Tag: direction cosines and direction ratios

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Multiple choice direction cosines and direction ratios three dimensional geometry maths

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

  1. $a=-40$
  2. $a=40$
  3. $a=20$
  4. $None\ of\ these$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

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


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

Multiple choice direction cosines and direction ratios three dimensional geometry maths

 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.

  1. True

  2. False
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Three points with position vectors a, b, and c are collinear if (b-a) is a scalar multiple of (c-a). Given a = i - 2j + 3k, b = 2i + 3j - 4k, and c = 0i - 7j + 10k: b-a = i + 5j - 7k; c-a = -i - 5j + 7k. Since c-a = -1(b-a), the vectors are collinear.

Multiple choice direction cosines and direction ratios three dimensional geometry maths

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

  1. collinear

  2. coplanar but not collinear

  3. non-coplanar

  4. none
Reveal answer Fill a bubble to check yourself
C Correct answer
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.

Multiple choice direction cosines and direction ratios three dimensional geometry maths

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 $\lambda$.

  1. True

  2. False
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

For three points with position vectors to be collinear, the vectors connecting them must be parallel. The vectors a+b and a-b are not generally collinear with a+lambda*b unless specific conditions are met for lambda, and they are certainly not collinear for all real values of lambda.

Multiple choice direction cosines and direction ratios three dimensional geometry maths

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 $\sin \theta =$

  1. $\sqrt{\dfrac{2}{3}}$
  2. $\dfrac{\sqrt{3}}{2}$
  3. $\dfrac{4\sqrt{2}}{3}$
  4. $\dfrac{2\sqrt{2}}{3}$
Reveal answer Fill a bubble to check yourself
B Correct answer
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.

Multiple choice direction cosines and direction ratios three dimensional geometry maths

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

  1. $a = -40$
  2. $a = 40$
  3. $a = 20$

  4. None of these
Reveal answer Fill a bubble to check yourself
A Correct answer
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 $-55=-11K$
$\Rightarrow a=-100+60=-40$

Multiple choice direction cosines and direction ratios three dimensional geometry maths

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

  1. $(2,-3)$
  2. $(-2,3)$
  3. $(-2,-3)$
  4. $(2,3)$
Reveal answer Fill a bubble to check yourself
A Correct answer
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 $x=2$,  $y=-3$

Multiple choice direction cosines and direction ratios three dimensional geometry maths

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 

  1. $1$
  2. $2$
  3. $3$
  4. $4$
Reveal answer Fill a bubble to check yourself
C Correct answer
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 
Multiple choice direction cosines and direction ratios three dimensional geometry maths

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

  1. $\dfrac{1}{2}$
  2. $\dfrac{7}{2}$

  3. 2

  4. -1
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

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

Multiple choice direction cosines and direction ratios three dimensional geometry maths

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) $

  1. collinear

  2. skew-lines

  3. co-planar lines

  4. parallel lines
Reveal answer Fill a bubble to check yourself
A Correct answer
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.