Mathematics

Three Dimensional Geometry Planes

198 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

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} $

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

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

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

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

The equation of plane passing through $(-1,0,-1)$ parallel to $xz$ plane is

  1. $y=-2$
  2. $y=0$
  3. $-x-z=0$

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

Given that the plane is parallel to $xz$ plane and the plane passes through $(-1,0,-1)$

Since the plane is parallel to $xz$ plane , the $y$ coordinate should be constant
Given that it passes through point $(-1,0,-1)$ , therefore the plane lies on $xz$ plane
Therefore the equation of plane is $y=0$
The correct options are $B$

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

The planes $2x-y+4z=5$ and $5x-2.5y+10z=6$ are

  1. Parallel

  2. Perpendicular

  3. Intersect
  4. intersect $x$ axis
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Planes are $2x-y+4z=5$ 

and $5x-2.5y+10z=6$
Multiply both sides by 2 to the second equation
$\Rightarrow 10x-5y+20=12$
Now divide both sides by $2$
$\Rightarrow 2x-y+4z=\dfrac{12}{5}$

Clearly both planes are parallel 

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

In a three-dimensional space, the equation $3x - 4y = 0$ represents.

  1. A plane containing $Z-axis$
  2. A plane containing $X-axis$
  3. A plane containing $Y-axis$
  4. Passing through $(0, 0)$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

If we consider the z-part also, then the equation is $3x-4y+0z=0$

Which means on putting any value of z equation will have no effect as $0\times z=0$
$\therefore$ it will be a plane containing $Z-axis$.($\because$ it will pass through all points z if it satisfy condition for$ (x,y) $)
(D) is not right because its plane and not a line so it will pass through $(0,0,0)$ not $(0,0)$.
Hence, $(A)$


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

Which of the following is true for a plane?

  1. A locus is called a plane if the line joining any two arbitrary points on the locus is also a part of the locus.
  2. Value of $y$ in a $zx$ plane is non-zero.
  3. Value of $z$ in a $xy$ plane is zero.

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

Option A and C are correct 

A locus is called a plane if the line joining any two arbitrary points on the locus is also a part of the locus. and also Value of z in a xy plane is zero.

Multiple choice maths vectors: lines in two and three dimensions distance from a point to line perpendicular distance of a point from a plane the distance from a point to a line

Perpendicular distance between the plane $  2 x-y+2 z=1  $ and origin is 

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

The perpendicular distance of a plane and origin is give as 

$\dfrac{|d|}{\sqrt {a^2+b^2+c^2}}\\dfrac{1}{\sqrt {2^2+1^2+2^2}}\\dfrac 1{\sqrt 9}=\dfrac 13$

Multiple choice maths vectors: lines in two and three dimensions distance from a point to line perpendicular distance of a point from a plane the distance from a point to a line

A line is perpendicular to the plane $x+2y+2z=0$ and passes through $(0, 1, 0)$. The perpendicular distance of this line from the origin is

  1. $\displaystyle \frac {\sqrt 5}{3}$
  2. $\displaystyle \frac {\sqrt 7}{3}$
  3. $\displaystyle \frac {2}{3}$
  4. $3$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Since the line is perpendicular to the plane $x+2y+2z=0$, so directon of line will be $(1,2,2)$ and it passes through $(0,1,0)$,
Therefore equation of line is $\dfrac{x}{1}=\dfrac{y-1}{2}=\dfrac{z}{2}$
So general point on the line is $(r,2r+1,2r)$
Now direction ratio of line passing through this point and origin which perpendicular to given line is $(r,2r+1,2r)$
Since they are perpendicular, so dot product will be $0$ .
$r+2\times (2r+1)+2\times 2r{=}0$
$r+4r+2+4r{=}0$
$\therefore r{=}$$\dfrac{-2}{9}$
So points on line which perpendicular from origin is $(-2/9,5/9,-4/9)$
Now applying distance formula,
${=}$ $\sqrt{{(-2/9)}^{2}+{(5/9)}^{2}+{(-4/9)}^{2}}$
${=}$ $\sqrt{4/81+25/81+16/81}$
${=}$ $\dfrac{\sqrt{5}}{3}$

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
$(2(1)+3(2)-2(3)-k)(2(2)+3(-1)-2(0)-k)<0$
$(2-k)(1-k)<0\implies (k-1)(k-2)<0$
$\implies 1<k<2$