Tag: point of intersection of a line and a plane

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

Statement-I: The point $A(3,1,6)$ is the mirror image of the point $B(1,3,4)$ in the plane $x-y+z=5$.
Statement-2: The plane $x-y+z=5$ bisects the line segment joining $A(3,1,6)$ and $B(1,3,4)$.

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

  1. (1 ) StatementI is true. Statement-1 is true: Statement--2 is a correct explanation for Statement-1.

  2. (2) StatementI is true, Statement-2 is true: Statement-9 is not a correct explanation for statement-1.

  3. (3) Statement--I is true, Statement-2 is false.

  4. (4) StatementI is false. Statement-2 is true.

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Correct Option: A
Explanation:

Mid-point of AB=$\begin{array}{l} = \left( {\dfrac{{3 + 1}}{2},\dfrac{{1 + 3}}{2},\dfrac{{4 + 6}}{2}} \right)\ = (2,2,5)\end{array}$
lies on the plane as it satisfies the equation of the plane
and DR s of AB $ = (2, - 2,2)$
DR s of normal to the plane $= (1, - 1,1)$
AB is the perpendicular bisector.
Hence, A is the image of 2 
 

If the points $(1,2,3)$ and $(2,-1,0)$ lie on the opposite sides of the plane $2x+3y-2z=k$, then

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

  1. $k< 1$

  2. $k> 2$

  3. $k< 1$ or $k> 2$

  4. $1< k< 2$

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Correct Option: D
Explanation:

 Given plane equation is $2{x}+3{y}-2{z}-k=0$

$(1,2,3)$ and $(2,-1,0)$ lies on the opposite sides of the plane
$(2(1)+3(2)-2(3)-k)(2(2)+3(-1)-2(0)-k)<0$
$(2-k)(1-k)<0\implies (k-1)(k-2)<0$
$\implies 1<k<2$

If the planes $x - cy - bz = 0,cx - y + az = 0\,$ and $bx + ay - z = 0$ pass through a stright line,then the value of ${a^2} + {b^2} + {c^2} + 2abc\,$ is:

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. $3$

  4. none of these

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Correct Option: A
Explanation:
If the planes $(x - cy - bz = 0), (cx - y + az = 0)$ and $(bx + ay - z = 0)$ are in same line.
$\therefore$ They must be collinear.
$\begin{vmatrix}1 & -c & -b\\ c & -1 & a\\ b & a & -1\end{vmatrix} = 0$
$\Rightarrow 1(1 - a^2) + c(-c - ab) -b(ac + b) = 0$
$\Rightarrow 1 - a^2 - c^2 - abc - abc - b^2 = 0$
$\therefore a^2 + b^2 + c^2 + 2abc = 1$
Option A is correct

The point where the line through $A=(3, -2, 7)$ and $B= (13, 3, -8)$ meets the xy-plane

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

  1. $(\cfrac { 23 }{ 3 } ,\cfrac { 1 }{ 3 } ,0)$

  2. $(\cfrac { 23 }{ 6 } ,\cfrac { 1 }{ 6 } ,0)$

  3. $(\cfrac { 23 }{ 3 } ,\cfrac { 1 }{ 3 } , 1)$

  4. $(\cfrac { 23 }{ 3 } ,\cfrac { 1 }{ 3 } , 3)$

Show answer
Correct Option: A
Explanation:

Equation of line through $A(3,-2,7)$ and $B(13,3,-8)$ is:

$\cfrac { x-3 }{ 10 } =\cfrac { y+2 }{ 5 } =\cfrac { z-7 }{ -15 }$
When the line meets $x-y$ plane $\Rightarrow z=0$
$\therefore \cfrac { x-3 }{ 10 } =\cfrac { y+2 }{ 5 } =\cfrac { 7 }{ 15 } \quad \quad \Rightarrow x=\cfrac { 23 }{ 3 } ,y=\cfrac { 1 }{ 3 } \quad \quad \Rightarrow (x,y,z)=(\cfrac { 23 }{ 3 } ,\cfrac { 1 }{ 3 } ,0)$

The ratio in which the plane $4x+5y-3z=8$ divides the line joining the points $(-2,1,5)$ and $(3,3,2)$ is

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. $1 : 2$

  3. $-2 : 1$

  4. $3 : 2$

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Correct Option: A
Explanation:

We know that the ratio in which the plane $ax+by+cz+d=0$ divides the line segment joining (${x _1},{y _1},{z _1}$) and (${x _2},{y _2},{z _2}$) is

$\begin{array}{l} \dfrac { { -\left( { a{ x _{ 1 } }+b{ y _{ 1 } }+c{ z _{ 1 } }+d } \right)  } }{ { a{ x _{ 2 } }+b{ y _{ 2 } }+c{ z _{ 2 } }+d } }  \ a=4;b=5;c=-3;d=-8;{ x _{ 1 } }=-2;{ y _{ 1 } }=1;{ z _{ 1 } }=5;{ x _{ 2 } }=3;{ y _{ 2 } }=3;{ z _{ 2 } }=2 \ so,\, the\, required\, ratio=\dfrac { { -\left( { 4\left( { -2 } \right) +5\left( 1 \right) -3\left( 5 \right) -8 } \right)  } }{ { 4\left( 3 \right) +5\left( 3 \right) -3\left( 2 \right) -8 } }  \ =\dfrac { { -\left( { -8+5-15-8 } \right)  } }{ { 12+15-6-8 } }  \ =\dfrac { { 26 } }{ { 13 } }  \ =\dfrac { 2 }{ 1 } \ or\ 2:1 \end{array}$

Let the equations of a line and a plane be $\dfrac {x+3}{2}=\dfrac {y-4}{3}=\dfrac {z+5}{2}$ and $4x-2y-z=1$, respectively, then

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

  1. the line is parallel to the plane.

  2. the line is perpendicular to the plane.

  3. the line lies in the plane.

  4. none of these.

Show answer
Correct Option: A
Explanation:

Direction ratios of the line is $2i+3j+2k$
and normal of plane is along $4i-2j-k$
Now, $(2i+3j+2k).(4i-2j-k)=8-6-2=0$
Therefore, line is parallel to plane

Ans: A

The ratio in which the plane $r.\left( \hat { i } -2\hat { j } +2\hat { k }  \right) =17$ 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:

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

  1. $3:5$

  2. $1:10$

  3. $3:10$

  4. $1:5$

Show answer
Correct Option: C
Explanation:

Let the plane $r.(i-2j+3k)=17$ divide the line joining the points. 

$-2i+4j+7k$ and $2i-5j+8k$ in the ratio $t:1$ at the point $P.$

$\therefore P$ is $\displaystyle \dfrac { 3t-1 }{ t+1 } i+\dfrac { -5t+4 }{ t+1 } j+\dfrac { 8t+7 }{ t+1 } k.$

This lies on the given plane, 

$\displaystyle \therefore \dfrac { 3t-2 }{ t+1 } .1+\dfrac { -5t+4 }{ t+1 } \left( 2 \right) +\dfrac { 8t+7 }{ t+1 } \left( 3 \right) =17$

$\Rightarrow 3t-2+10t-8+24t+21=17t+17$

$\displaystyle \therefore 20t=17-21+10=6\Rightarrow =\dfrac { 6 }{ 20 } =\dfrac { 3 }{ 10 } $

$\therefore$ required ratio is $3:10$.

Line $\vec r=\vec a+\lambda \vec b$ will not meet the plane $\vec r\cdot \vec n=q$, if-

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

  1. $\vec b\cdot \vec n=0, \vec a\cdot \vec n=q$

  2. $\vec b\cdot \vec n\neq 0, \vec a\cdot \vec n\neq q$

  3. $\vec b\cdot \vec n=0, \vec a\cdot \vec n\neq q$

  4. $\vec b\cdot \vec n\neq 0, \vec a\cdot \vec n=q$

Show answer
Correct Option: C
Explanation:

Given line is $\overrightarrow { r } =\overrightarrow { a } +\lambda \overrightarrow { b } $

Substitute it in plane equation $\overrightarrow { r } .\overrightarrow { n } =q$
We get $(\overrightarrow { a } +\lambda \overrightarrow { b } ).\overrightarrow { n } =q$
$\Rightarrow \overrightarrow { a } .\overrightarrow { n } +\lambda (\overrightarrow { b } .\overrightarrow { n } )=q$
If $\overrightarrow { b } .\overrightarrow { n } =0$ and $\overrightarrow { a } .\overrightarrow { n } \neq q$ then the line will not meet the plane
Therefore the correct option is $C$

The ratio in which the plane $\vec r\cdot (\vec i-2\vec j+3\vec k)=17$ divides the line joining the points $-2\vec i+4\vec j+7\vec k$ and $3\vec i-5\vec j+8\vec k$ is-

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

  1. $1:5$

  2. $1:10$

  3. $3:5$

  4. $3:10$

Show answer
Correct Option: D
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 $3:10$ 

The plane $\vec r\cdot \vec n=q$ will contain the line $\vec r=\vec a+\lambda \vec b$, if-

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

  1. $\vec b\cdot \vec n\neq 0, \vec a\cdot \vec n\neq q$

  2. $\vec b\cdot \vec n=0, \vec a\cdot \vec n\neq q$

  3. $\vec b\cdot \vec n=0, \vec a\cdot \vec n=q$

  4. $\vec b\cdot \vec n\neq 0, \vec a\cdot \vec n=q$

Show answer
Correct Option: C
Explanation:

Normal of the plane $\vec { r } \cdot \vec { n } =q$ is $\vec { n } $ and direction ratio of the line $\vec { r } =\vec { a } +\lambda \vec { b } $ is $\vec { b } $
Since, line lies in the plane, normal and direction ratios should be perpendicular.
Therefore, $\vec { b } \cdot \vec { n } =0$
Also position vector $\vec { a } $ should lie on plane $\vec { r } \cdot \vec { n } =q$
Therefore, $\vec { a } \cdot \vec { n } =q$

Ans: C