Tag: three dimensional geometry

Questions Related to three dimensional geometry

If the given planes $ax+by+cz+d=0$ and $ax+by+cz+d=0$ be mutually perpendicular, then 

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

  1. $\dfrac{a}{a}=\dfrac{b}{b}=\dfrac{c}{c}$

  2. $\dfrac{a}{a}+\dfrac{b}{b}+\dfrac{c}{c}=0$

  3. $aa+bb+cc+dd=0$

  4. $aa+bb+cc=0$

Show answer
Correct Option: D
Explanation:
$ax+by+cz+d=0$ and ${ a }^{ 1 }x+{ b }^{ 1 }y+{ c }^{ 1 }z+{ d }^{ 1 }=0$ are mutually perpendicular then their respective norm also will be perpendicular too.
Hence by the perpendicular condition,
${ aa }^{ 1 }+{ bb }^{ 1 }+{ cc }^{ 1 }=0$
Where, $\left< a,b,c \right> $ are the direction ratio of the normal to the plane $ax+by+cz+d=0$ and $\left< { a }^{ 1 },{ b }^{ 1 },{ c }^{ 1 } \right> $ are the direction ratio of the normal to the plane ${ a }^{ 1 }x+{ b }^{ 1 }y+{ c }^{ 1 }z+{ d }^{ 1 }=0$
Correct option will be $(D)$
(But there's a formating error. In the question both plane equation are same, which contradicts the fact that the planes are perpendicular).

The ratio in which the joint of $(2, 1, 5), (3, 4, 3)$ is divided by the plane $2x + 2y - 2z - 1 = 0$

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

  1. $5 : 12$

  2. $12 : 5$

  3. $5 : 7$

  4. $7 : 5$

Show answer
Correct Option: C
Explanation:

Let P point lie on a plane $2x+2y-2z-1=0 $and divide $(2, 1, 5), (3, 4, 3)$ in a ratio of  $\lambda :1$
by section formula
P=($\frac { 3\lambda +2 }{ \lambda +1 } ,\frac { 4\lambda +1 }{ \lambda +1 } ,\frac { 3\lambda +5 }{ \lambda +1 } $)
put co-ordinate of P in plane equation
2$\left(\cfrac { 3\lambda +2 }{ \lambda +1 } +\cfrac { 4\lambda +1 }{ \lambda +1 } -\cfrac { 3\lambda +5 }{ \lambda +1 } \right)$=1
by solving above equation,
we get,
 $\lambda=\dfrac{5}{7}$

A straight line $\overline { r } =\overline { a } +\lambda \overline { b } $ meets the plane $\overline { r } .\overline { n } =0$ at a point $p$. 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. $\overline { a } +\left( \cfrac { \overline { a } .\overline { n } }{ \overline { b } .\overline { n } } \right) \overline { b } $

  2. $\overline { a } -(\overline { b } .\overline { n } )\overline { b } $

  3. $\overline { a } -\left( \cfrac { \overline { a } .\overline { n } }{ \overline { b } .\overline { n } } \right) \overline { b } $

  4. $\overline { a } +(\overline { b } .\overline { n } )\overline { b } $

Show answer
Correct Option: C
Explanation:
$\rightarrow \ $ Intersection of $\vec r=\vec a+\lambda \vec b$ and $\vec r.\vec n=0$ is $(\vec a+\lambda \vec b),\vec n=0$
$\Rightarrow \ \vec a.\vec n+\lambda \vec b.\vec n=0\ \Rightarrow \lambda =-\dfrac {\vec a.\vec n}{\vec b.\vec n}$
Putting $\lambda $ in $\vec r=\vec a+\lambda \vec b$
$\vec p=\vec a-\left (\dfrac {\vec a.\vec n}{\vec b.\vec n}\right)\vec b\ \Rightarrow \ (c)$


The distance of the point $(-1,-5,-10)$ from the point of intersection of the line $\dfrac{x-2}{2}=\dfrac{y+1}{4}=\dfrac{z-2}{12}$ and the plane $x-y+z=5$ is

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

  1. $2\sqrt{11}$

  2. $\sqrt{126}$

  3. $13$

  4. $14$

Show answer
Correct Option: C
Explanation:

Given line is, $\dfrac{x-2}{2} = \dfrac{y+1}{4} = \dfrac{z-12}{2} = k$ (say)


So any point on this line is given by, $(2k+2, 4k-1, 12k+2)$

Now line intersects the plane $x-y+z=5$

$\Rightarrow 2k+2-(4k-1)+12k+2=5 \Rightarrow k = 0$

Thus point of intersection is, $(2, -1, 2)$  

Therefore required distance is $= \sqrt{(2+1)^2+ (-1+5)^2 + (2+10)^2} = 13$

Hence, option 'C' is correct.

The point of intersection of the line joining the points $(2,0,2)$ and $(3,-1,3)$ and the plane $x-y+z=1$ is

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

  1. $(3,2,0)$

  2. $(-1,1,3)$

  3. $(1,1,1)$

  4. $(4,2,-1)$

Show answer
Correct Option: A

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

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