Tag: three dimensional geometry - ii

Questions Related to three dimensional geometry - ii

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

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. $29$

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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$