Tag: three dimensional geometry - ii

Questions Related to three dimensional geometry - ii

If $\overline {c}$ is perpendicular to $\overline {a}$ and $\overline {b}$ , $\left| \overline {a} \right| =3,\ \left| \overline {b} \right|=4,\ \left| \overline {c} \right|=5$ and the angle between $\overline {a}$ and $\overline {b}$ is $\dfrac{\pi}{6}$ then $[\overline {a}\ \ \ \overline {b}\ \ \ \overline {c}]=$

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

  1. $30\sqrt{3}$

  2. $30$

  3. $15$

  4. $15\sqrt{3}$

Show answer
Correct Option: B
Explanation:

We have,

$\begin{matrix} \left[ { \overline { a } \, \, \overline { b } \, \, \overline { c }  } \right] =\overline { c } \times \left( { \overline { a } \times \overline { b }  } \right)  \ =\overline { c } \times \left( { \overline { a } \times \overline { b }  } \right) \cos { 0^{ 0 } }  \ =5\times 3\times 4\times \sin  \frac { \pi  }{ 6 }  \  \end{matrix}$
$ = 5 \times 3 \times 4 \times \frac{1}{2}$
$ = 30$
Then,
Option $B$ is correct answer.

An angle between the plane , $x+y+z=5$ and the line of intersection of the planes, $3x+4y+x-1=0$ and $5x+8y+2z+14=0$

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

  1. $\sin^{-1}(\sqrt{3/17})$

  2. $\cos^{-1}(\sqrt{3/17})$

  3. $\cos^{-1}(3/\sqrt{17})$

  4. $\sin^{-1}(3/\sqrt{17})$

Show answer
Correct Option: A
Explanation:

We have

$\overrightarrow {{\pi _1}} :x + y + z = 5$
$\overrightarrow {{r _1}} :3x + 4y + z - 1 = 0$
$\overrightarrow {{r _2}} :5x + 8y + 2z + 14 = 0$
Line of intersection of planes $\parallel \,\,to\,\,\overrightarrow {{r _1}}  \times \overrightarrow {{r _1}} $
By the helps of determinate
$\left| { \begin{array} { *{ 20 }{ c } }{ \widehat { i }  } & { \widehat { j }  } & { \widehat { k }  } \ 3 & 4 & 1 \ 4 & 8 & 2 \end{array} } \right| $
$ = \widehat i\left( 0 \right) - \widehat j\left( {6 - 5} \right) + \widehat k\left( {24 - 20} \right)$
$ =  - \widehat j + 4\widehat k$
Now,
$\sin \theta  = \frac{{ - 1 + 4}}{{\sqrt 3 \sqrt {17} }} = \sqrt {\frac{3}{{17}}} $
$\therefore \theta  = {\sin ^{ - 1}}\sqrt {\frac{3}{{17}}} $
Hence the option $A$ is the correct answer.

Read the following statement carefully and identify the true statement
(a) Two lines parallel to a third line are parallel
(b) Two lines perpendicular to a third line are parallel
(c) Two lines parallel to a plane are parallel
(d) Two lines perpendicular to a plane are parallel
(e) Two lines either intersect or are parallel

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

  1. a & b

  2. a & d

  3. d & e

  4. a

Show answer
Correct Option: B,D
Explanation:
$(a)$ Two lines parallel to a third line are parallel-True

$(b)$ Two lines perpendicular to a third line are parallel-False. 

The $x-$ and $y-$axes are both perpendicular to the $z-$axis, yet the $x-$ and $y-$axes are not parallel.

$(c)$ Two lines parallel to a plane are parallel-False. 

The $x-$ and $y-$axes are not parallel, yet they are both parallel to the plane $z = 1.$

$(d)$ Two lines perpendicular to a plane are parallel-True

$(e)$ Two lines either intersect or are parallel-False. They can be skew.
Only the statements $(a)$ and $(d)$ are true.

The line $\dfrac {x - 2}{3} = \dfrac {y - 3}{4} = \dfrac {z - 4}{5}$ is parallel to the plane.

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

  1. $3x + 4y + 5z = 7$

  2. $2x + y - 2z=0$

  3. $x + y - z = 2$

  4. $2x + 3y$

Show answer
Correct Option: B
Explanation:

Direction ratios of given line are $3,4,5$ direction ratios of the perpendicular to the  plane $2x+y-2z=0$ are $2,1,-2$


Now
$\begin{array}{l} 2\times 3+1\times 4+\left( { -2 } \right) \times 5 \ =6+4-10 \ =0 \end{array}$

perpendicular to the plane is perpendicular to the given line
so, the plane $2x+y-2z$ is parallel to the given line.

Hence, the correct option is $B$

If the projection of point P$(\vec{p})$ on the plane $\vec{r}\cdot \vec{n}=q$ is the points $S(\vec{s})$, then.

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

  1. $\vec{s}=\dfrac{(q-\vec{p}\cdot \vec{n})\vec{n}}{|\vec{n}|^2}$

  2. $\vec{s}=\vec{p}+\dfrac{(q-\vec{p}\cdot \vec{n})\vec{n}}{|\vec{n}|^2}$

  3. $\vec{s}=\vec{p}-\dfrac{(\vec{p}\cdot \vec{n})\vec{n}}{|\vec{n}|^2}$

  4. $\vec{s}=\vec{p}-\dfrac{(\vec{q}-\vec{p}\cdot \vec{n})\vec{n}}{|\vec{n}|^2}$

Show answer
Correct Option: B
Explanation:

Consider the problem 

Let,
$\pi : \vec r.\vec n=q$ be the plane.
Now $S$ will lie on $\pi $. 
Consider $\vec S$ will joins $O(0,0,0)$ and  $S$. And as $P$ is projection on plane, 
then 
$\vec {PS}$ is perpendicular to plane.

$\vec P=\vec {OP}$
Now In triangle $OSP$, by triangle law of vector addition.

$\vec {OS}+\vec {SP}=\vec {OP}$  

$|\vec {SP}|=$ distance between $P$ and plane 
$=\dfrac{\vec p.\vec n-q}{|\vec n|}$

$\vec {SP}=|\vec {SP}|.\hat {SP}$

Now, 
$\hat {SP}=\hat n=\dfrac{\vec n}{|\vec n|}$

$\vec {SP}=\dfrac{|\vec {SP}|\vec n}{|\vec n|}$

$\vec {SP}=\dfrac{(\vec p.\vec n-q)\vec n}{|\vec n|^2}$

$\vec {OS}+\vec {OP}=\vec {OP}$
$\vec S+\vec {SP}=\vec P$

$\vec S=\vec p-\vec {SP}$
$=\vec p-(\dfrac{\vec p.\vec n-q}{|\vec n|^2})\vec n$

$\vec S=\vec p+\dfrac{(q-\vec p.\vec n)\vec n}{|\vec n|^2}$

The line $\cfrac{x+3}{3}=\cfrac{y-2}{-2}=\cfrac{z+1}{1}$ and the plane $4x+5y+3z-5=0$ intersect at a point

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

  1. $(3,1,-2)$

  2. $(3,-2,1)$

  3. $(2,-1,3)$

  4. $(-1,-2,-3)$

Show answer
Correct Option: B
Explanation:

Let $\dfrac{x+3}{3}=\dfrac{y-2}{-2}=\dfrac{z+1}{1}=k$


On solving, we get

$\Rightarrow x=3k-3$

$\Rightarrow y=-2k+2$ 

$\Rightarrow z=k-1$

On substituing these values in the given plane equation we get,

$\Rightarrow 4x+5y+3z-5=0$

$\Rightarrow 4(3k-3)+5(-2k+2)+3(k-1)-5=0$

On simpliying we get,

$\Rightarrow 5k=10$

$\Rightarrow k=2$

Substituting this value of $k$ in equations of $x,y,z$ we get

$\Rightarrow x=3,y=-2,z=1$

Hence point of intersection is $(3,-2,1)$

If $a,b$ and $c$ are three unit vectors equally inclined to each other at angle $\theta$. Then, angle between $a$ and the plane of $b$ and $c$ is

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

  1. $\cos ^{ -1 }{ \left( \cfrac { \cos { \theta } }{ \cos { \left( \theta /2 \right) } } \right) } $

  2. $\sin ^{ -1 }{ \left( \cfrac { \sin { \theta } }{ \sin { \left( \theta /2 \right) } } \right) } $

  3. $\sin ^{ -1 }{ \left( \cfrac { \cos { \theta } }{ \cos { \left( \theta /2 \right) } } \right) } $

  4. $\cos ^{ -1 }{ \left( \cfrac { \sin { \theta } }{ \sin { \left( \theta /2 \right) } } \right) } $

Show answer
Correct Option: A

If the line $\cfrac{x-1}{2}=\cfrac{y+3}{1}=\cfrac{z-5}{-1}$ is parallel to the plane $px+3y-z+5=0$, then the value of $p$

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

  1. $2$

  2. $-2$

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

  4. $\cfrac{1}{3}$

Show answer
Correct Option: B
Explanation:
line $11$ plane 
$\therefore$ line $\bot$ normal to plane 
$\therefore (2)(P)+(1)(3)+(-1)(-1)=0$  
$\therefore 2p + 3 + 1 =0$
$\therefore P=-2$

The angle between the plane $2 x - y + z = 6$ and a perpendiculars to the planes $x + y + 2 z = 7$ and $x - y = 3$ is

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

  1. $\frac { \pi } { 4 }$

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

  3. $\frac { \pi } { 6 }$

  4. $\frac { \pi } { 2 }$

Show answer
Correct Option: D

Statement 1: Line $\dfrac {x-1}{1}=\dfrac {y-0}{2}=\dfrac {z+2}{-1}$ lies in the plane $2x-3y-4z-10=0$.
Statement 2: If line $\vec r=\vec a+\lambda \vec b$ lies in the planar $\vec r\cdot \vec c=n$ (where n is scalar), then $\vec b\cdot \vec c=0$.

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

  1. Both the statements are true, and Statement 2 is the correct explanation for Statement 1.

  2. Both the statements are true, but Statement 2 is not the correct explanation for Statement 1.

  3. Statement 1 is true and Statement 2 is false.

  4. Statement 1 is false and Statement 2 is true.

Show answer
Correct Option: A
Explanation:

If line $\vec r=\vec a+\lambda \vec b$ lies in the planar $\vec r\cdot \vec c=n$ (where n is scalar), then $\vec b\cdot \vec c=0$ &  $\vec a\cdot \vec c=n$
Therefore, statement 2 is true.
Since, line $\dfrac {x-1}{1}=\dfrac {y-0}{2}=\dfrac {z+2}{-1}$ lies in the plane $2x-3y-4z-10=0$
Then, $2(1)-3(0)-4(-2)-10=0$
          $\Rightarrow 0=0$
and $(i+2j-k).(2i-3j-4k)=0$
       $\Rightarrow 2-6+4=0$
       $\Rightarrow 0=0$
Therefore, statement 1 is true.

Ans: A