Tag: three dimensional geometry

Questions Related to three dimensional geometry

 Consider a point $P (1, 2, 3)$, plane $ \pi : x + y + z = 11 $ and the line $ L : \displaystyle \frac{x+1}{1}=\displaystyle \frac{y-12}{-2} = \displaystyle \frac{z-7}{2} $ The foot of the $ \perp  $ drawn from the point P meet the plane $ \pi $ at M, then co-ordinate of M is

maths three dimensional geometry - ii point of intersection of a line and a plane line and a plane three dimensional geometry

  1. $ \left ( \displaystyle \frac{8}{3},:\displaystyle \frac{-11}{3},:\displaystyle \frac{14}{3} \right ) $

  2. $ \left ( \displaystyle \frac{-8}{3},:\displaystyle \frac{-11}{3},:\displaystyle \frac{-14}{3} \right ) $

  3. $ \left ( \displaystyle \frac{8}{3},:\displaystyle \frac{11}{3},:\displaystyle \frac{14}{3} \right ) $

  4. None of these

Show answer
Correct Option: C
Explanation:
$P = (1,2,3) \rightarrow$ Point
$\Rightarrow\pi  : x+y+z = 11$
     $L : \dfrac { x+1 }{ 1 } =\dfrac { y-12 }{ -2 } =\dfrac { z-7 }{ 2 }$
Foot of perpendicular to the plane :
$\Rightarrow\dfrac { h-{ x } _{ 1 } }{ a } =\dfrac { k-{ y } _{ 1 } }{ b } =\dfrac { l-{ z } _{ 1 } }{ c } =\dfrac { -(a{ x } _{ 1 }+b{ y } _{ 1 }+c{ z } _{ 1 }) }{ { a }^{ 2 }+{ b }^{ 2 }+{ c }^{ 2 } }$
     $\dfrac { h-1 }{ 1 } =\dfrac { k-2 }{ 1 } =\dfrac { l-3 }{ 1 } =\dfrac { 1+2+3 }{ 3 }$
     $h-1 = k-2= l-3= -2$
     $h= -1, k=0, l=1$
The required point is
$(h+\dfrac { 11 }{ 3 } , k+\dfrac { 11 }{ 3 } , l+\dfrac { 11 }{ 3 } )$
$(\dfrac { 8 }{ 3 } , \dfrac { 11 }{ 3 } ,\dfrac { 14 }{ 3 } )$
Hence the answer is $(\dfrac { 8 }{ 3 } , \dfrac { 11 }{ 3 } ,\dfrac { 14 }{ 3 } ).$

The point of intersection of the line $\dfrac{x-1}{3}=\dfrac{y+2}{4}=\dfrac{z-3}{-2}$ and the plane $2x-y+3z-1=0$, is

maths three dimensional geometry - ii point of intersection of a line and a plane line and a plane three dimensional geometry

  1. $(-10, 10, 3)$

  2. $(10, 10, -3)$

  3. $(10, -10, 3)$

  4. $(10, -10, -3)$

Show answer
Correct Option: B

Find the point where the line of intersection of the planes $x-2y+z=1$ and $x+2y-2z=5$ intersects the plane $3x+2y+z+6=0$.

maths three dimensional geometry - ii point of intersection of a line and a plane line and a plane three dimensional geometry

  1. $P\left( 1,-2,-4 \right) $

  2. $P\left( 1,2,-4 \right) $

  3. $P\left( 1,-2,4 \right) $

  4. None of these

Show answer
Correct Option: A
Explanation:

${ P } _{ 1 }\equiv x-2y+z=1$

${ P } _{ 2 }\equiv x+2y-2z=5$
Let $Dr's$ of intersection of planes be $a,b,c$
$\Rightarrow a-2b+c=0$ and $a+2b-2c=0$
$\displaystyle\Rightarrow \frac { a }{ 4-2 } =\frac { b }{ 1+2 } =\frac { c }{ 2+2 } \Rightarrow \frac { a }{ 2 } =\frac { b }{ 3 } =\frac { c }{ 4 } $
Calculating a point which lies on both plane ${P} _{1}$ and ${P} _{2}$,
$x-2y+z=1\Rightarrow2y=x+z-1$ and $x+2y-2z=5\Rightarrow2y=-x+2z+5$
$\Rightarrow x+z-1=-x+2z+5\Rightarrow 2x=z+6$
Now, $z=0\Rightarrow x=3\Rightarrow y=1$
$\therefore$ equation of line passing through $\left( 3,1,0 \right) $ and has $dr's$  $2,3,4$ is
$\displaystyle\therefore \frac { x-3 }{ 2 } =\frac { y-1 }{ 3 } =\frac { z }{ 4 } =\lambda $(say)
$\therefore$ general point on it is $P\left( 2\lambda +3,3\lambda +1,4\lambda  \right) $
Now, solving with plane $2x+2y+z+6=0$
$\Rightarrow 2\left( 2\lambda +3 \right) +2\left( 3\lambda +1 \right) +4\lambda +6=0\ \Rightarrow 4\lambda +6+6\lambda +2+4\lambda +6=0\ \Rightarrow 14\lambda +14=0\Rightarrow \lambda =-1$
$\Rightarrow P\left( 1,-2,-4 \right) $ is the required point of intersection.

The line joining the points $\left (2, -3, 1  \right )$ and $\left (3, -4, -5  \right )$ cuts a coordinate plane at the point.

maths three dimensional geometry - ii point of intersection of a line and a plane line and a plane three dimensional geometry

  1. $\left (0, -1, 13 \right )$

  2. $\left ( 0, 0, 1 \right )$

  3. $\left ( -1, 0, 19 \right )$

  4. $\left ( 8, -9, 0 \right )$

Show answer
Correct Option: A,C
Explanation:

Equation of the line is $\displaystyle \frac { x-2 }{ 2-3 } =\frac { y+3 }{ -3+4 } =\frac { z-1 }{ 1+5 } $ or $\displaystyle \frac { x-2 }{ -1 } =\frac { y+3 }{ 1 } =\frac { z-1 }{ 6 } $
Any point on the line $\left( -r+2,r-3,6r+1 \right) $ which cuts the $yz$-plane at the point where $-r+2=0$ or $r=2$
and the point of intersection is $\left( 0,-1,13 \right) $ 
Similarly it cuts the $zx$-plane at $\left( -1,0,19 \right) $ and $xy$ plane at $\displaystyle \left( \frac { 13 }{ 6 } ,\frac { -19 }{ 6 } ,0 \right) $.

Let line L: $\displaystyle \frac{x-1}{2} = \frac{y - 1}{1} = \frac{z - 0}{4} $ & Plane P: $x + 2y - z = 3$
Then which of the following is true?

maths three dimensional geometry - ii point of intersection of a line and a plane line and a plane three dimensional geometry

  1. Line is perpendicular to plane

  2. Line is neither parallel nor perpendicular to plane

  3. Plane contains the line

  4. Line and plane do not intersect

Show answer
Correct Option: C
Explanation:

Given line L: $\displaystyle \frac{x-1}{2} = \frac{y - 1}{1} = \frac{z - 0}{4} $

Plane P: $x + 2y - z = 3$
So from the below option c $i.e.$ plane contains the line is correct

If a line which passes through the point $A(0,\,1,\,2)$ and makes angle $\displaystyle\frac{\pi}{4},\,\displaystyle\frac{\pi}{4},\,\displaystyle\frac{\pi}{2}$ with $x,\,y,\,&\,z$ axes respectively. The line meets the plane $x+y+z=0$ at point $B$. The length $\sqrt{2}AB$ is equal to

maths three dimensional geometry - ii point of intersection of a line and a plane line and a plane three dimensional geometry

  1. $3$

  2. $-3$

  3. $4$

  4. $3 \sqrt {2}$

Show answer
Correct Option: A
Explanation:

The D.C. of the line are $\begin{pmatrix}\displaystyle\frac{1}{\sqrt{2}},\,\displaystyle\frac{1}{\sqrt{2}},\,0\end{pmatrix}$
any point on the line at a distance $\lambda$ from $A(0,\,1,\,2)$
is $\begin{pmatrix}0+\displaystyle\frac{\lambda}{\sqrt{2}},\,1+\displaystyle\frac{\lambda}{\sqrt{2}},\,2+\lambda.0\end{pmatrix}$
which lies on $x+y+z=0$
$\therefore\;\displaystyle\frac{\lambda}{\sqrt{2}}+\begin{pmatrix}1+\displaystyle\frac{\lambda}{\sqrt{2}}\end{pmatrix}+2=0$
$\Rightarrow\;\displaystyle\frac{2\lambda}{\sqrt{2}}=-3\;\;\;\;\Rightarrow\;\;\;\;\lambda=\displaystyle\frac{-3}{\sqrt{2}}$
$\therefore\;B=\begin{pmatrix}-\displaystyle\frac{3}{2},\,-\displaystyle\frac{1}{2},\,2\end{pmatrix}$
$A=(0,\,1,\,2)\;&\;B\equiv\;\begin{pmatrix}-\displaystyle\frac{3}{2},\,-\displaystyle\frac{1}{2},\,2\end{pmatrix}$
$AB=\sqrt{\displaystyle\frac{9}{4}+\displaystyle\frac{9}{4}}=\displaystyle\frac{3\sqrt{2}}{2}=\displaystyle\frac{3}{\sqrt{2}}$

The ratio in which the plane $\vec{r}.(\hat{i}-2\hat{j}+3\hat{k})=17$ divides the line joining the points $(-2\hat{i}+4\hat{j}+7\hat{k})$ and $(3\hat{i}-5\hat{j}+8\hat{k})$ is

maths three dimensional geometry - ii point of intersection of a line and a plane line and a plane three dimensional geometry

  1. $1 : 5$

  2. $1 : 10$

  3. $3 : 5$

  4. $3 : 10$

Show answer
Correct Option: D
Explanation:

Equation of plane in cartesian form is, $x-2y+3z-17=0.....(1)$
Assume this plane (1) divide the line segment joining the points $(-2,4,7)$ and $(3,-5,8)$ in $m:n$ ratio.
Therefore,  $\dfrac{m}{n} = \dfrac{-2-2(4)+3(7)-17}{3-2(-5)+3(8)-17}= \dfrac{-3}{10} < 0$
Hence plane (1) divides the given line segment externally  $3:10$.  

A straight line $\overrightarrow { r } =\overrightarrow { a } +\lambda \overrightarrow { b } $ meets the plane $\overrightarrow { r } .\overrightarrow { n } =0$ in $P$. The position vector of $P$ is

maths three dimensional geometry - ii point of intersection of a line and a plane line and a plane three dimensional geometry

  1. $\displaystyle \overrightarrow { a } +\dfrac { \overrightarrow { a } .\overrightarrow { n }  }{ \overrightarrow { b } .\overrightarrow { n }  } \overrightarrow { b } $

  2. $\displaystyle \overrightarrow { a } -\dfrac { \overrightarrow { a } .\overrightarrow { n }  }{ \overrightarrow { b } .\overrightarrow { n }  } \overrightarrow { b } $

  3. $\displaystyle \dfrac { \overrightarrow { a } .\overrightarrow { n }  }{ \overrightarrow { b } .\overrightarrow { n }  } \overrightarrow { b } $

  4. None of these

Show answer
Correct Option: B
Explanation:

A straight line $\overrightarrow { r } =\overrightarrow { a } +\lambda \overrightarrow { b } $ meets the plane $\overrightarrow { r } .\overrightarrow { n } =0$ at $P$ for which $\lambda$ is given by,

$\displaystyle \left( \overrightarrow { a } +\lambda \overrightarrow { b }  \right) .\overrightarrow { n } =0\Rightarrow \lambda =-\dfrac { \overrightarrow { a } .\overrightarrow { n }  }{ \overrightarrow { b } .\overrightarrow { n }  } $

Thus, the position vector of $P$ is

$\displaystyle \overrightarrow { r } =\overrightarrow { a } -\dfrac { \overrightarrow { a } .\overrightarrow { n }  }{ \overrightarrow { b } .\overrightarrow { n }  } \overrightarrow { b } $  $[$ putting the value of $\lambda$ in $\overrightarrow { r } =\overrightarrow { a } +\lambda \overrightarrow { b } ]$

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

maths three dimensional geometry - ii point of intersection of a line and a plane line and a plane three dimensional geometry

  1. $7$

  2. $-7$

  3. no real value

  4. $4$

Show answer
Correct Option: A
Explanation:

Given equation of straight line $\displaystyle \dfrac { x-4 }{ 1 } =\dfrac { y-2 }{ 1 } =\dfrac { z-k }{ 2 } $


Since, the line lies in the plane $2x-4y+z=7$

$\therefore $ Point $\left( 4,2,k \right) $ must satisfy the plane.

$\Rightarrow 8-8+k=7 $

$\Rightarrow k=7$

The plane $x-2y+z-6=0$ and the line $\displaystyle\frac{x}{1}=\displaystyle\frac{y}{2}=\displaystyle\frac{z}{3}$ are related as.

maths three dimensional geometry - ii point of intersection of a line and a plane line and a plane three dimensional geometry

  1. Parallel to the plane

  2. At right angle to the plane

  3. Lies in the plane

  4. Meets the plane obliquely

Show answer
Correct Option: A
Explanation:

Given line is $\displaystyle\frac{x}{1}=\frac{y}{2}=\frac{z}{3}$ any point on the line is $(t, 2t, 3t)$. It lies in the given plane, if $t-2(2t)+3t-6=0$
i.e., $0\cdot t=6$, which is not true for any real $t$. So, the line and plane do not meet. i.e., the line is parallel to the plane.