Tag: direction cosines and direction ratios

Questions Related to direction cosines and direction ratios

The points with position vectors $60\hat{i}+3\hat{j}$, $40\hat{i}-8\hat{j}$, $a\hat{i}-52\hat{j}$  are collinear if

direction cosines and direction ratios three dimensional geometry maths

  1. $a=-40$

  2. $a=40$

  3. $a=20$

  4. $None\ of\ these$

Show answer
Correct Option: A
Explanation:

suppose ${60i + 3j}$ , ${40i - 8j}$ and ${ai - 52j}$ is the three position of vector $A,B,C$


$\begin{array}{l} \overrightarrow { AB } =\left( { 40i-8j } \right) -\left( { 60i+3j } \right)  \ \overrightarrow { AB } =-20i-11j \ \overrightarrow { BC } =\left( { ai-52j } \right) -\left( { 40i-8j } \right)  \ \overrightarrow { BC } =\left( { a-40 } \right) i-44j \ \left( { a-40 } \right) i-44j=m\left( { -20i-11j } \right)  \ \left( { a-40 } \right) i-44j=-20im-11jm \ -44=-11m \ m=\frac { { -44 } }{ { -11 } }  \ m=4 \ a-40=-20m \ a-40=-20\left( 4 \right)  \ a=-80+40 \ a=-40 \end{array}$

 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

Show answer
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

Show answer
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

Show answer
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. $(2,3)$

Show answer
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.