Tag: applications of vector algebra

Questions Related to applications of vector algebra

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

  2. $\dfrac 27$

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

  4. $1$

Show answer
Correct Option: B
Explanation:

Given plane is $2x−3y+6z−11=0$


We know $\sin\theta=\dfrac {\vec b . \vec n}{|\vec b| |\vec n|}$


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

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

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

Hence, B is correct option

The angle between the line $\displaystyle x = y = z$ and the plane $\displaystyle 4x - 3y + 5z = 2$ is

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

  1. $\displaystyle \cos^{-1} \frac{\sqrt{6}}{5}$

  2. $\displaystyle \sin ^{-1} \frac{\sqrt{6}}{5}$

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

  4. $\displaystyle \sin ^{-1} \frac{1}{\sqrt{6}}$

Show answer
Correct Option: B
Explanation:

Direction ratios of $x=y=z$ are $(1,1,1)$....(1)

Let angle between the above line and $4x -3y+5z=2$ is $ \theta$.

$ \Rightarrow $ angle between the line and normal to the plane is $90^{o} - \theta$...(2)

Direction ratios of normal to the given plane are $(4, -3, 5)$....(3)

From (1), (2) and (3), $ \cos {(90^{o}- \theta)}=\dfrac{1 \times 4 + 1 \times (-3)+ 1 \times 5}{\sqrt{1^2+1^2+1^2}{ \sqrt {4^2+3^2+5^2}}}=\dfrac{6}{\sqrt{3} \times 5 \times \sqrt{2}}$

$ \Rightarrow \sin {\theta}= \dfrac{\sqrt{6}}{5} $

$\Rightarrow \theta = \sin^{-1}{\dfrac{\sqrt{6}}{5}}$

Given the line $\displaystyle L:\frac { x-1 }{ 3 } =\frac { y+1 }{ 2 } =\frac { z-3 }{ -1 } $ and the plane $\pi :x-2y=0$. Of the following assertion, 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 $\bot$ to $\pi$

  2. $L$ lies in $\pi$

  3. $L$ is parallel to $\pi$

  4. None of these

Show answer
Correct Option: B
Explanation:

Since $3\left( 1 \right) +2\left( -2 \right) +\left( -1 \right) \left( -1 \right) =3-4+1=0,$
$\therefore$ given line is $\bot $ to the normal to the plane i.e., given line is parallel to the given plane.
Also $\left( 1,-1,3 \right) $ lies on the plane $x-2y-z=0$, if $1-2\left( -1 \right) -3=0$ i.e., $1+2-3=0$
which is true 

$\therefore L$ lies in plane $\pi .$

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. $1$

Show answer
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

Show answer
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 } $