Tag: three dimensional geometry - ii

Questions Related to three dimensional geometry - ii

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 -

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

  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. $\vec { r } _ { 1 } = \vec { r } _ { 0 } + \left( \dfrac { \vec { r } _ { 0 } \cdot \vec { n } } { | \vec { n } | } \right) \vec { n }$

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Correct Option: A

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

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

  1. $\displaystyle 1$

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

  3. $\displaystyle \frac{5}{7}$

  4. no possible value

Show answer
Correct Option: B
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.

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

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

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

  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

Show answer
Correct Option: D
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 $P \equiv \left(-\cfrac{3}{61}, -\cfrac{4}{61}, \cfrac{6}{61}\right)$

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

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

  1. $1$

  2. $-2$

  3. $2$

  4. None of these

Show answer
Correct Option: B
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 $(1 , -2 , 7)$

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

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

  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. $A{\alpha }'+B{\beta }'+C{\gamma }'=0\ and\ Al+Bm+Cn= 0$

Show answer
Correct Option: C
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 $A\alpha^1+B\beta^1+C\gamma^1+D=0$

 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