Tag: three dimensional geometry

Questions Related to three dimensional geometry

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

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

  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. $(\dfrac{1}{3a}, \dfrac{1}{3b}, \dfrac{1}{3c})$

Show answer
Correct Option: D
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})$

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;

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

  1. $\sqrt{29}$

  2. $\sqrt{41}$

  3. $\sqrt{61}$

  4. None of these

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Correct Option: C
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.

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

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

  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. $\displaystyle \left ( \frac{8}{3},\frac{1}{3},-\frac{16}{3}\right )$

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Correct Option: A
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 $\left (\dfrac{8}{3},\dfrac{1}{3},\dfrac{16}{3}\right)$

Let the line $\displaystyle \frac{x-2}{3}= \frac{y-1}{-5}= \frac{z+2}{2}$ lie in the plane $x+3y-\alpha z+\beta = 0$. Then $\left ( \alpha ,\beta  \right )$ equals :

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

  1. $\left ( -6,7 \right )$

  2. $\left ( 5,-15 \right )$

  3. $\left ( -5,5 \right )$

  4. $\left ( 6,-17 \right )$

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Correct Option: A
Explanation:

The line is $\displaystyle \frac{x-2}{3}=\frac{y-1}{-5}=\frac{z+2}2{}$


The direction ratios of the line are $(3,-5,2)$

As the line is in the plane $x+3y-az+ \beta =0$,

We have $\left ( 3 \right )\left ( 1 \right )+\left ( -5 \right )\left ( 3 \right )+2\left ( -\alpha  \right )=0$

$\Rightarrow-12-2 \alpha =0$

$ \therefore \alpha = -6$

Again $(2,1,-2)$ lies on the plane

$\Rightarrow 2+3+2 \alpha + \beta =0$

$\Rightarrow \beta = -2 \alpha -5=12-5=7$

Hence, $\left ( \alpha ,\beta  \right )$ is $\left ( -6,7 \right )$

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

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

  1. $\left ( 3, 2, 3 \right )$

  2. $\left ( 5, 2, 1 \right )$

  3. $\left ( 2, 5, 1 \right )$

  4. $\left ( 3, 4, 1 \right )$

Show answer
Correct Option: C
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.

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

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

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

  1. $3$

  2. $14$

  3. $28$

  4. $29$

Show answer
Correct Option: B
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 $\alpha^2+\beta^2+\gamma^2=1^2+2^2+3^2$
$=14$

A line with positive direction cosines passes through the point $\displaystyle P\left ( 2,-1,2 \right )$ and makes equal angles with the coordinates axis. The line meet the plane $\displaystyle 2x+y+z=9$ at ponit $Q$.
The length of the line segment $PQ$ equals.

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

  1. $1$

  2. $\displaystyle \sqrt{2}$

  3. $\displaystyle \sqrt{3}$

  4. $2$

Show answer
Correct Option: C
Explanation:

Equation of a line through the point $\displaystyle P\left ( 2,-1,2 \right )$, equally inclined to the axes is given by,
$\displaystyle \frac{x-2}{1}=\frac{y+1}{1}=\frac{z-2}{1}=k$ (say)
Any point on the line is $Q ( k+2,k-1, k+2  )$ which lies on the plane $\displaystyle 2x+y+z=9$
 $\Rightarrow \displaystyle 2\left ( k+2 \right )+k-1+k+2=9 \Rightarrow k=1$,
For this values of $k$, the coordinates of $Q$ are $\displaystyle \left ( 3,0,3 \right )$.
So, $\displaystyle PQ=\sqrt{\left ( 3-2 \right)^2+\left ( 0+1 \right )^2+\left ( 3-2 \right )^2}=\sqrt{3}$

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

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

  1. $(1, -2, -4)$

  2. $(0,0,-6)$

  3. $(1,0,-8)$

  4. $(-1,-1,-2)$

Show answer
Correct Option: A

The line passing through the points $(5, 1,  a)$ and $(3, b, 1)$ crosses the $yz$-plane at the point $\left (0,\dfrac{17}{2},\dfrac{-13}{2}\right)$. Then,

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

  1. $a = 2, b = 8$

  2. $a = 4, b = 6$

  3. $a = 6, b = 4$

  4. $a = 8, b = 2$

Show answer
Correct Option: C
Explanation:

Equation of line passing through $(5, 1, a)$ and $(3, b, 1)$ is
$\dfrac{x-3}{5-3}= \dfrac{y-b}{1-b}= \dfrac{z-1}{a-1}$   ...(i)


Point $\left ( 0, \dfrac{17}{2}, -\dfrac{13}{2} \right )$ satisfies equation (i), we get

$-\dfrac{3}{2} = \dfrac{\dfrac{17}{2} -b}{1-b} = \dfrac{-\dfrac{13}{2}-1}{a-1}$

$\Rightarrow  a-1 = \dfrac{\left ( -\dfrac{15}{2} \right )}{\left ( -\dfrac{3}{2} \right )} = 5$
$\Rightarrow  a = 6$


Also,  $-3\left ( 1 - b \right )= 2 \left ( \dfrac{17}{2} - b\right )$

$\Rightarrow  3b - 3 = 17 - 2b$

$\Rightarrow  5b = 20   $

$  \Rightarrow  b = 4$

The points with position vectors $60\hat{i}+3\hat{j}$, $40\hat{i}-8\hat{j}$, $a\hat{i}-52\hat{j}$  are collinear if

direction cosines and direction ratios three dimensional geometry maths

  1. $a=-40$

  2. $a=40$

  3. $a=20$

  4. $None\ of\ these$

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Correct Option: A
Explanation:

suppose ${60i + 3j}$ , ${40i - 8j}$ and ${ai - 52j}$ is the three position of vector $A,B,C$


$\begin{array}{l} \overrightarrow { AB } =\left( { 40i-8j } \right) -\left( { 60i+3j } \right)  \ \overrightarrow { AB } =-20i-11j \ \overrightarrow { BC } =\left( { ai-52j } \right) -\left( { 40i-8j } \right)  \ \overrightarrow { BC } =\left( { a-40 } \right) i-44j \ \left( { a-40 } \right) i-44j=m\left( { -20i-11j } \right)  \ \left( { a-40 } \right) i-44j=-20im-11jm \ -44=-11m \ m=\frac { { -44 } }{ { -11 } }  \ m=4 \ a-40=-20m \ a-40=-20\left( 4 \right)  \ a=-80+40 \ a=-40 \end{array}$