Tag: three dimensional geometry

Questions Related to three dimensional geometry

$\bar a,\bar b,\bar c$ are three non-zero vectors such that any two of them are non-collinear. If  $\bar a+\bar b$ is collinear with  $\bar c$ and  $\bar b+\bar c$ is collinear with $\bar a$, then what is their sum?

direction cosines and direction ratios three dimensional geometry maths

  1. $-1$

  2. $0$

  3. $1$

  4. $2$

Show answer
Correct Option: B
Explanation:

We have

$\bar a+\bar b =t\bar c$ ----$(1)$
$\bar b+\bar c =s\bar a$ ----$(2)$
From $(1)$ and $(2)$
$\bar a+\bar b=t(s\bar a-\bar b)$
Since no two of them are collinear, comparing coeffficients gives
$st=1$ and $t=-1$
$\Rightarrow s=-1$ and $t=-1$
From $(1)$
$\therefore \bar a+\bar b+\bar c=0$
Hence, option $B$.

The line passes through the points $\left ( 5,1,a \right )$ & $\left ( 3,b,1 \right )$ crosses the $yz$ plane at the point $\displaystyle \left ( 0,\frac{17}{2},-\frac{13}{2} \right )$ ,then

direction cosines and direction ratios three dimensional geometry maths

  1. $a= 4, b= 6$

  2. $a= 6, b= 4$

  3. $a= 8, b= 2$

  4. $a= 2, b= 8$

Show answer
Correct Option: B
Explanation:

Equation of the line through the points $( 5,1,a  ) $ & $( 3,b,1)$ is
$\displaystyle \frac{x-5}{3-5}=\frac{y-1}{b-1}=\frac{z-a}{1-a}=\lambda $
Now it passes through $\displaystyle \left ( 0,\frac{17}{2},\frac{-13}2{} \right )$
$\displaystyle \therefore \frac{0-5}{-2}=\frac{17/2-1}{b-1}=\frac{-13/2-a}{1-a}=\lambda  :$

$ \Rightarrow \lambda =\dfrac{5}{2}$
$\displaystyle \therefore \frac{17/2-1}{b-1}=\frac{5}{2} $
$\Rightarrow b=4$
and $\displaystyle \frac{-13/2-a}{1-a}=\frac{5}{2} $
$\Rightarrow a=6$

If the three points with position vectors $\displaystyle \bar{a}-2\bar{b}+3\bar{c}, \ 2\bar{a}+\lambda \bar{b}-4\bar{c}, \ -7\bar{b}+10\bar{c} $ are collinear, then $\displaystyle \lambda= $

direction cosines and direction ratios three dimensional geometry maths

  1. $1$

  2. 2

  3. $3$

  4. none of these

Show answer
Correct Option: C
Explanation:

The given vectors are collinear, so $l(\bar{a} - 2\bar{b} + 3\bar{c}) + k(2\bar{a} + \lambda\bar{b} - 4\bar{c}) = (l + k)(-7\bar{b} + 10\bar{c})$
Comparing the coefficients of $\bar{a} \rightarrow l + 2k = 0 $
$\bar{b} \rightarrow -2l + \lambda k = -7l -7k$
$\bar{c} \rightarrow 3l - 4k = 10l + 10k$
$\Rightarrow l = -2k$ and so $\lambda = 3$

The vectors $2\hat i + 3\hat j, \ 5\hat i + 6\hat j$ and $8\hat i + \lambda \hat j$ have their initial points at $(1,1)$. The value of $\lambda$ so that the vectors terminate on one straight line is

direction cosines and direction ratios three dimensional geometry maths

  1. 9

  2. 6

  3. 3

  4. 0

Show answer
Correct Option: A

For what value of $m$, the points $(3,5)$, $(m,6)$ and $\begin{pmatrix} \dfrac { 1 }{ 2 },\dfrac {15 }{ 2 } \end{pmatrix}$ are collinear?

direction cosines and direction ratios three dimensional geometry maths

  1. $9$

  2. $5$

  3. $3$

  4. $2$

Show answer
Correct Option: D
Explanation:

As the points are collinear, the slope of the line joining any two points, should be same as the slope of the line joining two other points. 
Slope of the line passing through points $\left( { x } _{ 1 },{ y } _{ 1 } \right) $ and $\left( { x } _{ 2 },{ y } _{ 2 } \right)$ $ $=$ $ $\dfrac { { y } _{ 2 }-{ y } _{ 1 } }{ { x } _{ 2 }-x _{ 1 } } $
So, slope of the line joining $ (3,5) , (m,6) = $ Slope of the line joining $ (3,5) $ and $\left  (\dfrac {1}{2}, \dfrac {15}{2}\right ) $ 

Therefore, $ \dfrac { 6 - 5 }{ m - 3 } = \dfrac { \frac {15}{2} - 5 }{ \frac {1}{2} - 3 } $
$\Rightarrow  \dfrac { 1 }{ m - 3 } = -1 $

$\Rightarrow  m - 3 = -1 $

$\Rightarrow  m = 2 $

If the points $(p,0)$, $(0,q)$ and $(1,1)$ are collinear, then $\dfrac { 1 }{ p }+\dfrac { 1 }{ q }$ is equal to:

direction cosines and direction ratios three dimensional geometry maths

  1. $-1$

  2. $1$

  3. $2$

  4. $0$

Show answer
Correct Option: B
Explanation:

As the points are collinear, the slope of the line joining

any two points, should be same as the slope of the line joining two other

points.

Slope of the line passing through points $\left( { x } _{ 1 },{ y } _{ 1 }

\right) $ and $\left( { x } _{ 2 },{ y } _{ 2 } \right)$ $ = $ $\dfrac { { y

} _{ 2 }-{ y } _{ 1 } }{ { x } _{ 2 }-x _{ 1 } } $

So, slope of the line joining $ (p,0) , (0,q) = $ Slope of the line joining

$ (0,q) $ and $ (1,1) $

$ \dfrac { q - 0 }{ 0 - p } = \dfrac { 1 - q }{ 1 - 0 } $

$ - \dfrac { q }{ p } = 1 - q $

Dividing both sides by $q$,
$ - \dfrac { 1 }{ p } =  \dfrac { 1 }{ q } - 1 $

$ => \dfrac { 1 }{ p } +  \dfrac { 1 }{ q } = 1 $

Determine if the points $(1,5)$ $(2,3)$ and $(-2,-11)$ are collinear.

direction cosines and direction ratios three dimensional geometry maths

  1. True

  2. False

Show answer
Correct Option: B
Explanation:

The given points are $A(1,5)$, $B(2,3)$ and $C(-2,-11)$.


Let us calculate the distance : $AB$, $BC$ and $CA$ by using distance formula.

$AB =\sqrt { (2-1)^{ 2 }+(3-5)^{ 2 } } =\sqrt { (1)^{ 2 }+(-2)^{ 2 } } $

$=\sqrt {1+4} = \sqrt{ 5 }$ units

$BC =\sqrt { (-2-2)^{ 2 }+(-11-3)^{ 2 } }=\sqrt { (-4)^{ 2 }+(-14)^{ 2 } }$

$=\sqrt {16+196} =\sqrt {212} = 2\sqrt{53}$ units

$CA =\sqrt { (-2-1)^{ 2 }+(-11-5)^{ 2 } }$

$=\sqrt { (-3)^{ 2 }+(-16)^{ 2 } } =\sqrt {9+256} = \sqrt {265 }$ 

$=\sqrt {5}\times\sqrt {53}$ units

From the above we see that : $AB+BC\neq CA$

Hence, the above stated points $A(1,5)$, $B(2,3)$ and $C(-2,-11)$ are not collinear.

In each of the following find the value of $k$, for which the points are collinear.
(i) $(7,-2)$, $(5,1)$, $(3,k)$
(ii) $(8,1)$, $(k,-4)$, $(2,-5)$

direction cosines and direction ratios three dimensional geometry maths
  1. (i) $k = 4$

  2. (i) $k = 5$

  3. (ii) $k = 3$

  4. (ii) $k = 2$

Show answer
Correct Option: A,C
Explanation:

Since the given points are collinear, they do not form a triangle, which means area of the triangle is Zero.

Area of a triangle with vertices $({ x } _{ 1 },{ y } _{ 1 })$ ; $({ x } _{ 2 },{ y

} _{ 2 })$  and $({ x } _{ 3 },{ y } _{ 3 })$  is $ \left| \dfrac { {

x } _{ 1 }({ y } _{ 2 }-{ y } _{ 3 })+{ x } _{ 2 }({ y } _{ 3 }-{ y } _{ 1 })+{ x } _{

3 }({ y } _{ 1 }-{ y } _{ 2 }) }{ 2 }  \right| $


1) Substituting the points $({ x } _{ 1 },{ y } _{ 1 }) = (7,-2) $ ; $({ x

} _{ 2 },{ y } _{ 2 }) = (5,1) $  and $({ x } _{ 3 },{ y } _{ 3 }) = (3,k)$

In the area formula, we get

$ \left| \dfrac { 7(1-k) + 5(k+2) + 3(-2-1) }{ 2 }  \right|  =

0 $

$ \left| \dfrac { 7 -7k + 5k + 10 - 9 }{ 2 }  \right|  =

0 $

$ \left| \dfrac { 8 -2k }{ 2 }  \right|  =

0 $

$ \Rightarrow  8 - 2k = 0 $

$ \Rightarrow  k = 4 $

2) Substituting the points $({ x } _{ 1 },{ y } _{ 1 }) = (8,1) $ ; $({ x

} _{ 2 },{ y } _{ 2 }) = (k,-4) $  and $({ x } _{ 3 },{ y } _{ 3 }) = (2,-5)$ in the area formula, we get


$ \left| \dfrac { 8(-4+5) + k(-5-1) + 2(1+4) }{ 2 }  \right|  =

0 $

$ \left| \dfrac { 8 -6k +10 }{ 2 }  \right|  =

0 $

$ \left| \dfrac { 18 -6k }{ 2 }  \right|  =

0 $

$ \Rightarrow  18 - 6k = 0 $

$ \Rightarrow  k = 3 $

Are the points (1, 1), (2, 3) and (8, 11) collinear ?

direction cosines and direction ratios three dimensional geometry maths

  1. collinear

  2. Non collinear

  3. coplaner

  4. None of above

Show answer
Correct Option: B
Explanation:

Area of triangle formed by these vertices is 
$\displaystyle \Delta =\frac { 1 }{ 2 } \begin{vmatrix} 1 & 1 & 1 \ 2 & 3 & 1 \ 8 & 11 & 1 \end{vmatrix}$
Applying ${ R } _{ 2 }\rightarrow { R } _{ 2 }-{ R } _{ 1 },{ R } _{ 3 }\rightarrow { R } _{ 3 }-{ R } _{ 1 }$
$\displaystyle \Delta =\frac { 1 }{ 2 } \begin{vmatrix} 1 & 1 & 1 \ 1 & 2 & 0 \ 7 & 10 & 0 \end{vmatrix}=\frac { 1 }{ 2 } \left( 10-14 \right) =2$
Hence points are non collinear 

If $\vec{a},\vec{b},\vec{c}$ are the position vectors of points lie on a line, then $\vec{a}\times \vec{b}+\vec{b}\times \vec{c}+\vec{c}\times \vec{a}=$

direction cosines and direction ratios three dimensional geometry maths

  1. $0$

  2. $ \vec{b}$

  3. $1$

  4. $\vec{a}$

Show answer
Correct Option: A