Tag: three dimensional geometry

Questions Related to three dimensional geometry

 The points with position vectors $\vec {a}=\hat {i}-2\hat {j}+3\hat {k}, \vec {b}=2\hat {i}+3\hat {j}-4\hat {k}$ & $-7\hat {j}+10\hat {k}$ are collinear.

direction cosines and direction ratios three dimensional geometry maths

  1. True

  2. False

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Correct Option: A

The points $i + j + k, \, i + 2j, \, 2i+2j+k,\, 2i+3j+2k$ are

direction cosines and direction ratios three dimensional geometry maths

  1. collinear

  2. coplanar but not collinear

  3. non-coplanar

  4. none

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Correct Option: C
Explanation:

$\begin{matrix} A& B& C& D\i+j+k, &i+2j, &2i+2j+k,&2i+3j+2k \end{matrix}$
$\overline{AC} = (2-1)i + (2-1)j + k-k$
$=i+j$
$\overline{AB} = o + j - \overline{k} = j - \overline{k}$
$\overline{AD} = i + 2j + k$
$\begin{vmatrix} 1&1&0 \0&2 &1\end{vmatrix} = 1(1+2)-1(0+1)$
$=3-1 = 2 \neq 0$
Non coplanar.

If $\vec a, \, \vec b$ are two non-collinear vectors, then the position vector $\vec a + \vec b, \, \vec a - \vec b, \,and \, \vec a + \lambda {\vec b}$ are collinear for some real values of $\lambda$.

direction cosines and direction ratios three dimensional geometry maths

  1. True

  2. False

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Correct Option: B

If $\bar {a}, \bar {b}$ and $\bar {c}$ are non-zero non collinear vectors and $\theta(\neq 0 , \pi)$ is the angle between $\bar {b}$ and $\bar {c}$ if $(\bar {a}\times \bar {b}) \times \bar {c}=\dfrac {1}{2} |\bar {b}|\bar {c}|\bar {a}$. then $\sin \theta =$

direction cosines and direction ratios three dimensional geometry maths

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

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

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

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

Show answer
Correct Option: B
Explanation:

We have

$\left( {\overrightarrow a  \times \overrightarrow b } \right) \times \overrightarrow c  = \frac{1}{2}\left| {\overrightarrow b } \right|\left| {\overrightarrow c } \right|\overrightarrow a $
$\overrightarrow c  \times \left( {\overrightarrow a  \times \overrightarrow b } \right) = \frac{1}{2}\left| {\overrightarrow b } \right|\left| {\overrightarrow c } \right|\overrightarrow a $
$ - \left[ {\left( {\overrightarrow c .\overrightarrow b } \right)\overrightarrow a  - \left( {\overrightarrow c .\overrightarrow a } \right)\overrightarrow b } \right] = \frac{1}{2}\left| {\overrightarrow b } \right|\left| {\overrightarrow c } \right|\overrightarrow a $
$\left( {\overrightarrow c .\overrightarrow a } \right)\overrightarrow b  - \left( {\overrightarrow c .\overrightarrow b } \right)\overrightarrow a  = \frac{1}{2}\left| {\overrightarrow b } \right|\left| {\overrightarrow c } \right|\overrightarrow a $
$\overrightarrow c .\overrightarrow a  = 0$
$\overrightarrow c .\overrightarrow a  = \frac{{ - 1}}{2}\left| {\overrightarrow b } \right|\left| {\overrightarrow c } \right|$
$\cos \theta  = \frac{{ - 1}}{2}$
$ \Rightarrow \theta  = \frac{{2\pi }}{3}$
$\therefore \sin \theta  = \frac{{\sqrt 3 }}{2}$
Hence, $B$is the correct answer.

The points with position vectors $ 60i + 3j,  40i -8j$ and $ ai -52j $ 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:

Denoting $a,b,c$ by the given vectors respectively
These vectors will be collinear if there is some constant $k$ such that $c-a=K\left( b-a \right) $
$\Rightarrow a-60=-20K$ and $-55=-11K$
$\Rightarrow a=-100+60=-40$

The three points $ABC$ have position vectors $(1,x,3),(3,4,7)$ and $(y,-2,-5)$ are collinear then $(x,y)=$

direction cosines and direction ratios three dimensional geometry maths

  1. $(2,-3)$

  2. $(-2,3)$

  3. $(-2,-3)$

  4. $(2,3)$

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Correct Option: A
Explanation:

$(1,x,3)=\lambda(3,4,7) + \mu (y,-2,-5)$
$1=3\lambda +\mu y$
$x= 4\lambda +(-2\mu)$
$3 = 7\lambda -5 \mu$
$2-x= (-3-y)\mu$
So only $x=2$,  $y=-3$

If the three points  $A(\overline a),B(\overline  b),C(\overline c) $ are collinear ,the line passing through them is

$\overline r=\overline a+\lambda(\overline b-\overline a)$ then value of $\lambda $ is 

direction cosines and direction ratios three dimensional geometry maths

  1. $1$

  2. $2$

  3. $3$

  4. $4$

Show answer
Correct Option: C
Explanation:
Given line
$\vec{r}=\vec{a}+\lambda(\vec{b}-\vec{a})$
$\vec{r}=(1-\lambda)\vec{a}+\lambda\vec{b}$
if $a$ and $b$ are collinear then 
$xa+yb=0$
$x=1-\lambda$
$y=\lambda$
if we pass line through c then 
$\vec{r}=1\neq0$
SO $\lambda=3$ to satisfy eq 

If points (1,2), (3 , 5) and (0 , b ) are collinear the value of b is  

direction cosines and direction ratios three dimensional geometry maths

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

  2. $\dfrac{7}{2}$

  3. 2

  4. -1

Show answer
Correct Option: A
Explanation:

$Area=\dfrac{1}{2}| 1(5-b)+3(b-2)+0(2-5)|$
As points are collinear , so area =0
$\therefore \dfrac{1}{2}| 1(5-b)+3(b-2)+0(2-5)|=0$
$\Rightarrow 5-b+3b-6=0$
$\Rightarrow=1=2b$
$\therefore b=\dfrac{1}{2}$

The following lines are $\hat { r } =\left( \hat { i } +\hat { j }  \right) +\lambda \left( \hat { i } +2\hat { j } -\hat { k }  \right) +\mu \left( -\hat { i } +\hat { j } -\hat { 2k }  \right) $

direction cosines and direction ratios three dimensional geometry maths

  1. collinear

  2. skew-lines

  3. co-planar lines

  4. parallel lines

Show answer
Correct Option: A
Explanation:

Condition for three lines $\vec { { r } _{ 1 } } $ , $\vec { { r } _{ 2 } } $ , and $\vec { { r } _{ 3 } } $ to be collinear is:

$\vec { { r } _{ 1 } } +\lambda \vec { { r } _{ 2 } } +\vec { { \mu r } _{ 3 } } =0$
where $\vec { { r } _{ 1 } } =\left( \vec { i } +\vec { j }  \right) $
$\vec { { r } _{ 2 } } =\left( \vec { i } +2\vec { j } -\vec { k }  \right) $
$\vec { { r } _{ 3 } } =\left( -\vec { i } +\vec { j } -2\vec { k }  \right) $
and $\lambda $ and $\mu $ are scalars
Hence, the answer is collinear.

If the lines $x=1+a,y=-3-\lambda a,z=1+\lambda a$ and $x=\cfrac { b }{ 2 } ,y=1+b,z=2-b$ are coplanar, then $\lambda$ is equal to

direction cosines and direction ratios three dimensional geometry maths

  1. $-3$

  2. $2$

  3. $1$

  4. $-2$

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Correct Option: D
Explanation:
The given lines are $\cfrac { x-1 }{ 1 } =\cfrac { y+3 }{ -\lambda  } =\cfrac { z-1 }{ \lambda  } =\left( a \right) $ and $\cfrac { x-0 }{ 1/2 } =\cfrac { y-1 }{ 1 } =\cfrac { z-2 }{ -1 } =(b)$
$\therefore$ coplanarity, we must have
$\begin{vmatrix} -1 & 4 & 1 \\ 1 & -\lambda  & \lambda  \\ 1/2 & 1 & -1 \end{vmatrix}=0$
$\Rightarrow -1\left( \lambda -\lambda  \right) -4\left( -1-\cfrac { \lambda  }{ 2 }  \right) +\left( 1+\cfrac { \lambda  }{ 2 }  \right) =0$
$4+2\lambda +1+\cfrac { \lambda  }{ 2 } =0\quad \Rightarrow 5+\cfrac { 5\lambda  }{ 2 } =0\quad \therefore \lambda =-2$