of a point (x₁, y₁, z₁) from the plane ax + by + cz + d = 0 is given by:

\(d = \frac{|ax_1 + by_1 + cz_1 + d|}{\sqrt{a^2 + b^2 + c^2}}\)

Calculation:

Given:

Equation of the plane: x - 3y + 4z - 6 = 0

Point: Origin (0, 0, 0)

a = 1, b = -3, c = 4, d = -6

x₁ = 0, y₁ = 0, z₁ = 0

 \(d = \frac{|1(0) - 3(0) + 4(0) - 6|}{\sqrt{1^2 + (-3)^2 + 4^2}}\)

⇒ \(d = \frac{|-6|}{\sqrt{1 + 9 + 16}}\)

⇒ \(d = \frac{6}{\sqrt{26}}\)

Hence, option 2) \(\frac{6}{\sqrt{26}}\) is the correct answer.

Concept:

Dot product of two perpendicular lines is zero.

Distance between two points (x₁, y_1, z₁) and (x₂, y₂, z₂) is given by, \(\rm \sqrt{(x_2-x_1)^2+(y_2-y_1)^2+(z_2-z_1)^2}\)

Calculation:

Let M be the foot of perpendicular drawn from the the point P(2, 3, 4)

Let , \(\rm \dfrac{x-0}{1}=\dfrac{y-0}{0}=\dfrac{z-0}{0} =k\)

x = k, y  = 0, z = 0

So M = (k, 0, 0)

Now direction ratios of PM = (2 - k, 3 - 0, 4 - 0) = (2- k, 3, 4) and direction ratios of given line are 1, 0, 0

PM is perepedicular to the given line so,

(2 - k) (1) + 3(0) + 4 (0) = 0

∴ k = 2

M = (2, 0, 0)

Perpendicular distance PM =

 \(\rm \sqrt {(2-2)^2+(0-3)^2+(0-4)^2}\\ =\sqrt{9+16}\\ =5\)

Hence, option (2) is correct. 

Concept:

Perpendicular Distance of a Point from a Plane 

Let us consider a plane given by the Cartesian equation, Ax + By + Cz = d and a point whose coordinate is, (x1, y1, z1). Then the distance between the point and the plane is given by \(\left| {\frac{{A{x_1}\; + \;B{y_1}\; + \;C{z_1} - \;d}}{{\sqrt {{A^2}\; + \;\;{B^2}\; + \;{C^2}} }}} \right|\)

Calculation:

We have to find the distance of the point (2, 3, -5) from the plane x + 2y - 2z = 9 

As we know that,  the distance between the point and the plane is given by \(\left| {\frac{{A{x_1}\; + \;B{y_1}\; + \;C{z_1} - \;d}}{{\sqrt {{A^2}\; + \;\;{B^2}\; + \;{C^2}} }}} \right|\)

Here, A = 1, B = 2, C = -2, d = 9, x1 = 2, y1 = 3 and z1 = - 5

So, the distance of the given point form the given plane \(= \left| {\frac{{2\; \times \;1\; + \ {2}\ \times \;3\; + \;2 \times 5\; - \;9}}{{\sqrt {{1^2}\; + \;\;{{\left( {-2} \right)}^2}\; + \;{{\left( 2\right)}^2}} }}} \right|\)

\(= \;\left| {\frac{9}{{\sqrt {9} }}} \right|=3\)

So, the distance between the plane and the point is 3 units

Hence, option D is the correct answer.

Concept:

The distance of a point (p, q, r) from the plane ax + by + cz + d = 0

D = \(\rm \left|ap+bq+cr + d\over \sqrt{a^2+b^2+c^2}\right|\)

Calculation:

Given the equation of the plane is 

2x - 3y + 6z - 42 = 0

The distance of the plane from the origin (0, 0, 0)

D = \(\rm \left|2\times0-3\times 0+6\times0 -42\over \sqrt{2^2+(-3)^2+6^2}\right|\)

D = \(\rm \left|-42\over \sqrt{4+9+36}\right|\) = \(\left|-42\over7\right|\)

D = 6

Concept:

For a plane having intercept a, b and c then equation of the plane is given by:

\(\rm {x\over a}+{y\over b}+{z\over c} = 1\)

The distance of a point (p, q, r) from the plane ax + by + cz + d = 0

D = \(\rm \left|ap+bq+cr + d\over \sqrt{a^2+b^2+c^2}\right|\)

Calculation:

Given intercepts are 3, 4 and -1

The equation of the plane is 

\(\rm {x\over 3}+{y\over 4}+{z\over -1} = 1\)

4x + 3y - 12z - 12 = 0

The distance of the plane from the origin (0, 0, 0)

D = \(\rm \left|4\times0+3\times 0-12\times0 -12\over \sqrt{4^2+3^2+(-12)^2}\right|\)

D = \(\rm \left|-12\over \sqrt{9+16+144}\right|\) = \(\boldsymbol{12\over 13}\)

Concept:

The distance of the origin (0, 0, 0) from the plane ax + by + cz + d = 0 is given by \(\rm \left|\frac {(a)(0) +(b)(0) +(c)(0) + d }{\sqrt {a^2+b^2+c^2}}\right|\)

Calculation:

We know that the distance of the origin (0, 0, 0) from the plane ax + by + cz + d = 0 is given by \(\rm \left|\frac {(a)(0) +(b)(0) +(c)(0) + d }{\sqrt {a^2+b^2+c^2}}\right|\)

⇒ The distance of the origin from the plane 2x + 6y - 3z + 7 = 0

\(=\rm |\frac {(2)(0) +(6)(0) +(-3)(0) + 7 }{\sqrt {2^2+6^2+(-3)^2}}|\)

\(=\rm |\frac { 7 }{\sqrt {49}}|\)

= 1

Hence, the distance of the origin from the plane 2x + 6y - 3z + 7 = 0 is 1.

Concept:

Perpendicular Distance of a Point from a Plane

Let us consider a plane given by the Cartesian equation, Ax + By + Cz = d

And a point whose coordinate is, (x₁, y₁, z₁)

Now, distance = \(\left| {\frac{{{\rm{A}}{{\rm{x}}_1}{\rm{\;}} + {\rm{\;B}}{{\rm{y}}_1}{\rm{\;}} + {\rm{\;C}}{{\rm{z}}_1} - {\rm{\;d}}}}{{\sqrt {{{\rm{A}}^2}{\rm{\;}} + {\rm{\;\;}}{{\rm{B}}^2}{\rm{\;}} + {\rm{\;}}{{\rm{C}}^2}} }}} \right|\)

Calculation:

We know that normal is always perpendicular to the plane,

Given: Equation of the plane is x + 2y – 2z = 9

⇒ x + 2y – 2z - 9 = 0

Now we have to find the distance from the origin (0, 0, 0)

We know that distance = \(\left| {\frac{{{\rm{A}}{{\rm{x}}_1}{\rm{\;}} + {\rm{\;B}}{{\rm{y}}_1}{\rm{\;}} + {\rm{\;C}}{{\rm{z}}_1} - {\rm{\;d}}}}{{\sqrt {{{\rm{A}}^2}{\rm{\;}} + {\rm{\;\;}}{{\rm{B}}^2}{\rm{\;}} + {\rm{\;}}{{\rm{C}}^2}} }}} \right|\)

∴ Distance = \(\left| {\frac{0 + 0 - 0 - 9}{{\sqrt {{{\rm{1}}^2}{\rm{\;}} + {\rm{\;\;}}{{\rm{2}}^2}{\rm{\;}} + {\rm{\;}}{{\rm{(-2)}}^2}} }}} \right| = \frac{9}{3} = 3\)

Given:

Point = (2, 5, -3) 

Equation of the plane :  \(\vec{r}\).\((6̂{i}-3̂{j}+2̂{k})\)= 6

Concept:

Equation of a plane in cartesian form can be written by putting \(\vec{r} = x̂{i} + ŷ{j} + ẑ{k}\) in the vector equation of the plane.

Formula:

The distance of a point (p, q, r) from a plane ax + by + cz = d is given by,

\(D = |\frac{ap + bq + rc -d}{\sqrt{a^2 + b^2 + c^2 }}|\)

Solution:

Writing the  equation of a  plane in the  cartesian form :

\(\vec{r}\).\((6̂{i}-3̂{j}+2̂{k})\)= 6

⇒ (xî + yĵ + zk̂).(6î - 3ĵ + 2k̂) = 6

⇒ 6x -3y + 2z = 6

Now, using distance formula - 

D = \(|\frac{6(2) - 3(5) + 2(-3) -6}{\sqrt{6^2 + (-3)^2 + 2^2 }}|\)

⇒ D = 15/7

Concept:

• The distance between a point with position vector \(\rm \vec a\) and a plane with vector equation \(\rm \vec r.\vec n=d\), where is \(\rm \vec n\) is the normal vector to the plane, is given by the formula: Distance = \(\rm \dfrac{|\vec a.\vec n-d|}{|\vec n|}\).
• \(\rm \vec a.\vec b=|\vec a||\vec b|\cos \theta\).

Calculation:

We have \(\rm \vec a=2\vec i-\vec j-3\vec k\), \(\rm \vec n=\vec i+2\vec j-4\vec k\) and d = 9.

Using the formula Distance = \(\rm \dfrac{|\vec a.\vec n-d|}{|\vec n|}\):

Distance = \(\rm \dfrac{|(2\vec i-\vec j-3\vec k).(\vec i+2\vec j-4\vec k)-9|}{|\vec i+2\vec j-4\vec k|}\)

= \(\rm \dfrac{|2-2+12-9|}{\sqrt{1^2+2^2+(-4)^2}}\)

= \(\rm \dfrac{3}{\sqrt{21}}=\dfrac{\sqrt 3\times\sqrt3}{\sqrt3\times\sqrt7}=\sqrt{\dfrac{3}{7}} \).

CONCEPT:

Perpendicular distance of a plane ax + by + cz + d = 0 from a point P (x₁, y₁, z₁) is given by: \(D = \left| {\frac{{a{x_1} + b{y_1} + c{z_1}+d}}{{\sqrt {{a^2} + {b^2} + {c^2}} }}} \right|\)

CALCULATION:

Here, we have to find the distance of the point (2, 3, 4) from the plane 3x - 6y + 2z + 11 = 0

As we know that, the perpendicular distance of a plane ax + by + cz + d = 0 from a point P (x1, y1, z1) is given by: \(D = \left| {\frac{{a{x_1} + b{y_1} + c{z_1}+d}}{{\sqrt {{a^2} + {b^2} + {c^2}} }}} \right|\)

Here, x₁ = 2, y₁ = 3, z₁ = 4, a = 3, b = - 6 and c = 2

So, the required distance between the given point and plane is \(D = \left| {\frac{{3 \times {2} - 6 \times {3} + 2 \times {4} + 11}}{{\sqrt {{3^2} + {(-6)^2} + {2^2}} }}} \right|\)

⇒ D = 1

Hence, option A is the correct answer.

Concept:

The equation of a plane passing through the line of intersection of two planes a1x + b1y + c1z + d1 = 0 and a2x + b2y + c2z + d2 = 0 is given by:

(a1x + b1y + c1z + d1) + λ (a2x + b2y + c2z + d2) = 0 where λ ∈ R.

If the angle between the lines \(\frac{{x - {x_1}}}{{{a_1}}} = \frac{{y - {y_1}}}{{{b_1}}} = \frac{{z - {z_1}}}{{{c_1}}}\;and\frac{{x - {x_2}}}{{{a_2}}} = \frac{{y - {y_2}}}{{{b_2}}} = \frac{{z - {z_2}}}{{{c_2}}}\) where a1, b1, c1, a2, b2 and c2 are the direction ratios is 90° then

a1 ⋅ a2 + b1 ⋅ b2 + c1 ⋅ c2 = 0

The perpendicular distance of a plane ax + by + cz + d = 0 from a point P (x₁, y₁, z₁) is given by: \(\left| {\frac{{a{x_1} + b{y_1} + c{z_1} + d}}{{\sqrt {{a^2} + {b^2} + {c^2}} }}} \right|\)

Calculation:

Given: Plane passes through the line of intersection of the planes x - 2y + z = 1 and 2x + y + z = 8 and is parallel to the line with direction ratios 1, 2, 1

As we know that, equation of 

plane passing through the line of intersection of two planes a1x + b1y + c1z + d1 = 0 and a2x + b2y + c2z + d2 = 0 is given by:

(a1x + b1y + c1z + d1) + λ (a2x + b2y + c2z + d2) = 0 where λ ∈ R.

So, the equation of plane passing through the line of intersection of the planes x - 2y + z = 1 and 2x + y + z = 8 is given by:

⇒ (x - 2y + z - 1) + λ (2x + y + z - 8) = 0

⇒ (1 + 2λ) x + (λ - 2) y + (λ + 1) z - (1 + 8λ) = 0------------(1)

∵ It is given that the plane which passes through the line of intersection of the planes x - 2y + z = 1 and 2x + y + z = 8 is  parallel to the line with direction ratios 1, 2, 1.

Then the normal to the plane represented by (1) is perpendicular to the line having direction ratios 1, 2, 1.

So, the direction ratios of the normal of the plane represented by (1) are: 1 + 2λ, λ - 2, λ + 1

As we know that if two lines are perpendicular then a1 ⋅ a2 + b1 ⋅ b2 + c1 ⋅ c2 = 0 where a1, b1, c1, a2, b2 and c2 are the direction ratios

⇒ (1 + 2λ) + 2 (λ - 2) + (λ + 1) = 0

⇒ λ = 2/5

By substituting λ = 2/5 in equation (1), we get

⇒ 9x - 8y + 7z - 21 = 0

So, the equation of the required plane is: 9x - 8y + 7z - 21 = 0

As we know that, the perpendicular distance of a plane ax + by + cz + d = 0 from a point P (x1, y1, z1) is given by: \(\left| {\frac{{a{x_1} + b{y_1} + c{z_1} + d}}{{\sqrt {{a^2} + {b^2} + {c^2}} }}} \right|\)

So, the distance between the plane 9x - 8y + 7z - 21 = 0 and the point (1, 1, 1) is:

\(\Rightarrow \left| {\frac{{9 \times 1 - 8 \times 1 + 7 \times 1 - 21}}{{\sqrt {{9^2} + {{\left( { - \;8} \right)}^2} + {7^2}} }}} \right| = \frac{{13}}{{\sqrt {194} }}\;units\)