Mathematics

Three Dimensional Geometry Planes

254 Questions

Three dimensional geometry planes involve calculating intercepts, angles between planes, and lines of intersection. These concepts are essential for advanced mathematics assessments. The questions cover spatial relationships of planar surfaces and vector equations.

Plane interceptsLine of intersectionAngle between planesVector equations of planesPoint and plane relationships

Three Dimensional Geometry Planes Questions

Multiple choice angle between a line and a plane three dimensional geometry - ii product of vectors applications of vector algebra maths

An angle between the plane , $x+y+z=5$ and the line of intersection of the planes, $3x+4y+x-1=0$ and $5x+8y+2z+14=0$

  1. $\sin^{-1}(\sqrt{3/17})$
  2. $\cos^{-1}(\sqrt{3/17})$
  3. $\cos^{-1}(3/\sqrt{17})$
  4. $\sin^{-1}(3/\sqrt{17})$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

We have

$\overrightarrow {{\pi _1}} :x + y + z = 5$
$\overrightarrow {{r _1}} :3x + 4y + z - 1 = 0$
$\overrightarrow {{r _2}} :5x + 8y + 2z + 14 = 0$
Line of intersection of planes $\parallel \,\,to\,\,\overrightarrow {{r _1}}  \times \overrightarrow {{r _1}} $
By the helps of determinate
$\left| { \begin{array} { *{ 20 }{ c } }{ \widehat { i }  } & { \widehat { j }  } & { \widehat { k }  } \ 3 & 4 & 1 \ 4 & 8 & 2 \end{array} } \right| $
$ = \widehat i\left( 0 \right) - \widehat j\left( {6 - 5} \right) + \widehat k\left( {24 - 20} \right)$
$ =  - \widehat j + 4\widehat k$
Now,
$\sin \theta  = \frac{{ - 1 + 4}}{{\sqrt 3 \sqrt {17} }} = \sqrt {\frac{3}{{17}}} $
$\therefore \theta  = {\sin ^{ - 1}}\sqrt {\frac{3}{{17}}} $
Hence the option $A$ is the correct answer.

Multiple choice angle between a line and a plane three dimensional geometry - ii product of vectors applications of vector algebra maths

The line $\dfrac {x - 2}{3} = \dfrac {y - 3}{4} = \dfrac {z - 4}{5}$ is parallel to the plane.

  1. $3x + 4y + 5z = 7$
  2. $2x + y - 2z=0$
  3. $x + y - z = 2$
  4. $2x + 3y$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Direction ratios of given line are $3,4,5$ direction ratios of the perpendicular to the  plane $2x+y-2z=0$ are $2,1,-2$


Now
$\begin{array}{l} 2\times 3+1\times 4+\left( { -2 } \right) \times 5 \ =6+4-10 \ =0 \end{array}$

perpendicular to the plane is perpendicular to the given line
so, the plane $2x+y-2z$ is parallel to the given line.

Hence, the correct option is $B$

Multiple choice angle between a line and a plane three dimensional geometry - ii product of vectors applications of vector algebra maths

The line $\cfrac{x+3}{3}=\cfrac{y-2}{-2}=\cfrac{z+1}{1}$ and the plane $4x+5y+3z-5=0$ intersect at a point

  1. $(3,1,-2)$
  2. $(3,-2,1)$
  3. $(2,-1,3)$
  4. $(-1,-2,-3)$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Let $\dfrac{x+3}{3}=\dfrac{y-2}{-2}=\dfrac{z+1}{1}=k$


On solving, we get

$\Rightarrow x=3k-3$

$\Rightarrow y=-2k+2$ 

$\Rightarrow z=k-1$

On substituing these values in the given plane equation we get,

$\Rightarrow 4x+5y+3z-5=0$

$\Rightarrow 4(3k-3)+5(-2k+2)+3(k-1)-5=0$

On simpliying we get,

$\Rightarrow 5k=10$

$\Rightarrow k=2$

Substituting this value of $k$ in equations of $x,y,z$ we get

$\Rightarrow x=3,y=-2,z=1$

Hence point of intersection is $(3,-2,1)$

Multiple choice angle between a line and a plane three dimensional geometry - ii product of vectors applications of vector algebra maths

If $a,b$ and $c$ are three unit vectors equally inclined to each other at angle $\theta$. Then, angle between $a$ and the plane of $b$ and $c$ is

  1. $\cos ^{ -1 }{ \left( \cfrac { \cos { \theta } }{ \cos { \left( \theta /2 \right) } } \right) } $
  2. $\sin ^{ -1 }{ \left( \cfrac { \sin { \theta } }{ \sin { \left( \theta /2 \right) } } \right) } $
  3. $\sin ^{ -1 }{ \left( \cfrac { \cos { \theta } }{ \cos { \left( \theta /2 \right) } } \right) } $
  4. $\cos ^{ -1 }{ \left( \cfrac { \sin { \theta } }{ \sin { \left( \theta /2 \right) } } \right) } $
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

This is a standard vector geometry problem involving the angle between a vector and a plane defined by two other vectors. The formula derived for the angle between a unit vector and the plane of two other unit vectors equally inclined at theta is indeed cos^-1(cos(theta) / cos(theta/2)).

Multiple choice angle between a line and a plane three dimensional geometry - ii product of vectors applications of vector algebra maths

If the line $\cfrac{x-1}{2}=\cfrac{y+3}{1}=\cfrac{z-5}{-1}$ is parallel to the plane $px+3y-z+5=0$, then the value of $p$

  1. $2$
  2. $-2$
  3. $\cfrac{1}{2}$
  4. $\cfrac{1}{3}$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation
line $11$ plane 
$\therefore$ line $\bot$ normal to plane 
$\therefore (2)(P)+(1)(3)+(-1)(-1)=0$  
$\therefore 2p + 3 + 1 =0$
$\therefore P=-2$
Multiple choice angle between a line and a plane three dimensional geometry - ii product of vectors applications of vector algebra maths

The angle between the plane $2 x - y + z = 6$ and a perpendiculars to the planes $x + y + 2 z = 7$ and $x - y = 3$ is

  1. $\frac { \pi } { 4 }$
  2. $\frac { \pi } { 3 }$
  3. $\frac { \pi } { 6 }$
  4. $\frac { \pi } { 2 }$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

The normal to the plane 2x - y + z = 6 is n1 = (2, -1, 1). The perpendiculars to the other two planes are their normals: n2 = (1, 1, 2) and n3 = (1, -1, 0). The cross product n2 x n3 gives the direction vector of the line perpendicular to both planes, which is (2, 2, -2). The dot product of n1 and (2, 2, -2) is 4 - 2 - 2 = 0, meaning the plane is parallel to the line, but the question asks for the angle between the plane and the perpendiculars. Given the orthogonality, the angle is pi/2.

Multiple choice angle between a line and a plane three dimensional geometry - ii product of vectors applications of vector algebra maths

Statement 1: Line $\dfrac {x-1}{1}=\dfrac {y-0}{2}=\dfrac {z+2}{-1}$ lies in the plane $2x-3y-4z-10=0$.
Statement 2: If line $\vec r=\vec a+\lambda \vec b$ lies in the planar $\vec r\cdot \vec c=n$ (where n is scalar), then $\vec b\cdot \vec c=0$.

  1. Both the statements are true, and Statement 2 is the correct explanation for Statement 1.

  2. Both the statements are true, but Statement 2 is not the correct explanation for Statement 1.

  3. Statement 1 is true and Statement 2 is false.

  4. Statement 1 is false and Statement 2 is true.
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

If line $\vec r=\vec a+\lambda \vec b$ lies in the planar $\vec r\cdot \vec c=n$ (where n is scalar), then $\vec b\cdot \vec c=0$ &  $\vec a\cdot \vec c=n$
Therefore, statement 2 is true.
Since, line $\dfrac {x-1}{1}=\dfrac {y-0}{2}=\dfrac {z+2}{-1}$ lies in the plane $2x-3y-4z-10=0$
Then, $2(1)-3(0)-4(-2)-10=0$
          $\Rightarrow 0=0$
and $(i+2j-k).(2i-3j-4k)=0$
       $\Rightarrow 2-6+4=0$
       $\Rightarrow 0=0$
Therefore, statement 1 is true.

Ans: A

Multiple choice angle between a line and a plane three dimensional geometry - ii product of vectors applications of vector algebra maths

If $\theta$ denotes the acute angle between the line $\bar{r} = (\bar{i} + 2\bar{j} - \bar{k}) + \lambda  (\bar{i} - \bar{j} + \bar{k})$ and the plane $\bar{r} = (2\bar{i} - \bar{j} + \bar{k}) = 4$, then $\sin \theta + \sqrt 2 \cos \theta$

  1. $\dfrac{1}{\sqrt 2}$
  2. $1$
  3. $\sqrt 2$
  4. $1 + \sqrt 2$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The sine of the angle between a line with direction vector v and a plane with normal vector n is given by |v.n| / (|v||n|). Here v = (1, -1, 1) and n = (2, -1, 1). The dot product is 2 + 1 + 1 = 4, and the magnitudes are sqrt(3) and sqrt(6). Thus sin(theta) = 4 / (sqrt(3)*sqrt(6)) = 4 / (3*sqrt(2)) = 2*sqrt(2)/3. Using cos(theta) = sqrt(1 - sin^2(theta)), we calculate the expression.

Multiple choice angle between a line and a plane three dimensional geometry - ii product of vectors applications of vector algebra maths

Gives the line $\displaystyle L:\frac { x-1 }{ 3 } =\frac { y+1 }{ 2 } =\frac { z-3 }{ -1 } $ and the plane $\pi :x-2y=0$. Of the following assertions, the only one that is always true is:

  1. $L$ is $\bot$ to $\pi$
  2. $L$ lies in $\pi$
  3. $L$ is parallel to $\pi$

  4. none of these
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Since $3\left( 1 \right) +2\left( -2 \right) +\left( -1 \right) \left( -1 \right) =3-4+1=0$

$\therefore$ given line is $\bot$ to the normal to the plane i.e. given line is parallel to the given plane.
Also, $(1,-1,3)$ lies on the plane $x-2y-z=0$
$1-2\left( -1 \right) -3=0\Rightarrow 1+2-3=0$
which is true
$\therefore L$ lies in plane $\pi$

Multiple choice angle between a line and a plane three dimensional geometry - ii product of vectors applications of vector algebra maths

Consider a plane $x + y - z = 1$ and the point $A(1, 2, -3)$. A line $L$ has the equation $x = 1 + 3r$, $y = 2 - r$, $z = 3 + 4r$

The coordinate of a point $B$ of line $L$, such that $AB$ is parallel to the plane, is

  1. $(10, -1, 15)$
  2. $(-5, 4, -5)$
  3. $(4, 1, 7)$
  4. $(-8, 5, -9)$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Let $\vec { OB } =\left( 1+3r \right)\hat i+\left( 2-r \right)\hat j+\left( 3+4r \right)\hat k$
$\vec { AB } =\vec { OB } -\vec { OA } =\left( 1+3r \right)\hat i+\left( 2-r \right)\hat j+\left( 3+4r \right)\hat k-\hat i-2\hat j+3\hat k=3r\hat i-r\hat j+\left( 6+4r \right)\hat k$
Since, $\vec { AB }$ is parallel to $x+y-z=1$
Therefore, $\vec { AB } .\left(\hat i+\hat j-\hat k \right) =0$
$\Rightarrow \left( 3r\hat i-r\hat j+\left( 6+4r \right)\hat k \right) .\left(\hat i+\hat j-\hat k \right)=0 $
$\Rightarrow 3r-r-6-4r=0$
$\Rightarrow r=-3$
Therefore, $\vec { OB } =-8i+5j-9k$

Ans: D

Multiple choice angle between a line and a plane three dimensional geometry - ii product of vectors applications of vector algebra maths

If the angle between the line $x=\dfrac{y-1}{2}=\dfrac{z-3}{\lambda}$ and the plane $x+2y+3z=4$ is $\cos ^{ -1 }{ \left( \sqrt { 5/14 }  \right)  } $ then $\lambda$=

  1. $\dfrac{3}{2}$
  2. $\dfrac{5}{3}$
  3. $\dfrac{2}{3}$
  4. $\dfrac{2}{5}$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The angle theta between a line with direction (1, 2, lambda) and a plane with normal (1, 2, 3) satisfies sin(theta) = |(1, 2, lambda).(1, 2, 3)| / (sqrt(1+4+lambda^2) * sqrt(1+4+9)). Given cos(theta) = sqrt(5/14), then sin(theta) = sqrt(1 - 5/14) = 3/sqrt(14). Solving |5 + 3*lambda| / (sqrt(5+lambda^2) * sqrt(14)) = 3/sqrt(14) leads to lambda = 3/2.

Multiple choice angle between a line and a plane three dimensional geometry - ii product of vectors applications of vector algebra maths

Consider plane containing line $\dfrac{x+1}{-3} = \dfrac{y-3}{2} = \dfrac{z+2}{-1}$ and passing through the point $(1, -1, 0)$ . The angle made by the plane with x-axis is 

  1. $tan^{-1} \sqrt{2}$
  2. $cot^{-1} \sqrt{2}$
  3. $\dfrac{\pi}{6}$

  4. none of these
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

This is a variation of the previous question with a corrected direction vector for the line. The steps involve finding the plane equation using the point and line, then calculating the angle between the plane's normal and the x-axis vector.

Multiple choice angle between a line and a plane three dimensional geometry - ii product of vectors applications of vector algebra maths

If the plane $2x-3y+6z-11=0$ makes an angle $\sin^{-1}(k)$ with x-axis, then $k$ is equal to:

  1. $\cfrac {\sqrt{3}}{2}$
  2. $\dfrac 27$
  3. $\dfrac {\sqrt{2}}{3}$
  4. $1$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Given plane is $2x−3y+6z−11=0$


We know $\sin\theta=\dfrac {\vec b . \vec n}{|\vec b| |\vec n|}$


Here $\vec n$ is the normal vector to the plane

$\vec n= 2\vec i -3 \vec j+6\vec k$

and $\vec b $ is along x axis

$\therefore \sin\theta=\dfrac {(2\vec i-3\vec j+6\vec k).\vec i}{{\sqrt{2^2+(-3)^2+6^2\sqrt12}}}$

$\therefore \dfrac{2}{\sqrt49}=\dfrac{2}{7}$

Hence, B is correct option

Multiple choice angle between a line and a plane three dimensional geometry - ii product of vectors applications of vector algebra maths

The angle between the line $\displaystyle x = y = z$ and the plane $\displaystyle 4x - 3y + 5z = 2$ is

  1. $\displaystyle \cos^{-1} \frac{\sqrt{6}}{5}$
  2. $\displaystyle \sin ^{-1} \frac{\sqrt{6}}{5}$
  3. $\displaystyle \frac{\pi }{2}$
  4. $\displaystyle \sin ^{-1} \frac{1}{\sqrt{6}}$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Direction ratios of $x=y=z$ are $(1,1,1)$....(1)

Let angle between the above line and $4x -3y+5z=2$ is $ \theta$.

$ \Rightarrow $ angle between the line and normal to the plane is $90^{o} - \theta$...(2)

Direction ratios of normal to the given plane are $(4, -3, 5)$....(3)

From (1), (2) and (3), $ \cos {(90^{o}- \theta)}=\dfrac{1 \times 4 + 1 \times (-3)+ 1 \times 5}{\sqrt{1^2+1^2+1^2}{ \sqrt {4^2+3^2+5^2}}}=\dfrac{6}{\sqrt{3} \times 5 \times \sqrt{2}}$

$ \Rightarrow \sin {\theta}= \dfrac{\sqrt{6}}{5} $

$\Rightarrow \theta = \sin^{-1}{\dfrac{\sqrt{6}}{5}}$