Tag: three dimensional geometry - ii

Quizzes (6) Questions (94)

Questions Related to three dimensional geometry - ii

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

The expression in the vector form for the point  $\vec { r } _ { 1 }$  of intersection of the plane  $\vec { r } \cdot \vec { n } = d$  and the perpendicular line  $\vec { r } = \vec { r } _ { 0 } + \hat { n }$  where  $t$  is a parameter given by -

  1. $\vec { r _ { 1 } } = \vec { r } _ { 0 } + \left( \dfrac { d - \vec { r } _ { 0 } \cdot \vec { n } } { \vec { n } ^ { 2 } } \right) \vec { n }$
  2. $\vec { r } _ { 1 } = \vec { r } _ { 0 } - \left( \dfrac { \vec { r } _ { 0 } \cdot \vec { n } } { \vec { n } ^ { 2 } } \right) \vec { n }$
  3. $\vec { r } _ { 1 } = \vec { r } _ { 0 } - \left( \dfrac { \vec { r } _ { 0 } \cdot \vec { n } - d } { | \vec { n } | } \right) \vec { n }$
  4. $\vec { r } _ { 1 } = \vec { r } _ { 0 } + \left( \dfrac { \vec { r } _ { 0 } \cdot \vec { n } } { | \vec { n } | } \right) \vec { n }$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The intersection of a line r = r0 + tn and a plane r dot n = d is found by substituting the line equation into the plane equation: (r0 + tn) dot n = d. Solving for t gives t = (d - r0 dot n) / (n dot n). Substituting t back into the line equation gives r1 = r0 + ((d - r0 dot n) / n^2) n.

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

If the line $\displaystyle \frac{x - 1}{1} = \frac{y + 1}{-2} = \frac{z + 1}{\lambda}$ lies in the plane $\displaystyle 3x - 2y + 5z = 0$ then $\displaystyle \lambda$ is

  1. $\displaystyle 1$
  2. $\displaystyle -\frac{7}{5}$
  3. $\displaystyle \frac{5}{7}$

  4. no possible value
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation
We have equation of plane,
$3x-2y+5z=0.......(1)$

We have line,

$\dfrac{x-1}{1}=\dfrac{y+1}{-2}=\dfrac{z+1}{\lambda}=\mu......(2)$
General point on line is,
$P=(\mu+1,-2\mu-1,\lambda\mu-1)$
Since line (2) lies on line plane (1),so point P satisfy equation (1)
Therefore,
$3(\mu+1)-2(2\mu-1)+5(\lambda\mu-1)=0$
$3\mu+3+4\mu+2+5\mu\lambda-5=0$
$7\mu+5\mu\lambda=0$
$\lambda=\dfrac{-7}{5}$ 
Therefore option (B) 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

The Foot of the $\displaystyle \perp$ from origin to the plane $\displaystyle 3x + 4y - 6z + 1 = 0$ is

  1. $\displaystyle - \frac {3}{61}, \frac {4}{61}, \frac {6}{61}$
  2. $\displaystyle \frac {-3}{61}, \frac {-4}{61}, \frac {-6}{61}$
  3. $\displaystyle \frac {4}{61}, \frac {-3}{61}, \frac {5}{61}$

  4. None of these
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Clearly direction ratios of perpendicular drawn from origin to the given plane are $3,4,-6$
Hence equation of perpendicular line to the given plane and  passing through origin is given by,
$\cfrac{x}{3}=\cfrac{y}{4}=\cfrac{z}{-6}=k$ (say)
Now let foot of perpendicular be $P(3k, 4k, -6k)$
Also this point lie in the given plane $\Rightarrow 3(3k)+4(4k)-6(-6k)+1=0\Rightarrow k = -\cfrac{1}{61}$
Hence $P \equiv \left(-\cfrac{3}{61}, -\cfrac{4}{61}, \cfrac{6}{61}\right)$

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

The co-ordinate of a point where the line $(2, -3, 1)$ and $(3, -4, -5)$ cuts the plane $2x + y + z = 7$ are $(1, k, 7)$ then value of $k$ equals

  1. $1$
  2. $-2$
  3. $2$

  4. None of these
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

We know that the equation of the line passing through the point $(x _1,y _1,z _1) , (x _2,y _2,z _2) $


$ \dfrac{x-x _1}{x _2-x _1} = \dfrac {y-y _1}{y _2-y _1} =  \dfrac {z-z _1}{z _2-z _1}$

Now the given line passes through the points $(2,−3,1) , (3,−4,−5)$


Hence the equation of the line is $ \dfrac{x-2}{3-2} = \dfrac {y+3}{-4+3} =  \dfrac {z-1}{-5-1}$
`
= $ \dfrac{x-2}{1} = \dfrac {y+3}{-1} =  \dfrac {z-1}{-6}$

Let the above equation be equal to :

$ \dfrac{x-2}{1} = 1$ .... (1)

$\dfrac {y+3}{-1} = a$ .....(2)

$\dfrac {z-1}{-6} = 7$ .....(3)

$\Rightarrow$ from eqn (2) 

$y=−4+a$

cube cuts the plane $2x+y+z=7$ ... (4)

substitute the $(1, k ,7)$ in eqn (4)

$2(1) + (-4+a) + 7 = 7$

$a = 2$

substitute a value in $y=−4+a$
we get $y = -4+2 = -2$

hence the ans will be $(1 , -2 , 7)$

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

The condition that the line $\displaystyle \frac{x-{\alpha }'}{l}=\frac{y -{\beta   }'}{m}=\frac{z-{\gamma  }'}{n}$ in the plane $Ax + By + Cz + D = 0$ is

  1. $A{\alpha }'+B{\beta }'+C{\gamma }'+D=0\ and\ Al+Bm+Cn\neq 0$
  2. $A{\alpha }'+B{\beta }'+C{\gamma }'+D\neq0\ and\ Al+Bm+Cn= 0$
  3. $A{\alpha }'+B{\beta }'+C{\gamma }'+D=0\ and\ Al+Bm+Cn= 0$
  4. $A{\alpha }'+B{\beta }'+C{\gamma }'=0\ and\ Al+Bm+Cn= 0$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

The line $\dfrac{x- \alpha^1}{l}=\dfrac{y-\beta^1}{m}=\dfrac{z-\gamma^1}{n}$ in the plane $Ax+By+Cz+D=0$

then multiplication sum id corresponding direction ratio will be zero so here $Al+Bm+Cn=0$
 and  one thing more line passes through $(\alpha^1,\beta^1,\gamma^1)$
So, plane Also passe through $(\alpha^1,\beta^1,\gamma^1)$
hence $A\alpha^1+B\beta^1+C\gamma^1+D=0$

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