Tag: three dimensional geometry - ii

Questions Related to three dimensional geometry - ii

The angle between the line $\overrightarrow { r } =\left( -\hat { i } +3\hat { j } +3\hat { k }  \right) +t\left( 2\hat { i } +3\hat { j } +6\hat { k }  \right) $ and the plane $\overrightarrow { r } .\left( -\hat { i } +\hat { j } +\hat { k }  \right) $ is
angle between a line and a plane three dimensional geometry - ii product of vectors applications of vector algebra maths

  1. $\displaystyle\sin ^{ -1 }{ \dfrac { 1 }{ \sqrt { 3 }  }  } $

  2. $\displaystyle\sin ^{ -1 }{ \dfrac { 1 }{ \sqrt { 2 }  }  } $

  3. $\displaystyle\sin ^{ -1 }{ \dfrac { 2 }{ \sqrt { 3 }  }  } $

  4. $\displaystyle\sin ^{ -1 }{ \dfrac { 3 }{ \sqrt { 2 }  }  } $

Show answer
Correct Option: A
Explanation:
Angle between the line and the plane is given by
$\displaystyle \sin { \theta  } =\dfrac { \left( 2\hat { i } +3\hat { j } +6\hat { k }  \right) .\left( -\hat { i } +\hat { j } +\hat { k }  \right)  }{ \sqrt { 4+9+36 } \sqrt { 1+1+1 }  } $
$\displaystyle =\dfrac { -2+3+6 }{ 7\times \sqrt { 3 }  } =\dfrac { 7 }{ 7\sqrt { 3 }  } =\dfrac { 1 }{ 3 } $
$\displaystyle \Rightarrow \theta =\sin ^{ -1 }{ \left( \dfrac { 1 }{ \sqrt { 3 }  }  \right)  } $

If the plane $2x - 3y + 6z - 11 = 0$ makes an angle $sin^{-1}(k)$ with x-axis, then k is equal to

angle between a line and a plane three dimensional geometry - ii product of vectors applications of vector algebra maths

  1. $\displaystyle \frac{\sqrt{3}}{2}$

  2. $\displaystyle \frac{2}{7}$

  3. $\displaystyle \frac{\sqrt{2}}{7}$

  4. $1$

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Correct Option: B
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.

$\therefore  sin\theta=\dfrac{(2\vec i-3\vec j +6\vec k).\vec i}{\sqrt {2^2+(-3)^2+6^2.}\sqrt12}$

$\therefore \dfrac{2}{\sqrt49}=\dfrac{2}{7}$.

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?

angle between a line and a plane three dimensional geometry - ii product of vectors applications of vector algebra maths

  1. L is $\perp$ to II

  2. L lies in II

  3. L is parallel to II

  4. None of these

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Correct Option: C
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.

Plane $2x+3y+6z=15=0$ makes angle of measure ________ with Y-axis.

angle between a line and a plane three dimensional geometry - ii product of vectors applications of vector algebra maths

  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. $\cos^{-1}\left(\dfrac{3}{7}\right)$

Show answer
Correct Option: A
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 $\alpha =\sin^{-1}\left(\dfrac{3}{7}\right)$.

If the angle bwteen a line $x=\dfrac{y-1}{2}=\dfrac{z-3}{\lambda}$ and normal to the plane $x+2y+3z=4$ is $\cos^{-1}{\sqrt{\dfrac{5}{14}}}$, then possible value(s) of $\lambda$ is/are

angle between a line and a plane three dimensional geometry - ii product of vectors applications of vector algebra maths

  1. $\dfrac{5}{2}$

  2. $\dfrac{2}{5}$

  3. 00

  4. $\dfrac{2}{3}$

Show answer
Correct Option: A
Explanation:

Consider the given line equation $x=\dfrac{y-1}{2}=\dfrac{z-3}{\lambda }$ and plane equation$x+2y+3z=4$.

Let $\theta $ be the angle between the line and normal to plane converting the given equations into normal form, we have

  $ \overrightarrow{r}=0.\widehat{i}+\widehat{j}+3\widehat{k}+\beta \left( \widehat{i}+2\widehat{j}+\lambda \widehat{k} \right) $

 $ \overrightarrow{r}=\widehat{i}+2\widehat{j}+3.\widehat{k}=3 $

Now,

  $ \overrightarrow{b}=\left( \widehat{i}+2\widehat{j}+\lambda \widehat{k} \right) $

 $ \overrightarrow{n}=\widehat{i}+2\widehat{j}+3.\widehat{k}=3 $

We know that,

  $ \cos \theta =\left| \dfrac{\widehat{b}.\widehat{n}}{\left| \widehat{b} \right|\left| \widehat{n} \right|} \right| $

 $ =\left| \dfrac{\left( \widehat{i}+2\widehat{j}+\lambda \widehat{k} \right).\left( \widehat{i}+2\widehat{j}+3.\widehat{k} \right)}{\left| \widehat{i}+2\widehat{j}+\lambda \widehat{k} \right|\left| \widehat{i}+2\widehat{j}+3.\widehat{k} \right|} \right| $

 $ \cos \theta =\left| \dfrac{1+4+3\lambda }{\sqrt{{{1}^{2}}+{{2}^{2}}+{{\lambda }^{2}}}\sqrt{{{1}^{2}}+{{2}^{2}}+{{3}^{2}}}} \right|=\left| \dfrac{5+\lambda }{\sqrt{5+{{\lambda }^{2}}}\sqrt{14}} \right| $

But given that $\theta ={{\cos }^{-1}}\left( \sqrt{\dfrac{5}{14}} \right)$ ,so

  $ \cos {{\cos }^{-1}}\sqrt{\dfrac{5}{14}}=\left| \dfrac{5+\lambda }{\sqrt{5+{{\lambda }^{2}}}\sqrt{14}} \right| $

 $ \sqrt{\dfrac{5}{14}}=\left| \dfrac{5+\lambda }{\sqrt{5+{{\lambda }^{2}}}\sqrt{14}} \right| $

Taking square both sides ,we get

  $ \dfrac{5}{14}={{\left| \dfrac{5+\lambda }{\sqrt{5+{{\lambda }^{2}}}\sqrt{14}} \right|}^{2}}=\dfrac{{{\left( 5+\lambda  \right)}^{2}}}{\left( 5+{{\lambda }^{2}} \right)\left( 14 \right)} $

 $ \dfrac{{{\left( 5+\lambda  \right)}^{2}}}{\left( 5+{{\lambda }^{2}} \right)}=5 $

 $ 25+{{\lambda }^{2}}+10\lambda =25+5{{\lambda }^{2}} $

 $ 4{{\lambda }^{2}}-10\lambda =0 $

 $ 2\lambda \left( 2\lambda -5 \right)=0 $

 $ \lambda \left( 2\lambda -5 \right)=0 $

 $ \lambda =0,\lambda =\dfrac{5}{2} $

Ignore $\lambda =0$ as it is in denominator. Therefore,

$\lambda =\dfrac{5}{2}$

This is the answer .

If the angle between the line $x=\dfrac { y-1 }{ 2 } =\dfrac { z-3 }{ \lambda}$ and the plane $x+2y+3z=4$ is $\cos ^{ -1 }{ \sqrt { \dfrac { 5 }{ 14 }  }   },$ then $\lambda$ equals:

angle between a line and a plane three dimensional geometry - ii product of vectors applications of vector algebra maths

  1. $\dfrac { 2 }{ 5 } $

  2. $\dfrac { 5 }{ 3 } $

  3. $\dfrac { 2 }{ 3 } $

  4. $\dfrac { 3 }{ 2 } $

Show answer
Correct Option: C
Explanation:
Line : $\dfrac{x}{1} = \dfrac{y - 1}{2} = \dfrac{z - 3}{\lambda}$
and the plane $x + 2y + 3z = 4$
$\therefore \sin\theta = \dfrac{1 + 4 + 3\lambda}{\sqrt{1 + 4 + \lambda^2}\sqrt{1 + 4 + 9}}$
$= \dfrac{5 + 3\lambda}{\sqrt{5 + \lambda^2}\sqrt{14}}$
$cos^{-1} \sqrt{\dfrac{5}{14}} = sin^{-1}\dfrac{3}{\sqrt{14}}$
$\Rightarrow \dfrac{3}{\sqrt{14}} = \dfrac{5 + 3\lambda}{\sqrt{5 + \lambda^2}
\sqrt{14}}$
$\Rightarrow 9(5 + \lambda^2) = (5 + 3\lambda)^2$
$\Rightarrow 45 + 9\lambda^2 = 25 + 9\lambda^2 + 30\lambda$
$\Rightarrow \lambda = \dfrac{2}{3}$

If the angle between the line $x=\cfrac{y-1}{2}=\cfrac{z-3}{\lambda}$ and the plane $x+2y+3z=4$ is $\cos ^{ -1 }{ \left( \sqrt { \cfrac { 5 }{ 14 }  }  \right)  } $, then $\lambda$ equals:

angle between a line and a plane three dimensional geometry - ii product of vectors applications of vector algebra maths

  1. $2/5$

  2. $5/3$

  3. $2/3$

  4. $3/2$

Show answer
Correct Option: C
Explanation:
Angle between line and normal to plane would be $\cfrac { \pi  }{ 2 } -\cos ^{ -1 }{ \sqrt { \cfrac { 5 }{ 14 }  }  } =\sin ^{ -1 }{ \sqrt { \cfrac { 5 }{ 14 }  }  } =\cos ^{ -1 }{ \cfrac { 3 }{ \sqrt { 14 }  }  } $
$\Rightarrow \cfrac { \left( \hat { i } +2\hat { j } +\lambda \hat { k }  \right) .\left( \hat { i } +2\hat { j } +3\hat { k }  \right)  }{ \left| \hat { i } +2\hat { j } +\lambda \hat { k }  \right| \left| \hat { i } +2\hat { j } +3\hat { k }  \right|  } =\cfrac { 3 }{ \sqrt { 14 }  } $
$\Rightarrow \cfrac { 1+4+3\lambda  }{ \sqrt { 5+{ \lambda  }^{ 2 } } \sqrt { 1+4+9 }  } =\cfrac { 3 }{ \sqrt { 14 }  } $
$\Rightarrow 3\lambda +5=3\sqrt { 5+{ \lambda  }^{ 2 } } \Rightarrow { \left( 3\lambda +5 \right)  }^{ 2 }=9\left( { \lambda  }^{ 2 }+5 \right) $
$\Rightarrow 9{ \lambda  }^{ 2 }+25+30\lambda =9{ \lambda  }^{ 2 }+45\Rightarrow 30\lambda =20\Rightarrow \lambda =\cfrac { 2 }{ 3 } $

If the angle between the line $x=\cfrac{y-1}{2}=\cfrac{z-3}{\lambda}$ and the plane $x+2y+3z=4$ is $\cos ^{ -1 }{ \left( \sqrt { \cfrac { 5 }{ 14 }  }  \right)  } $, then $\lambda$ equals

angle between a line and a plane three dimensional geometry - ii product of vectors applications of vector algebra maths

  1. $\cfrac{15}{2}$

  2. $\cfrac{3}{2}$

  3. $\cfrac{2}{5}$

  4. $\cfrac{5}{3}$

Show answer
Correct Option: A
Explanation:

equation of line $\dfrac{x-0}{1}=\dfrac{y-1}{2}=\dfrac{2-3}{\lambda}$ where $a _1=1, a _2=2, a _3=\lambda$

equation of plane $n+2y+32=4$
where $b _1=1, b _2=2, b _3=3$
angle between Line and plane
$\cos\theta=\dfrac{|a _1b _1+a _2b _2+a _3b _3|}{\sqrt{a _1^2+a _2^2+a _3^2}.\sqrt{b _1^2+b _2^2+b _3^2}}$
$\cos \theta=\dfrac{|1.1+2.2+3.\lambda|}{\sqrt{1+4+\lambda^2}.\sqrt{1+4+9}}$
$\cos \theta =\dfrac{5+3\lambda}{\sqrt{5+\lambda^2}.\sqrt{14}}\quad ---(1)$
given $\theta =\cos^{-1}\left(\sqrt{\dfrac{5}{14}}\right)$
$\cos \theta =\left(\sqrt{\dfrac{5}{14}}\right)\quad ----(2)$
By eqn $(1)$ & $(2)$
$\dfrac{5+3\lambda }{\sqrt{5+\lambda^2}\sqrt{14}}=\sqrt{\dfrac{5}{14}}$
$5+3\lambda=\sqrt{5}.\sqrt{5+\lambda^2}$
$(5+3\lambda)^2=5.(5+\lambda^2)$
$25+9\lambda^2+30\lambda =25+5\lambda^2$
$4\lambda^2+30\lambda =0$
$\lambda (2\lambda +15)=0\Rightarrow \lambda =0, \dfrac{15}{2}$
but $\lambda \neq 0$
So $\lambda =\dfrac{15}{2}$ Ans

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) $\displaystyle x-y+z-6=0$

angle between a line and a plane three dimensional geometry - ii product of vectors applications of vector algebra maths

  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

Show answer
Correct Option: C
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.

If the angle $\theta $ between the line $\displaystyle \frac{x+1}{1}=\frac{y-1}{2}=\frac{z-2}{2}$ and the plane $2x-y+\sqrt{\lambda} z+4=0$ is such that $\displaystyle \sin \theta =\frac{1}{3}$, then value of $\lambda $ is

angle between a line and a plane three dimensional geometry - ii product of vectors applications of vector algebra maths

  1. $\displaystyle -\frac{3}{5}$

  2. $\displaystyle \frac{5}{3}$

  3. $\displaystyle -\frac{4}{3}$

  4. $\displaystyle \frac{3}{4}$

Show answer
Correct Option: B
Explanation:

Angle between the line and plane is same as the angle between the line and normal to the plane
$\displaystyle \therefore \cos \left ( 90^{0}-\theta  \right )=\frac{a _{1}a _{2}+b _{1}b _{2}+c _{1}c _{2}}{\sqrt{a _{1}^{2}+b _{1}^{2}+c _{1}^{2}}\sqrt{a _{2}^{2}+b _{2}^{2}+c _{2}^{2}}}$
$\displaystyle \therefore\frac{1}{3}=\frac{\left ( 1\times 2+2\times \left ( -1 \right )+2\sqrt{\lambda } \right )}{\sqrt{1^{2}+2^{2}+2^{2}}\sqrt{2^{2}+1^{2}+\lambda }} $

$\therefore  \lambda =\dfrac{5}{3}$