distance between two parallel planes 2x y 2z8 and 4x 2y 4z

Distance between two parallel planes $$2x+y+2z=8$$ and $$4x+2y+4z+5=0$$ is :

A. $$\frac{9}{2}$$

B. $$\frac{5}{2}$$

C. $$\frac{7}{2}$$

D. $$\frac{3}{2}$$

Answer : $$\frac{7}{2}$$

Solution :
The planes are $$2x + y + 2z - 8 = 0.....(1)$$
and $$4x + 2y + 4z + 5 = 0$$
or $$2x + y + 2z + \frac{5}{2} = 0.....(2)$$
$$\therefore $$ Distance between (1) and (2)
$$ = \left| {\frac{{\frac{5}{2} + 8}}{{\sqrt {{2^2} + {1^2} + {2^2}} }}} \right| = \left| {\frac{{21}}{{2\sqrt 9 }}} \right| = \frac{7}{2}$$

Releted MCQ Question on
Geometry >> Three Dimensional Geometry

Releted Question 1

The value of $$k$$ such that $$\frac{{x - 4}}{1} = \frac{{y - 2}}{1} = \frac{{z - k}}{2}$$ lies in the plane $$2x - 4y + z = 7,$$ is :

A. $$7$$

B. $$ - 7$$

C. no real value

D. $$4$$

View Answer

Releted Question 2

If the lines $$\frac{{x - 1}}{2} = \frac{{y + 1}}{3} = \frac{{z - 1}}{4}$$ and $$\frac{{x - 3}}{1} = \frac{{y - k}}{2} = \frac{z}{1}$$ intersect, then the value of $$k$$ is :

A. $$\frac{3}{2}$$

B. $$\frac{9}{2}$$

C. $$ - \frac{2}{9}$$

D. $$ - \frac{3}{2}$$

View Answer

Releted Question 3

A plane which is perpendicular to two planes $$2x - 2y + z = 0$$ and $$x - y + 2z = 4,$$ passes through $$\left( {1,\, - 2,\,1} \right).$$ The distance of the plane from the point $$\left( {1,\,2,\,2} \right)$$ is :

A. $$0$$

B. $$1$$

C. $$\sqrt 2 $$

D. $$2\sqrt 2 $$

View Answer

Releted Question 4

Let $$P\left( {3,\,2,\,6} \right)$$ be a point in space and $$Q$$ be a point on the line $$\vec r = \left( {\hat i - \hat j + 2\hat k} \right) + \mu \left( { - 3\hat i + \hat j + 5\hat k} \right)$$
Then the value of $$\mu $$ for which the vector $$\overrightarrow {PQ} $$ is parallel to the plane $$x-4y+3z=1$$ is :

A. $$\frac{1}{4}$$

B. $$ - \frac{1}{4}$$

C. $$\frac{1}{8}$$

D. $$ - \frac{1}{8}$$

View Answer

Distance between two parallel planes $$2x+y+2z=8$$ and $$4x+2y+4z+5=0$$ is :

Releted MCQ Question on
Geometry >> Three Dimensional Geometry

The value of $$k$$ such that $$\frac{{x - 4}}{1} = \frac{{y - 2}}{1} = \frac{{z - k}}{2}$$ lies in the plane $$2x - 4y + z = 7,$$ is :

If the lines $$\frac{{x - 1}}{2} = \frac{{y + 1}}{3} = \frac{{z - 1}}{4}$$ and $$\frac{{x - 3}}{1} = \frac{{y - k}}{2} = \frac{z}{1}$$ intersect, then the value of $$k$$ is :

A plane which is perpendicular to two planes $$2x - 2y + z = 0$$ and $$x - y + 2z = 4,$$ passes through $$\left( {1,\, - 2,\,1} \right).$$ The distance of the plane from the point $$\left( {1,\,2,\,2} \right)$$ is :

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Three Dimensional Geometry

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Distance between two parallel planes $$2x+y+2z=8$$ and $$4x+2y+4z+5=0$$ is :

Releted MCQ Question on Geometry >> Three Dimensional Geometry

The value of $$k$$ such that $$\frac{{x - 4}}{1} = \frac{{y - 2}}{1} = \frac{{z - k}}{2}$$ lies in the plane $$2x - 4y + z = 7,$$ is :

If the lines $$\frac{{x - 1}}{2} = \frac{{y + 1}}{3} = \frac{{z - 1}}{4}$$ and $$\frac{{x - 3}}{1} = \frac{{y - k}}{2} = \frac{z}{1}$$ intersect, then the value of $$k$$ is :

A plane which is perpendicular to two planes $$2x - 2y + z = 0$$ and $$x - y + 2z = 4,$$ passes through $$\left( {1,\, - 2,\,1} \right).$$ The distance of the plane from the point $$\left( {1,\,2,\,2} \right)$$ is :

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Releted MCQ Question on
Geometry >> Three Dimensional Geometry

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Three Dimensional Geometry