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
The angle between the line $\overrightarrow { r } =\left( -\hat { i } +3\hat { j } +3\hat { k }  \right) +t\left( 2\hat { i } +3\hat { j } +6\hat { k }  \right) $ and the plane $\overrightarrow { r } .\left( -\hat { i } +\hat { j } +\hat { k }  \right) $ is
  1. $\displaystyle\sin ^{ -1 }{ \dfrac { 1 }{ \sqrt { 3 } } } $
  2. $\displaystyle\sin ^{ -1 }{ \dfrac { 1 }{ \sqrt { 2 } } } $
  3. $\displaystyle\sin ^{ -1 }{ \dfrac { 2 }{ \sqrt { 3 } } } $
  4. $\displaystyle\sin ^{ -1 }{ \dfrac { 3 }{ \sqrt { 2 } } } $
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation
Angle between the line and the plane is given by
$\displaystyle \sin { \theta  } =\dfrac { \left( 2\hat { i } +3\hat { j } +6\hat { k }  \right) .\left( -\hat { i } +\hat { j } +\hat { k }  \right)  }{ \sqrt { 4+9+36 } \sqrt { 1+1+1 }  } $
$\displaystyle =\dfrac { -2+3+6 }{ 7\times \sqrt { 3 }  } =\dfrac { 7 }{ 7\sqrt { 3 }  } =\dfrac { 1 }{ 3 } $
$\displaystyle \Rightarrow \theta =\sin ^{ -1 }{ \left( \dfrac { 1 }{ \sqrt { 3 }  }  \right)  } $
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. $\displaystyle \frac{\sqrt{3}}{2}$
  2. $\displaystyle \frac{2}{7}$
  3. $\displaystyle \frac{\sqrt{2}}{7}$
  4. $1$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Given plane is $2x−3y+6z−11=0$

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

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

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

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

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

  1. L is $\perp$ to II

  2. L lies in II

  3. L is parallel to II

  4. None of these
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation
$(1,-1,3)$ does not lie on the plane. Hence, $L$ cannot lie in the plane. 
The direction vector of line $L$ is $\vec{v}=3\hat{i}+2\hat{j}-\hat{k}$.
The normal vector of plane is $\vec{n}=\hat{i}-2\hat{j}-\hat{k}$.
Now, $\vec{n}\cdot\vec{v}=3-4+1=0$  
Hence, $\vec{n}$ and $\vec{v}$ are perpendicular and the plane and line cannot be perpendicular but are parallel.
Multiple choice angle between a line and a plane three dimensional geometry - ii product of vectors applications of vector algebra maths

Plane $2x+3y+6z=15=0$ makes angle of measure ________ with Y-axis.

  1. $\sin^{-1}\left(\dfrac{3}{7}\right)$
  2. $\sin^{-1}\left(\dfrac{2}{7}\right)$
  3. $\sin^{-1}\left(\dfrac{2}{\sqrt{7}}\right)$
  4. $\cos^{-1}\left(\dfrac{3}{7}\right)$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Direction of line $\bar{l}=(0, 1, 0)$
$\bar{n}=$ Normal of plane $=(2, 3, 6)$
$\therefore \alpha =$ Angle between line and plane.
$\therefore \sin\alpha =\left|\dfrac{\bar{l}\cdot \bar{n}}{|\bar{l}||\bar{n}|}\right|$
$=\dfrac{0(2)+1(3)+0(6)}{(1)\sqrt{4+4+36}}$
$=\dfrac{0+3+0}{\sqrt{49}}$
$=\dfrac{3}{7}$
$\therefore$ Measure of an angle between line and plane $\alpha =\sin^{-1}\left(\dfrac{3}{7}\right)$.

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

If the angle bwteen a line $x=\dfrac{y-1}{2}=\dfrac{z-3}{\lambda}$ and normal to the plane $x+2y+3z=4$ is $\cos^{-1}{\sqrt{\dfrac{5}{14}}}$, then possible value(s) of $\lambda$ is/are

  1. $\dfrac{5}{2}$
  2. $\dfrac{2}{5}$

  3. <span class="MathJax_Preview"><span class="MathJax"><span class="math"><span class="mrow"><span class="mn">0<span class="MJX_Assistive_MathML">0
  4. $\dfrac{2}{3}$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Consider the given line equation $x=\dfrac{y-1}{2}=\dfrac{z-3}{\lambda }$ and plane equation$x+2y+3z=4$.

Let $\theta $ be the angle between the line and normal to plane converting the given equations into normal form, we have

  $ \overrightarrow{r}=0.\widehat{i}+\widehat{j}+3\widehat{k}+\beta \left( \widehat{i}+2\widehat{j}+\lambda \widehat{k} \right) $

 $ \overrightarrow{r}=\widehat{i}+2\widehat{j}+3.\widehat{k}=3 $

Now,

  $ \overrightarrow{b}=\left( \widehat{i}+2\widehat{j}+\lambda \widehat{k} \right) $

 $ \overrightarrow{n}=\widehat{i}+2\widehat{j}+3.\widehat{k}=3 $

We know that,

  $ \cos \theta =\left| \dfrac{\widehat{b}.\widehat{n}}{\left| \widehat{b} \right|\left| \widehat{n} \right|} \right| $

 $ =\left| \dfrac{\left( \widehat{i}+2\widehat{j}+\lambda \widehat{k} \right).\left( \widehat{i}+2\widehat{j}+3.\widehat{k} \right)}{\left| \widehat{i}+2\widehat{j}+\lambda \widehat{k} \right|\left| \widehat{i}+2\widehat{j}+3.\widehat{k} \right|} \right| $

 $ \cos \theta =\left| \dfrac{1+4+3\lambda }{\sqrt{{{1}^{2}}+{{2}^{2}}+{{\lambda }^{2}}}\sqrt{{{1}^{2}}+{{2}^{2}}+{{3}^{2}}}} \right|=\left| \dfrac{5+\lambda }{\sqrt{5+{{\lambda }^{2}}}\sqrt{14}} \right| $

But given that $\theta ={{\cos }^{-1}}\left( \sqrt{\dfrac{5}{14}} \right)$ ,so

  $ \cos {{\cos }^{-1}}\sqrt{\dfrac{5}{14}}=\left| \dfrac{5+\lambda }{\sqrt{5+{{\lambda }^{2}}}\sqrt{14}} \right| $

 $ \sqrt{\dfrac{5}{14}}=\left| \dfrac{5+\lambda }{\sqrt{5+{{\lambda }^{2}}}\sqrt{14}} \right| $

Taking square both sides ,we get

  $ \dfrac{5}{14}={{\left| \dfrac{5+\lambda }{\sqrt{5+{{\lambda }^{2}}}\sqrt{14}} \right|}^{2}}=\dfrac{{{\left( 5+\lambda  \right)}^{2}}}{\left( 5+{{\lambda }^{2}} \right)\left( 14 \right)} $

 $ \dfrac{{{\left( 5+\lambda  \right)}^{2}}}{\left( 5+{{\lambda }^{2}} \right)}=5 $

 $ 25+{{\lambda }^{2}}+10\lambda =25+5{{\lambda }^{2}} $

 $ 4{{\lambda }^{2}}-10\lambda =0 $

 $ 2\lambda \left( 2\lambda -5 \right)=0 $

 $ \lambda \left( 2\lambda -5 \right)=0 $

 $ \lambda =0,\lambda =\dfrac{5}{2} $

Ignore $\lambda =0$ as it is in denominator. Therefore,

$\lambda =\dfrac{5}{2}$

This is the answer .

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 $\cos ^{ -1 }{ \sqrt { \dfrac { 5 }{ 14 }  }   },$ then $\lambda$ equals:

  1. $\dfrac { 2 }{ 5 } $
  2. $\dfrac { 5 }{ 3 } $
  3. $\dfrac { 2 }{ 3 } $
  4. $\dfrac { 3 }{ 2 } $
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation
Line : $\dfrac{x}{1} = \dfrac{y - 1}{2} = \dfrac{z - 3}{\lambda}$
and the plane $x + 2y + 3z = 4$
$\therefore \sin\theta = \dfrac{1 + 4 + 3\lambda}{\sqrt{1 + 4 + \lambda^2}\sqrt{1 + 4 + 9}}$
$= \dfrac{5 + 3\lambda}{\sqrt{5 + \lambda^2}\sqrt{14}}$
$cos^{-1} \sqrt{\dfrac{5}{14}} = sin^{-1}\dfrac{3}{\sqrt{14}}$
$\Rightarrow \dfrac{3}{\sqrt{14}} = \dfrac{5 + 3\lambda}{\sqrt{5 + \lambda^2}
\sqrt{14}}$
$\Rightarrow 9(5 + \lambda^2) = (5 + 3\lambda)^2$
$\Rightarrow 45 + 9\lambda^2 = 25 + 9\lambda^2 + 30\lambda$
$\Rightarrow \lambda = \dfrac{2}{3}$
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 $x=\cfrac{y-1}{2}=\cfrac{z-3}{\lambda}$ and the plane $x+2y+3z=4$ is $\cos ^{ -1 }{ \left( \sqrt { \cfrac { 5 }{ 14 }  }  \right)  } $, then $\lambda$ equals:

  1. $2/5$
  2. $5/3$
  3. $2/3$
  4. $3/2$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation
Angle between line and normal to plane would be $\cfrac { \pi  }{ 2 } -\cos ^{ -1 }{ \sqrt { \cfrac { 5 }{ 14 }  }  } =\sin ^{ -1 }{ \sqrt { \cfrac { 5 }{ 14 }  }  } =\cos ^{ -1 }{ \cfrac { 3 }{ \sqrt { 14 }  }  } $
$\Rightarrow \cfrac { \left( \hat { i } +2\hat { j } +\lambda \hat { k }  \right) .\left( \hat { i } +2\hat { j } +3\hat { k }  \right)  }{ \left| \hat { i } +2\hat { j } +\lambda \hat { k }  \right| \left| \hat { i } +2\hat { j } +3\hat { k }  \right|  } =\cfrac { 3 }{ \sqrt { 14 }  } $
$\Rightarrow \cfrac { 1+4+3\lambda  }{ \sqrt { 5+{ \lambda  }^{ 2 } } \sqrt { 1+4+9 }  } =\cfrac { 3 }{ \sqrt { 14 }  } $
$\Rightarrow 3\lambda +5=3\sqrt { 5+{ \lambda  }^{ 2 } } \Rightarrow { \left( 3\lambda +5 \right)  }^{ 2 }=9\left( { \lambda  }^{ 2 }+5 \right) $
$\Rightarrow 9{ \lambda  }^{ 2 }+25+30\lambda =9{ \lambda  }^{ 2 }+45\Rightarrow 30\lambda =20\Rightarrow \lambda =\cfrac { 2 }{ 3 } $
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 $x=\cfrac{y-1}{2}=\cfrac{z-3}{\lambda}$ and the plane $x+2y+3z=4$ is $\cos ^{ -1 }{ \left( \sqrt { \cfrac { 5 }{ 14 }  }  \right)  } $, then $\lambda$ equals

  1. $\cfrac{15}{2}$
  2. $\cfrac{3}{2}$
  3. $\cfrac{2}{5}$
  4. $\cfrac{5}{3}$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

equation of line $\dfrac{x-0}{1}=\dfrac{y-1}{2}=\dfrac{2-3}{\lambda}$ where $a _1=1, a _2=2, a _3=\lambda$

equation of plane $n+2y+32=4$
where $b _1=1, b _2=2, b _3=3$
angle between Line and plane
$\cos\theta=\dfrac{|a _1b _1+a _2b _2+a _3b _3|}{\sqrt{a _1^2+a _2^2+a _3^2}.\sqrt{b _1^2+b _2^2+b _3^2}}$
$\cos \theta=\dfrac{|1.1+2.2+3.\lambda|}{\sqrt{1+4+\lambda^2}.\sqrt{1+4+9}}$
$\cos \theta =\dfrac{5+3\lambda}{\sqrt{5+\lambda^2}.\sqrt{14}}\quad ---(1)$
given $\theta =\cos^{-1}\left(\sqrt{\dfrac{5}{14}}\right)$
$\cos \theta =\left(\sqrt{\dfrac{5}{14}}\right)\quad ----(2)$
By eqn $(1)$ & $(2)$
$\dfrac{5+3\lambda }{\sqrt{5+\lambda^2}\sqrt{14}}=\sqrt{\dfrac{5}{14}}$
$5+3\lambda=\sqrt{5}.\sqrt{5+\lambda^2}$
$(5+3\lambda)^2=5.(5+\lambda^2)$
$25+9\lambda^2+30\lambda =25+5\lambda^2$
$4\lambda^2+30\lambda =0$
$\lambda (2\lambda +15)=0\Rightarrow \lambda =0, \dfrac{15}{2}$
but $\lambda \neq 0$
So $\lambda =\dfrac{15}{2}$ Ans

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

How is the line $\displaystyle \frac{x-4}{4}=\frac{y-12}{12}=\frac{z-8}{8}$ related to the planes
(A) $\displaystyle x-y+z=0$
(B) $\displaystyle x-y+z-6=0$

  1. parallel to plane A but not B

  2. parallel to plane A and also lies in plane A but not parallel to B

  3. parallel to plane A and also lies in plane A

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

A line is inclined at Φ to a plane. The vector equation of the line is given by 

$\vec r =\vec a + \lambda \vec b $
Let $\theta$ be the angle between the line and the normal to the plane. Its value can be given by the following equation
$cos\theta=|\dfrac{\vec b . \vec n }{|\vec b|.|\vec n |}|$
Finding the value of the $Φ$ between the line and the plane we know that 
parallel to plane A and also lies in plane A.

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

If the angle $\theta $ between the line $\displaystyle \frac{x+1}{1}=\frac{y-1}{2}=\frac{z-2}{2}$ and the plane $2x-y+\sqrt{\lambda} z+4=0$ is such that $\displaystyle \sin \theta =\frac{1}{3}$, then value of $\lambda $ is

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

Angle between the line and plane is same as the angle between the line and normal to the plane
$\displaystyle \therefore \cos \left ( 90^{0}-\theta  \right )=\frac{a _{1}a _{2}+b _{1}b _{2}+c _{1}c _{2}}{\sqrt{a _{1}^{2}+b _{1}^{2}+c _{1}^{2}}\sqrt{a _{2}^{2}+b _{2}^{2}+c _{2}^{2}}}$
$\displaystyle \therefore\frac{1}{3}=\frac{\left ( 1\times 2+2\times \left ( -1 \right )+2\sqrt{\lambda } \right )}{\sqrt{1^{2}+2^{2}+2^{2}}\sqrt{2^{2}+1^{2}+\lambda }} $

$\therefore  \lambda =\dfrac{5}{3}$

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

If $\displaystyle \theta$ is the angle between the line 
$\vec r=2i+j-k+\left ( i+j+k \right )t$ and the plane
$\displaystyle \vec r\cdot \left ( 3i-4j+5k \right )=q$, then

  1. $\displaystyle \cos \theta =\frac{2\sqrt{6}}{15}$
  2. $\displaystyle \sin \theta =\frac{2\sqrt{6}}{15}$
  3. $\displaystyle \sin \theta =-\frac{11\sqrt{7}}{70}$
  4. $\displaystyle \cos \theta =-\frac{11\sqrt{7}}{70}$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

 $\theta$ is angle b/w $\xrightarrow [\gamma]{} =2\hat {  i}+j+k+(i+j+k)t$ and $\rightarrow.(3\hat { i }-4\hat { j }+5k)=q$

Angle b/w line and plane is given by 

$\sin\theta =\dfrac{4 _1a _2+b _1b _2+c _1c _2}{\sqrt{a _1^2+b _1^2+c _1^2}\sqrt{a _2^2+b _2^2+c _2^2}}$   

Where $(a _1,b _1,c _1)$ and $(a _2,b _2,c _2)$ are direction ratios of line and plane Respectively so here 

$a _1,b _1,c _1)=(1,1,1)$ and $(a _2,b _2,c _2)=(3,-4,5)$

So $\sin \theta=\dfrac{3-4+5}{\sqrt{1+1+1}\sqrt{9+16+25}}$

$\dfrac{4}{\sqrt{3}\sqrt{50}}=\dfrac{4}{\sqrt{3}5\sqrt{2}}=\dfrac{4}{\sqrt{6.5}}\times \dfrac{\sqrt{6}}{\sqrt{6}}=\dfrac{2\sqrt{6}}{5.3}=\dfrac{2\sqrt{6}}{15}$

so here $\sin\theta =\dfrac{2\sqrt{6}}{15} \Rightarrow \theta =\sin\dfrac{2\sqrt{6}}{15}$

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

The projection of line $\displaystyle\frac{x}{2}=\frac{y-1}{2}=\frac{z-1}{1}$ on a plane 'P' is $\displaystyle\frac{x}{1}=\frac{y-1}{1}=\frac{z-1}{-1}$. If the plane P passes through $(k, -2, 0)$, then k is greater than.

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

The projection of a line on a plane involves finding the plane that contains the original line and the projected line. By finding the normal to this plane and using the given point, the constant k can be determined.

Multiple choice maths introduction to three dimensional geometry coordinate axes and coordinate planes in 3d space coordinates in 3d introduction to 3d geometry

The foot of the perpendicular from the point $A(7, 14, 5)$ to the plane $2x+4y-z=2$ is?

  1. $(3, 1, 8)$
  2. $(1, 2, 8)$
  3. $(3, -3, 5)$
  4. $(5, -3, -4)$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Let N be the foot of the perpendicular drawn from the point $A(7, 14, 5)$ and perpendicular to the plane $2x+4y-z=2$

Then, the equation of the line PN is $\dfrac{x-7}{2}=\dfrac{y-14}{4}=\dfrac{z-5}{-1}=\lambda$ (say)
Let the coordinates of N be $N(2\lambda +7, 4\lambda +14, -\lambda +5)$
Since N lies on the plane $2x+4y-z=2$, so
$2(2\lambda +7)+4(4\lambda +14)-(-\lambda +5)=2$
$\Rightarrow 21\lambda =-63$
$\Rightarrow \lambda =-3$
$\therefore$ required foot of the perpendicular is
$N(-6+7, -12+14, 3+5)$, i.e., $N(1, 2, 8)$.

Multiple choice maths introduction to three dimensional geometry coordinate axes and coordinate planes in 3d space coordinates in 3d introduction to 3d geometry

An equation of sphere with centre at origin and radius $r$ can be represented as

  1. $x^2+y^2+z^2=r$
  2. $x^2+y^2+z^2=r^2$
  3. $x^2+y^2+z^2=2r^2$

  4. None of the above
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Sphere is locus of a point in 3D whose distance from a fixed point(center) is constant (radius)

$\Rightarrow \sqrt{(x-0)^2+(y-0)^2+(z-0)^2}=|r|$
$\Rightarrow x^2+y^2+z^2=r^2$, square both sides