given the line l x 1 3 y 1 2 z 3 1 and the plane pi x 2y z

Given the line $$L:\frac{{x - 1}}{3} = \frac{{y + 1}}{2} = \frac{{z - 3}}{{ - 1}}$$ and the plane $$\pi \,:x - 2y = z.$$ Of the following assertions, the only one that is always true is :

A. $$L$$ is $$ \bot $$ to $$\pi $$

B. $$L$$ lies in $$\pi $$

C. $$L$$ is parallel to $$\pi $$

D. None of these

Answer : $$L$$ lies in $$\pi $$

Solution :
Since $$3\left( 1 \right) + 2\left( { - 2} \right) + \left( { - 1} \right)\left( { - 1} \right) = 3 - 4 + 1 = 0$$
$$\therefore $$ given line is $$ \bot $$ to the normal to the plane i.e., given line is parallel to the given plane.
Also $$\left( {1,\, - 1,\,3} \right)$$ lies on the plane $$x - 2y - z = 0$$ if $$1 - 2\left( { - 1} \right) - 3 = 0$$ i.e. $$1 + 2 - 3 = 0$$ which is true.
$$\therefore \,L$$ lies in plane $$\pi .$$

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

Given the line $$L:\frac{{x - 1}}{3} = \frac{{y + 1}}{2} = \frac{{z - 3}}{{ - 1}}$$ and the plane $$\pi \,:x - 2y = z.$$ Of the following assertions, the only one that is always true 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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Given the line $$L:\frac{{x - 1}}{3} = \frac{{y + 1}}{2} = \frac{{z - 3}}{{ - 1}}$$ and the plane $$\pi \,:x - 2y = z.$$ Of the following assertions, the only one that is always true 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