Tag: three dimensional geometry

Quizzes (19) Questions (122)

Questions Related to three dimensional geometry

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

  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
Reveal answer Fill a bubble to check yourself
C Correct answer
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 $(\dfrac { 8 }{ 3 } , \dfrac { 11 }{ 3 } ,\dfrac { 14 }{ 3 } ).$

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

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

  1. $(-10, 10, 3)$
  2. $(10, 10, -3)$
  3. $(10, -10, 3)$
  4. $(10, -10, -3)$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

The line is (x-1)/3 = (y+2)/4 = (z-3)/-2 = t. So x=3t+1, y=4t-2, z=-2t+3. Substituting into 2x-y+3z-1=0 gives 2(3t+1) - (4t-2) + 3(-2t+3) - 1 = 0. 6t+2 - 4t+2 - 6t+9 - 1 = 0, so -4t + 12 = 0, t=3. Then x=10, y=10, z=-3.

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

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

  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
Reveal answer Fill a bubble to check yourself
A Correct answer
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.

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

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

  1. $\left (0, -1, 13 \right )$
  2. $\left ( 0, 0, 1 \right )$
  3. $\left ( -1, 0, 19 \right )$
  4. $\left ( 8, -9, 0 \right )$
Reveal answer Fill a bubble to check yourself
A,C Correct answer
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) $.

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

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?

  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
Reveal answer Fill a bubble to check yourself
C Correct answer
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

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

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

  1. $3$
  2. $-3$
  3. $4$
  4. $3 \sqrt {2}$
Reveal answer Fill a bubble to check yourself
A Correct answer
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 $x+y+z=0$
$\therefore\;\displaystyle\frac{\lambda}{\sqrt{2}}+\begin{pmatrix}1+\displaystyle\frac{\lambda}{\sqrt{2}}\end{pmatrix}+2=0$
$\Rightarrow\;\displaystyle\frac{2\lambda}{\sqrt{2}}=-3\;\;\;\;\Rightarrow\;\;\;\;\lambda=\displaystyle\frac{-3}{\sqrt{2}}$
$\therefore\;B=\begin{pmatrix}-\displaystyle\frac{3}{2},\,-\displaystyle\frac{1}{2},\,2\end{pmatrix}$
$A=(0,\,1,\,2)\;&\;B\equiv\;\begin{pmatrix}-\displaystyle\frac{3}{2},\,-\displaystyle\frac{1}{2},\,2\end{pmatrix}$
$AB=\sqrt{\displaystyle\frac{9}{4}+\displaystyle\frac{9}{4}}=\displaystyle\frac{3\sqrt{2}}{2}=\displaystyle\frac{3}{\sqrt{2}}$

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

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

  1. $1 : 5$
  2. $1 : 10$
  3. $3 : 5$
  4. $3 : 10$
Reveal answer Fill a bubble to check yourself
D Correct answer
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  $3:10$.  

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

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

  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
Reveal answer Fill a bubble to check yourself
B Correct answer
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 $\overrightarrow { r } =\overrightarrow { a } +\lambda \overrightarrow { b } ]$

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

 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 

  1. $7$
  2. $-7$

  3. no real value
  4. $4$
Reveal answer Fill a bubble to check yourself
A Correct answer
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 $

$\Rightarrow k=7$

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

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.

  1. Parallel to the plane

  2. At right angle to the plane

  3. Lies in the plane

  4. Meets the plane obliquely
Reveal answer Fill a bubble to check yourself
A Correct answer
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.