the distance of the point 1 5 9 from the plane x y z5

The distance of the point $$\left( {1,\, - 5,\,9} \right)$$ from the plane $$x-y+ z=5$$ measured along the line $$x=y=z$$ is :

A. $$\frac{{10}}{{\sqrt 3 }}$$

B. $$\frac{{20}}{3}$$

C. $$3\sqrt {10} $$

D. $$10\sqrt {3} $$

Answer : $$10\sqrt {3} $$

Solution :
Three Dimensional Geometry mcq solution image

$$\eqalign{ & {\text{Equation of PO :}}\,\,\frac{{x - 1}}{1} = \frac{{y + 5}}{1} = \frac{{z - 9}}{1} = \lambda \cr & \Rightarrow x = \lambda + 1\,;\,\,y = \lambda - 5\,;\,\,z = \lambda + 9 \cr} $$
Putting these in equation of plane :-
$$\eqalign{ & \lambda + 1 - \lambda + 5 + \lambda + 9 = 5 \cr & \Rightarrow \lambda = - 10 \cr & \Rightarrow {\text{O is }}\left( { - 9,\, - 15,\, - 1} \right) \cr & \Rightarrow {\text{distance OP }} = 10\sqrt 3 \cr} $$

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. $$4$$

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

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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. $$2\sqrt 2 $$

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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

The distance of the point $$\left( {1,\, - 5,\,9} \right)$$ from the plane $$x-y+ z=5$$ measured along the line $$x=y=z$$ 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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The distance of the point $$\left( {1,\, - 5,\,9} \right)$$ from the plane $$x-y+ z=5$$ measured along the line $$x=y=z$$ 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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