Tag: point of intersection of a line and a plane

Quizzes (0) Questions (42)

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

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

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 $B(1,3,4)$.

  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.
Reveal answer Fill a bubble to check yourself
A Correct answer
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 
 

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

  1. $k< 1$
  2. $k> 2$
  3. $k< 1$ or $k> 2$
  4. $1< k< 2$
Reveal answer Fill a bubble to check yourself
D Correct answer
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
$(2(1)+3(2)-2(3)-k)(2(2)+3(-1)-2(0)-k)<0$
$(2-k)(1-k)<0\implies (k-1)(k-2)<0$
$\implies 1<k<2$

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 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:

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

  4. none of these
Reveal answer Fill a bubble to check yourself
A Correct answer
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
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 where the line through $A=(3, -2, 7)$ and $B= (13, 3, -8)$ meets the xy-plane

  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. $(\cfrac { 23 }{ 3 } ,\cfrac { 1 }{ 3 } , 3)$
Reveal answer Fill a bubble to check yourself
A Correct answer
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 $\Rightarrow z=0$
$\therefore \cfrac { x-3 }{ 10 } =\cfrac { y+2 }{ 5 } =\cfrac { 7 }{ 15 } \quad \quad \Rightarrow x=\cfrac { 23 }{ 3 } ,y=\cfrac { 1 }{ 3 } \quad \quad \Rightarrow (x,y,z)=(\cfrac { 23 }{ 3 } ,\cfrac { 1 }{ 3 } ,0)$

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

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

  1. $2 : 1$
  2. $1 : 2$
  3. $-2 : 1$
  4. $3 : 2$
Reveal answer Fill a bubble to check yourself
A Correct answer
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

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

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

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

  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.
Reveal answer Fill a bubble to check yourself
A Correct answer
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

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

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:

  1. $3:5$
  2. $1:10$
  3. $3:10$
  4. $1:5$
Reveal answer Fill a bubble to check yourself
C Correct answer
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 $3:10$.

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

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

  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. $\vec b\cdot \vec n\neq 0, \vec a\cdot \vec n=q$
Reveal answer Fill a bubble to check yourself
C Correct answer
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 $C$

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

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-

  1. $1:5$
  2. $1:10$
  3. $3:5$
  4. $3:10$
Reveal answer Fill a bubble to check yourself
D Correct answer
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 $3:10$ 

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

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

  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. $\vec b\cdot \vec n\neq 0, \vec a\cdot \vec n=q$
Reveal answer Fill a bubble to check yourself
C Correct answer
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