Tag: three dimensional geometry

Questions Related to three dimensional geometry

If $\displaystyle \left ( 3, : \lambda, : \mu \right )$ is a point on the line then $\displaystyle 2x + y + z = 0 = x - 2y + z -1$ then

line of intersection of two planes line and a plane vectors, lines and planes three dimensional geometry maths

  1. $\displaystyle \lambda = \frac{-8}{3}, : \mu = - \frac{1}{3}$

  2. $\displaystyle \lambda = \frac{-1}{3}, : \mu = - \frac{8}{3}$

  3. $\displaystyle \lambda = \dfrac{-4}{3} : \mu = \dfrac{-14}{3}$

  4. $\displaystyle \lambda = -5, : \mu = -1$

Show answer
Correct Option: C
Explanation:

Since the point lies on the both planes we have, 

$6+\lambda+\mu=0$       ...(1)
$3-2\lambda+\mu-1=0$           ....(2) 
Subtracting equation 2 from 1 we get $\lambda=\cfrac{-4}{3}$. 

And substituting on equation 1 we get $\mu=\cfrac{-14}{3}$.

The variable plane $\displaystyle \left ( 2 \lambda + 1 \right )x + \left ( 3 - \lambda \right )y + z = 4$ always passes through the line 

line of intersection of two planes line and a plane vectors, lines and planes three dimensional geometry maths

  1. $\displaystyle \frac{x}{0} = \frac{y}{0} = \frac{x + 4}{1}$

  2. $\displaystyle \frac{x}{1} = \frac{y}{2} = \frac{z}{-3}$

  3. $\displaystyle \frac{x}{1} = \frac{y}{2} = \frac{z - 4}{-7}$

  4. none of these

Show answer
Correct Option: C
Explanation:

Given equation of plane is $ (2 \lambda+1)x+(3- \lambda)y+z=4$


Here $\lambda$ is a variable, we have to eliminate it.


The given equation can be written as $ \lambda(2x-y)+x+3y+z=4$

Now observe that $\lambda$ will be eliminated from the equation, if $2x-y=0$

$\Rightarrow 2x=y$

$\Rightarrow x=\dfrac{y}{2}$

The given equation becomes $x+3y+z=4$

$\Rightarrow 7x+z=4$

$\Rightarrow x=\dfrac{z-4}{-7}$

So, the given plane always passes through $ \dfrac{x}{1}=\dfrac{y}{2}=\dfrac{z-4}{-7}$ irrespective of $\lambda$

Therefore, option $C$ is correct.

The equation of the plane which contains the origin and the line of intersection of the planes $\vec r.\vec a=\vec p$ and $\vec r.\vec b=\vec q$ is

line of intersection of two planes line and a plane vectors, lines and planes three dimensional geometry maths

  1. $\vec r.\left( \vec p\vec a-\vec q\vec b \right) =0$

  2. $\vec r.\left(\vec  p\vec a+\vec q\vec b \right) =0$

  3. $\vec r.\left(\vec  q\vec a+\vec p\vec b \right) =0$

  4. $\vec r.\left( \vec q\vec a-\vec p\vec b \right) =0$

Show answer
Correct Option: D
Explanation:

Any plane through the inetersection of $\vec r.\vec a=\vec p$ and $\vec r.\vec b=\vec q$ is

$r.\left( \vec a-\lambda \vec b \right) =\vec p-\lambda \vec q$   ...(1)
Since it passes through the origin,
$\displaystyle \therefore 0.\left( \vec a-\lambda \vec b \right) =\vec p-\lambda \vec q$
$\Rightarrow \vec p-\lambda \vec q=0$
$\Rightarrow \lambda =\dfrac { \vec p }{ \vec q } $
Putting this value of $\lambda$ in (1), we get
$\displaystyle \vec r.\left( \vec a-\frac { \vec p }{\vec  q } \vec b \right) =\vec p-\frac {\vec  p }{ \vec q } \vec q=0\Rightarrow \vec r.\left( \vec a\vec q-\vec p\vec b \right) =0$
This is the required equation.

The distance of the point $(1, -2, 3)$ from the plane $x-y+z=5$ measured parallel to the line. $\frac { x }{ 2 } =\frac { y }{ 3 } =\frac { z }{ -6 } ,\quad is:$

line of intersection of two planes line and a plane vectors, lines and planes three dimensional geometry maths

  1. 1

  2. 6/7

  3. 7/6

  4. 1/6

Show answer
Correct Option: B
Explanation:
Given,

Let $P=(1,-2,3)$

given plane $x-y+z=5$

to find the distance of point $P=(1,-2,3)$ from the plane $x-y+z=5$ measured along the parallel line to

$\dfrac{x}{2}=\dfrac{y}{3}=\dfrac{z}{-6}$

equation of line passing through $(1,-2,3)$and having DR's $(2,3,-6)$

let $\dfrac{x-1}{2}=\dfrac{y+2}{3}=\dfrac{z-3}{-6}=\lambda $

$x=2\lambda+1,y=3\lambda-2,z=-6\lambda+3$

$\Rightarrow 2\lambda+1-3\lambda+2-6\lambda+3=2+3-6$

$-7\lambda =5-6$

$\therefore \lambda =\dfrac{1}{7}$

Therefore the coordinates of Q are

$\left ( \dfrac{2}{7}+1,\dfrac{3}{7}-2,-\dfrac{6}{7}+3 \right )$

$=\left ( \dfrac{9}{7},-\dfrac{11}{7},\dfrac{15}{7} \right )$

$PQ=\sqrt{\left ( \dfrac{9}{7}-1 \right )^2+\left ( -\dfrac{11}{7}+2 \right )^2+\left ( \dfrac{15}{7}-3 \right )^2}$

$=\sqrt{\dfrac{4}{49}+\dfrac{9}{49}+\dfrac{36}{49}}$

$=\sqrt{\dfrac{49}{49}}$

$=1$

Which of the following does not represent a straight line?

line of intersection of two planes line and a plane vectors, lines and planes three dimensional geometry maths

  1. $ax+by+cz+d=0,ax+b'y+cz+d=0(b\neq b')$

  2. $ax+by+cz+d=0,a'x+by+cz+d=0(a\neq a')$

  3. $ax+by+cz+d=0,ax+by+cz+d'=0(d\neq d')$

  4. $ax+by+cz+d=0,ax+by+c'z+d=0(c\neq c')$

Show answer
Correct Option: C
Explanation:

A. $ax+by+cz+d=0,ax+b'y+cz+d=0(b\neq b')$
Both the planes are different and are not parallel so they will definitely intersect on a line. Thus option A represents a line.
Similarly B and D represents a line. 
But C does not represents line. since $ax+by+cz+d=0,ax+by+cz+d'=0(d\neq d')$ represents two parallel planes which never intersects. 
Hence, option 'C' is correct choice.

Consider a plane $x+2y+3z=15$ and a line $\dfrac{x-1}{2}=\dfrac{y+1}{3}=\dfrac{z-2}{4}$ then find the distance of origin from point of intersection of line and plane.

line of intersection of two planes line and a plane vectors, lines and planes three dimensional geometry maths

  1. $\dfrac{1}{2}$

  2. $\dfrac{9}{2}$

  3. $\dfrac{5}{2}$

  4. $4$

Show answer
Correct Option: B
Explanation:

Let $\dfrac{x-1}{2}=\dfrac{y+1}{3}=\dfrac{z-2}{4}=\lambda$ $\Rightarrow x=2\lambda +1, y=3\lambda -1, z=4\lambda +2$
Now substitution in $x+2y+3z=15$
$\Rightarrow (2\lambda +1)+2(3\lambda -1)+3(4\lambda +2)=15$
$\Rightarrow 2\lambda +1+6\lambda -2+12\lambda +6=15$ $\Rightarrow 20\lambda +5=15$ $\Rightarrow \lambda =\dfrac{1}{2}$
Hence point of intersection is $(2, \dfrac{1}{2}, 4)$
Hence distance from origin is $\sqrt{4+\dfrac{1}{4}+16}=\sqrt{\dfrac{81}{4}}=\dfrac{9}{2}$.

Let $L$ be the line of intersection of the planes $2x+3y+z= 1$ and $x+3y+2z= 2$ . If $L$ makes an angle $\alpha $ with the positive $x$ -axis, then $\cos \alpha$ equals 

line of intersection of two planes line and a plane vectors, lines and planes three dimensional geometry maths

  1. $1$

  2. $\displaystyle \frac{1}{\sqrt{2}}$

  3. $\displaystyle \frac{1}{\sqrt{3}}$

  4. $\displaystyle \frac{1}{2}$

Show answer
Correct Option: C
Explanation:

Given planes are $2x+3y+z=1$ and $x+3y+2z=2$


Direction ratios of line $L$ through their intersection is given by the cross product of direction cosines of plane.


Let the direction ratios of line are represented by $\vec r$, then

$\vec { r } =(2i+3j+k)\times (i+3j+2k)$

$ \vec { r } =\left| \begin{matrix} i & j & k \\ 2 & 3 & 1 \\ 1 & 3 & 2 \end{matrix} \right| $

$ \vec { r } =i(6-3)-j(4-1)+k(6-3)$

$ \vec { r } =3i-3j+3k$

Direction ratio of $x$ axis is $\vec a =i$

$\vec { r } .\vec { a } =\left| \vec { r }  \right| \left| \vec { a }  \right| \cos { \alpha  } $

$ (3i-3j+3k).(i)=(3\sqrt { 3 } )(1)\cos { \alpha  } $

$ 3-0+0=3\sqrt { 3 } \cos { \alpha  } $

$ \cos { \alpha  } =\dfrac { 3 }{ 3\sqrt { 3 }  } =\dfrac { 1 }{ \sqrt { 3 }  } $

So, option C is correct.

The vector equation of the line of intersection of the planes $r.(i+2j+3k)=0$ and $r.(3i+2j+k)=0$ is

line of intersection of two planes line and a plane vectors, lines and planes three dimensional geometry maths

  1. $r=\lambda (i+2j+k)$

  2. $r=\lambda (i-2j+k)$

  3. $r=\lambda (i+2j-3k)$

  4. None of these

Show answer
Correct Option: B
Explanation:

The line of intersection of the planes $r.(i+2j+3k)=0$ and $r.(3i+2j+k)=0$ is parallel to the vector 

$\left( i+2j+3k \right) \times \left( 3i+2j+k \right) =-4i+8j-4k$
Since both the planes pass through the origin, therefore their line of intersection will also pass through the origin.
Thus, the required line passes through the origin and is parallel to the vector$-4i+8j-4k$
Hence, its equation is
$r=0+\lambda '\left( -4i+8j-4k \right) \Rightarrow r=\lambda \left( i-2j+k \right) $ where $\lambda =-4\lambda'$

The direction ratios of the line $x-y+z-5=0=x-3y-6$ are 

line of intersection of two planes line and a plane vectors, lines and planes three dimensional geometry maths

  1. $3,1,-2$

  2. $2,-4,1$

  3. $\displaystyle \dfrac { 3 }{ \sqrt { 14 }  } ,\dfrac { 1 }{ \sqrt { 14 }  } ,\dfrac { -2 }{ \sqrt { 14 }  } $

  4. $\displaystyle \dfrac { 2 }{ \sqrt { 14 }  } ,\dfrac { -4 }{ \sqrt { 14 }  } ,\dfrac { 1 }{ \sqrt { 14 }  } $

Show answer
Correct Option: A
Explanation:

If $l,m,n$ are the d.c's of the line, then

$1.l-1.m+1.n=0$

and $1/l-3.m+0.n=0$

$\displaystyle \therefore \dfrac { l }{ 0+3 } +\dfrac { m }{ 1-0 } =\dfrac { n }{ -3+1 } $

Hence, the dr's of the line are $3,1,-2$.

The line of intersection of the planes $\overrightarrow { r } .\left( 3i-j+k \right) =1$ and $\overrightarrow { r } .\left( i+4j-2k \right) =2$ is parallel to the vector:

line of intersection of two planes line and a plane vectors, lines and planes three dimensional geometry maths

  1. $2i+7j+13k$

  2. $-2i-7j+13k$

  3. $2i+7j-13k$

  4. $-2i+7j+13k$

Show answer
Correct Option: A
Explanation:

The line of intersection of the planes $\overrightarrow { r } .\left( 3i-k+k \right) =1$ and $\overrightarrow { r } .\left( i+4j-2k \right) =2$ is perpendicular to each, if the normal vector $\overrightarrow { { n } _{ 1 } } =3i-j+k$ and $\overrightarrow { { n } _{ 2 } } =i+4j-2k$.

$\therefore$ It is parallel to the vector, 
$\overrightarrow { { n } _{ 1 } } \times \overrightarrow { { n } _{ 2 } } =\left( 3i-j+k \right) \times \left( i+4j-2k \right) =2i+7j+13k$