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 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

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

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 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

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

Which of the following lines lie on the plane $x+2y-z=0$?

  1. $x-1=y-1=1$
  2. $x-y+z=2x+y-z=0$
  3. $\vec r=2\hat i-\hat j+4\hat k+\lambda (3\hat i+\hat j+5\hat k)$

  4. None of these.
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

consider $P: x+2y-z=0$
direction ratio of normal of $P$ is $(1,2,-1)$

a) for $L _{1}: \dfrac {x-1}{1}=\dfrac {y}{-1}=\dfrac {z-5}{-1}$
Since, $(1,0,5)$ does not lie in $P$ 
Therefore, $L _{1}$ does not lie in $P$

b) for $L _{2}: x-y+z=2x+y-z=0$
direction ratio of $L _{2}$ is $(i-j+k) \times (2i+j-k)=3(j+k)$
since, $(i+2j-k).(3j+3k)=6-3=3 \neq =0$
Therefore, $L _{2}$ does not lie on $P$

c) for $L _{3}:\vec r=2\hat i-\hat j+4\hat k+\lambda (3\hat i+\hat j+5\hat k)$
Since, $(2,-1,4)$ does not lie in $P$ 
Therefore, $L _{3}$  does not lie in $P$

Ans: D

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 given planes $ax+by+cz+d=0$ and $ax+by+cz+d=0$ be mutually perpendicular, then 

  1. $\dfrac{a}{a}=\dfrac{b}{b}=\dfrac{c}{c}$
  2. $\dfrac{a}{a}+\dfrac{b}{b}+\dfrac{c}{c}=0$
  3. $aa+bb+cc+dd=0$
  4. $aa+bb+cc=0$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation
$ax+by+cz+d=0$ and ${ a }^{ 1 }x+{ b }^{ 1 }y+{ c }^{ 1 }z+{ d }^{ 1 }=0$ are mutually perpendicular then their respective norm also will be perpendicular too.
Hence by the perpendicular condition,
${ aa }^{ 1 }+{ bb }^{ 1 }+{ cc }^{ 1 }=0$
Where, $\left< a,b,c \right> $ are the direction ratio of the normal to the plane $ax+by+cz+d=0$ and $\left< { a }^{ 1 },{ b }^{ 1 },{ c }^{ 1 } \right> $ are the direction ratio of the normal to the plane ${ a }^{ 1 }x+{ b }^{ 1 }y+{ c }^{ 1 }z+{ d }^{ 1 }=0$
Correct option will be $(D)$
(But there's a formating error. In the question both plane equation are same, which contradicts the fact that the planes are perpendicular).
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 of intersection of the line $\dfrac{x-1}{3}=\dfrac{y+2}{4}=\dfrac{z-3}{-2}$ and the plane $2x-y+3z-1=0$, is

  1. $(-10, 10, 3)$
  2. $(10, 10, -3)$
  3. $(10, -10, 3)$
  4. $(10, -10, -3)$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

The line is (x-1)/3 = (y+2)/4 = (z-3)/-2 = t. So x=3t+1, y=4t-2, z=-2t+3. Substituting into 2x-y+3z-1=0 gives 2(3t+1) - (4t-2) + 3(-2t+3) - 1 = 0. 6t+2 - 4t+2 - 6t+9 - 1 = 0, so -4t + 12 = 0, t=3. Then x=10, y=10, z=-3.

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

Find the point where the line of intersection of the planes $x-2y+z=1$ and $x+2y-2z=5$ intersects the plane $3x+2y+z+6=0$.

  1. $P\left( 1,-2,-4 \right) $
  2. $P\left( 1,2,-4 \right) $
  3. $P\left( 1,-2,4 \right) $

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

${ P } _{ 1 }\equiv x-2y+z=1$

${ P } _{ 2 }\equiv x+2y-2z=5$
Let $Dr's$ of intersection of planes be $a,b,c$
$\Rightarrow a-2b+c=0$ and $a+2b-2c=0$
$\displaystyle\Rightarrow \frac { a }{ 4-2 } =\frac { b }{ 1+2 } =\frac { c }{ 2+2 } \Rightarrow \frac { a }{ 2 } =\frac { b }{ 3 } =\frac { c }{ 4 } $
Calculating a point which lies on both plane ${P} _{1}$ and ${P} _{2}$,
$x-2y+z=1\Rightarrow2y=x+z-1$ and $x+2y-2z=5\Rightarrow2y=-x+2z+5$
$\Rightarrow x+z-1=-x+2z+5\Rightarrow 2x=z+6$
Now, $z=0\Rightarrow x=3\Rightarrow y=1$
$\therefore$ equation of line passing through $\left( 3,1,0 \right) $ and has $dr's$  $2,3,4$ is
$\displaystyle\therefore \frac { x-3 }{ 2 } =\frac { y-1 }{ 3 } =\frac { z }{ 4 } =\lambda $(say)
$\therefore$ general point on it is $P\left( 2\lambda +3,3\lambda +1,4\lambda  \right) $
Now, solving with plane $2x+2y+z+6=0$
$\Rightarrow 2\left( 2\lambda +3 \right) +2\left( 3\lambda +1 \right) +4\lambda +6=0\ \Rightarrow 4\lambda +6+6\lambda +2+4\lambda +6=0\ \Rightarrow 14\lambda +14=0\Rightarrow \lambda =-1$
$\Rightarrow P\left( 1,-2,-4 \right) $ is the required point of intersection.

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

Let line L: $\displaystyle \frac{x-1}{2} = \frac{y - 1}{1} = \frac{z - 0}{4} $ & Plane P: $x + 2y - z = 3$
Then which of the following is true?

  1. Line is perpendicular to plane

  2. Line is neither parallel nor perpendicular to plane

  3. Plane contains the line

  4. Line and plane do not intersect
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Given line L: $\displaystyle \frac{x-1}{2} = \frac{y - 1}{1} = \frac{z - 0}{4} $

Plane P: $x + 2y - z = 3$
So from the below option c $i.e.$ plane contains the line 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 plane $x-2y+z-6=0$ and the line $\displaystyle\frac{x}{1}=\displaystyle\frac{y}{2}=\displaystyle\frac{z}{3}$ are related as.

  1. Parallel to the plane

  2. At right angle to the plane

  3. Lies in the plane

  4. Meets the plane obliquely
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Given line is $\displaystyle\frac{x}{1}=\frac{y}{2}=\frac{z}{3}$ any point on the line is $(t, 2t, 3t)$. It lies in the given plane, if $t-2(2t)+3t-6=0$
i.e., $0\cdot t=6$, which is not true for any real $t$. So, the line and plane do not meet. i.e., the line is parallel to the plane.

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 ax + by + cz = 1 meets the coordinate axes in A, B and C. The centroid of $\triangle ABC$ is

  1. $(3a, 3b, 3c)$
  2. $(\dfrac{a}{3}, \dfrac{b}{3}, \dfrac{c}{3})$
  3. $(\dfrac{3}{a}, \dfrac{3}{b}, \dfrac{3}{c})$
  4. $(\dfrac{1}{3a}, \dfrac{1}{3b}, \dfrac{1}{3c})$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation
 Coordinates of A,B,C are $(\frac{1}{a},0),(\frac{1}{b},0),(\frac{1}{c},0) $respectively
  centriod of $ \Delta ABC = (\frac{1}{3a},\frac{1}{3b},\frac{1}{3c})$
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 $\frac{x}{y}+\frac{y}{3}+\frac{z}{4}$ =1 cutes the axes in A,B,C, then the are of the $\Delta ABC$ is;

  1. $\sqrt{29}$
  2. $\sqrt{41}$
  3. $\sqrt{61}$

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

We have,

$\dfrac{x}{2}+\dfrac{y}{3}+\dfrac{z}{4}=1$

Compare this equation of plane,

$\dfrac{x}{a}+\dfrac{y}{b}+\dfrac{z}{c}=1$

Then,

$a=2,\,b=3,\,c=4$

Then,

The coordinate of X-axis is $=A\left( a,0,0 \right)$$=A\left( 2,0,0 \right)$

The coordinate of Y-axis is $=B\left( 0,b,0 \right)$$=A\left( 0,3,0 \right)$

The coordinate of X-axis is $=C\left( 0,0,c \right)$$=C\left( 0,0,2 \right)$

We know that, the area of  \[\Delta ABC\] is

$ Area\,of\,\Delta ABC=\dfrac{1}{2}\sqrt{{{a}^{2}}{{b}^{2}}+{{b}^{2}}{{c}^{2}}+{{c}^{2}}{{a}^{2}}} $

$ =\dfrac{1}{2}\sqrt{{{2}^{2}}\times {{3}^{2}}+{{3}^{2}}\times {{4}^{2}}+{{4}^{2}}\times {{2}^{2}}} $

$ =\dfrac{1}{2}\sqrt{36+144+64} $

$ =\dfrac{1}{2}\sqrt{180+64} $

$ =\dfrac{1}{2}\sqrt{244} $

$ =\dfrac{1}{2}\sqrt{2\times 2\times 61} $

$ =\sqrt{61} $

Hence, this is the answer.

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

Perpendicular is drawn from the point $(0,3,4)$ to the plane $2x -2y + z + (-10) = 0$, then co-ordinates of the foot of the L's are

  1. $\displaystyle \left ( \frac{8}{3},\frac{1}{3},\frac{16}{3}\right )$
  2. $\displaystyle \left ( -\frac{8}{3},\frac{1}{3},\frac{16}{3}\right )$
  3. $\displaystyle \left ( \frac{8}{3},-\frac{1}{3},\frac{16}{3}\right )$
  4. $\displaystyle \left ( \frac{8}{3},\frac{1}{3},-\frac{16}{3}\right )$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

We have point $(0,3,4)$ and we have direction's ratio also of line so we can write equation of line
$\dfrac{x-0}{2}=\dfrac{y-3}{-2}=\dfrac{z-4}{1}=k$
So consider a point which is lied on line $(2k,-2k+3,k+4)$
This point lies on plane also so it will satisfy the equation of plane:
$\Rightarrow 2(2k)-2(-2k+3)+k+4-10=0$
$\Rightarrow k=\dfrac{4}{3}$
Substitute the value of $k$ in point $\left (\dfrac{8}{3},\dfrac{1}{3},\dfrac{16}{3}\right)$