Tag: point of intersection of a line and a plane

Questions Related to point of intersection of a line and a plane

The ratio in which the line segment joining the points whose position vectors are $2\hat i-4\hat j-7\hat k$ and $-3\hat i+5\hat j-8\hat k$ is divided by the plane whose equation is $\hat r\cdot (\hat i-2\hat j+3\hat k)=13$ is-

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

  1. $13:12$ internally

  2. $12:25$ externally

  3. $13:25$ internally

  4. $37:25$ internally

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Correct Option: B
Explanation:

Equation of plane in cartesian form is, $x-2y+3z-13=0$ .....(1)$
Assume this plane (1) divide the line segment joining the points
$(2,-4,-7)$ and $(-3,5,-8)$ in $m:n$ ratio
Therefore,  $\dfrac{m}{n} = \dfrac{2-2(-4)+3(-7)-13}{-3-2(5)+3(-8)-13}= \dfrac{-12}{25} < 0$
Henc,e plane (1) divides the given line segment externally  $12:25$.

Which of the following lines lie on the plane $x+2y-z=0$?

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

  1. $x-1=y-1=1$

  2. $x-y+z=2x+y-z=0$

  3. $\vec r=2\hat i-\hat j+4\hat k+\lambda (3\hat i+\hat j+5\hat k)$

  4. None of these.

Show answer
Correct Option: D
Explanation:

consider $P: x+2y-z=0$
direction ratio of normal of $P$ is $(1,2,-1)$

a) for $L _{1}: \dfrac {x-1}{1}=\dfrac {y}{-1}=\dfrac {z-5}{-1}$
Since, $(1,0,5)$ does not lie in $P$ 
Therefore, $L _{1}$ does not lie in $P$

b) for $L _{2}: x-y+z=2x+y-z=0$
direction ratio of $L _{2}$ is $(i-j+k) \times (2i+j-k)=3(j+k)$
since, $(i+2j-k).(3j+3k)=6-3=3 \neq =0$
Therefore, $L _{2}$ does not lie on $P$

c) for $L _{3}:\vec r=2\hat i-\hat j+4\hat k+\lambda (3\hat i+\hat j+5\hat k)$
Since, $(2,-1,4)$ does not lie in $P$ 
Therefore, $L _{3}$  does not lie in $P$

Ans: D

Find the ratio in which the segment joining $(1, 2, -1)$ and $(4, -5, 2)$ is divided by the plane $2x - 3y + z = 4$

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

  1. $2 : 1$

  2. $3 : 2$

  3. $3 : 7$

  4. $1 : 2$

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Correct Option: A

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.