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 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 joining the points $(2,0,2)$ and $(3,-1,3)$ and the plane $x-y+z=1$ is

  1. $(3,2,0)$
  2. $(-1,1,3)$
  3. $(1,1,1)$
  4. $(4,2,-1)$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

The equation of the line passing through (2,0,2) and (3,-1,3) can be written in parametric form. Substituting these parametric coordinates into the plane equation x - y + z = 1 allows us to solve for the parameter, which then gives the exact point of intersection as (1, 1, 1).

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

The condition that the line $\displaystyle \frac{x-{\alpha }'}{l}=\frac{y -{\beta   }'}{m}=\frac{z-{\gamma  }'}{n}$ in the plane $Ax + By + Cz + D = 0$ is

  1. $A{\alpha }'+B{\beta }'+C{\gamma }'+D=0\ and\ Al+Bm+Cn\neq 0$
  2. $A{\alpha }'+B{\beta }'+C{\gamma }'+D\neq0\ and\ Al+Bm+Cn= 0$
  3. $A{\alpha }'+B{\beta }'+C{\gamma }'+D=0\ and\ Al+Bm+Cn= 0$
  4. $A{\alpha }'+B{\beta }'+C{\gamma }'=0\ and\ Al+Bm+Cn= 0$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

The line $\dfrac{x- \alpha^1}{l}=\dfrac{y-\beta^1}{m}=\dfrac{z-\gamma^1}{n}$ in the plane $Ax+By+Cz+D=0$

then multiplication sum id corresponding direction ratio will be zero so here $Al+Bm+Cn=0$
 and  one thing more line passes through $(\alpha^1,\beta^1,\gamma^1)$
So, plane Also passe through $(\alpha^1,\beta^1,\gamma^1)$
hence $A\alpha^1+B\beta^1+C\gamma^1+D=0$

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

The 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

The line joining the points $\left (2, -3, 1  \right )$ and $\left (3, -4, -5  \right )$ cuts a coordinate plane at the point.

  1. $\left (0, -1, 13 \right )$
  2. $\left ( 0, 0, 1 \right )$
  3. $\left ( -1, 0, 19 \right )$
  4. $\left ( 8, -9, 0 \right )$
Reveal answer Fill a bubble to check yourself
A,C Correct answer
Explanation

Equation of the line is $\displaystyle \frac { x-2 }{ 2-3 } =\frac { y+3 }{ -3+4 } =\frac { z-1 }{ 1+5 } $ or $\displaystyle \frac { x-2 }{ -1 } =\frac { y+3 }{ 1 } =\frac { z-1 }{ 6 } $
Any point on the line $\left( -r+2,r-3,6r+1 \right) $ which cuts the $yz$-plane at the point where $-r+2=0$ or $r=2$
and the point of intersection is $\left( 0,-1,13 \right) $ 
Similarly it cuts the $zx$-plane at $\left( -1,0,19 \right) $ and $xy$ plane at $\displaystyle \left( \frac { 13 }{ 6 } ,\frac { -19 }{ 6 } ,0 \right) $.

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 ratio in which the plane $\vec{r}.(\hat{i}-2\hat{j}+3\hat{k})=17$ 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. $1 : 5$
  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 $x-2y+z-6=0$ 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 $\left (\dfrac{8}{3},\dfrac{1}{3},\dfrac{16}{3}\right)$

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

The line $x -2y + 4z + 4 = 0$, $x + y + z - 8 = 0$ intersects the plane $x - y + 2z + 1 = 0$ at the point

  1. $\left ( 3, 2, 3 \right )$
  2. $\left ( 5, 2, 1 \right )$
  3. $\left ( 2, 5, 1 \right )$
  4. $\left ( 3, 4, 1 \right )$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Given  lines
 $x -2y + 4z + 4 = 0$    ....(1)
 $x + y + z - 8 = 0$       .....(2)
Subtracting (2) from (1), we get
$\Rightarrow y-z=4$        .....(3)
Given equation of plane $x-y+2z+1=0$
Since, the line intersects the plane, so using (3), we get 
$\Rightarrow x+z=3$      ......(4)
Hence, $y=5$, $z=1 $ and $x=2$
Hence, option C 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

$L: \displaystyle \frac{x\, +\, 1}{2}= \frac{y\, +\, 1}{3}= \frac{z\, +\, 1}{4}$
$\pi _{1}:\, x\, +\, 2y\, +\, 3z= 14,\, \pi _{2}:\, 2x\, -\, y\, +\, 3z= 27$

If the line $L$ meets the plane $\pi _{1}$ in the point $P$, and the coordinates of $P$ are $\left ( \alpha ,\, \beta ,\, \gamma  \right )$, then $\alpha ^{2}\, +\, \beta ^{2}\, +\, \gamma ^{2}$ is equal to

  1. $3$
  2. $14$
  3. $28$
  4. $29$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation
$L=\dfrac{x-1}{2}=\dfrac{y+1}{3}=\dfrac{z+1}{4}$     ........(i) and plane
$\pi \Rightarrow x+2y+3z=14$        .......(ii)
Given that $L$ meets plane $\pi _1$ (means intersection points) so from $eq^n$ (i)
$\Rightarrow \dfrac{x+1}{2}=\dfrac{y+1}{3}=\dfrac{x+1}{4}=K$
$\Rightarrow x= 2k-1\,\, and\,\, y=3k-1,z=4k-1  $
putting this values in ..... (ii)
So $(2k-1)+2(3k-1)+3(4k-1)=14$
$\Rightarrow 2k-1+6k-2+12k-3=14$
$20k-6=14$
$\Rightarrow 20k=20 \rightarrow k=1$
so points $x=1,y=2,z=3$ inform of $\alpha ,\beta,\gamma= \alpha =1,\beta=2,\gamma =3$
so $\alpha^2+\beta^2+\gamma^2=1^2+2^2+3^2$
$=14$
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 $2x + 2y + z + 6 = 0$

  1. $(1, -2, -4)$
  2. $(0,0,-6)$
  3. $(1,0,-8)$
  4. $(-1,-1,-2)$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The line of intersection of two planes is found by solving the system of equations. Testing the point (1, -2, -4) in the first plane: 1 - 2(-2) + (-4) = 1 + 4 - 4 = 1 (Correct). In the second plane: 1 + 2(-2) - 2(-4) = 1 - 4 + 8 = 5 (Correct). In the third plane: 2(1) + 2(-2) + (-4) + 6 = 2 - 4 - 4 + 6 = 0 (Correct).