Tag: direction cosines and direction ratios

Questions Related to direction cosines and direction ratios

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
$\begin{vmatrix} -1 & 4 & 1 \\ 1 & -\lambda  & \lambda  \\ 1/2 & 1 & -1 \end{vmatrix}=0$
$\Rightarrow -1\left( \lambda -\lambda  \right) -4\left( -1-\cfrac { \lambda  }{ 2 }  \right) +\left( 1+\cfrac { \lambda  }{ 2 }  \right) =0$
$4+2\lambda +1+\cfrac { \lambda  }{ 2 } =0\quad \Rightarrow 5+\cfrac { 5\lambda  }{ 2 } =0\quad \therefore \lambda =-2$

If $\vec { a } ,\vec { b } ,\vec { c } $ are three non-zero vectors, no two of which are collinear and the vector $\vec { a } +\vec { b } $ is collinear with $\vec { c }, \vec { b } +\vec { c } $ is collinear with $\vec {a},$ then $\vec { a } +\vec { b } +\vec { c }$ is equal to -

direction cosines and direction ratios three dimensional geometry maths

  1. $\vec {a}$

  2. $\vec {b}$

  3. $\vec {c}$

  4. $none\ of\ these$

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Correct Option: D
Explanation:
$\bar a+\bar b=K _{1}\bar c $

$\bar b+\bar c=K _{2}\bar a $

$\bar a- \bar c=K _{1}c-K _{2}\bar a$

$(k _{2}+1) \bar a-\bar c(1+k _{1})$=0

$k _{2}=-1 $ and $k _{1}=-1 $

$\bar{a}+\bar{b}+\bar{c}=0$

If the points with position vectors $60\hat{i}+3\hat{j}, 40\hat{i}-8\hat{j}$ and $a\hat{i}-52j$ are collinear, then $a=?$

direction cosines and direction ratios three dimensional geometry maths

  1. $-40$

  2. $-20$

  3. $20$

  4. $40$

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Correct Option: A
Explanation:

Suppose, position vector $A=60\widehat i+3\widehat j$

position vector $B=40\widehat i-8\widehat j$

position vector $C=a\widehat i-52\widehat j$

Now, find vector AB and BC

$AB= -20\widehat i-11\widehat j$

$BC= (a-40)\widehat i-44\widehat j$

To be collinear,  angle between the vector AB and BC made by the given position vectors should be 0 or 180 degree.

That’s why the cross product of  the vectors should be zero

$ABXBC=(-20\widehat i-11\widehat j)X(a-40)\widehat i-44\widehat j$

$0\widehat i+0\widehat j+(880+11(a-40))=0$

$a-40= -80$

$a=-40$

Therefore, a should be $-40$ to be the given positions vectors collinear.


 Let $\overrightarrow{b}$ and  $\overrightarrow{c}$ be non collinear vectors.If $\overrightarrow{a}$ is a vector such that $\overrightarrow{a}.\left(\overrightarrow{b}+\overrightarrow{c}\right)=4$ and $\overrightarrow{a}\times\left(\overrightarrow{b}\times \overrightarrow{c}\right)=\left({x}^{2}-2x+6\right)\overrightarrow{b}+\sin{y} .\overrightarrow{c}$ then $\left(x,y\right)$ lies on the line

direction cosines and direction ratios three dimensional geometry maths

  1. $x+y=0 $

  2. $x-y=0$

  3. $x=1$

  4. $y=\dfrac{\pi}{2}$

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Correct Option: C,D
Explanation:

$\overrightarrow{a}\times \left(\overrightarrow{b}\times \overrightarrow{c}\right)=\left(\overrightarrow{a}.\overrightarrow{c}\right).\overrightarrow{b}-\left(\overrightarrow{a}.\overrightarrow{b}\right).\overrightarrow{c}$
$\therefore \overrightarrow{a}.\overrightarrow{c}={x}^{2}-2x+6=-\sin y$
$\overrightarrow{a}.\left(\overrightarrow{b}+\overrightarrow{c}\right)=4 \Rightarrow -\sin y+{x}^{2}-2x+6=4$
$\Rightarrow {x}^{2}-2x+2=\sin y$
$\Rightarrow {\left(x-1\right)}^{2}+1=\sin y$
Left side $\ge 1$, right side $\le 1$
$\therefore $ they are equal if 
${\left(x-1\right)}^{2}+1=\sin y=1$
$\therefore y=\dfrac{\pi}{2},x=1$

Three points whose position vectors are $x\bar{i}+y\bar{j}+z\bar{k}$, $\bar{i}+2\bar{j}$ and $-\bar{i}-\bar{j}$ are collinear, then relation between $x, y, z$ is?

direction cosines and direction ratios three dimensional geometry maths

  1. $x-2y=1, z=0$

  2. $z+y=1, z=0$

  3. $x-y=1, z=0$

  4. None of these

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Correct Option: D

If the points $(\alpha, - 1), (2, 1)$ and $(4, 5)$ are collinear, then find $\alpha $ by vector method.

direction cosines and direction ratios three dimensional geometry maths

  1. $4$

  2. $1$

  3. $8$

  4. None of these

Show answer
Correct Option: B
Explanation:
If there points are collinear then vectors from one to another will have scalar triple produced $0$.Point $\left(\alpha,-1\right), \left(2,1\right), \left(4,5\right)$
$\left( 2-\alpha  \right) \hat { i } +2\hat { j } -\bar { A }$
$2\hat { i } +4\hat { j } -\bar { B }$
$\left( 4-\alpha  \right) \hat { i } +6\hat { j } -\bar { C }$
$ \bar { A } .\left( \bar { B } \times \bar { C }  \right) =0$
$\left( \left( 2-\alpha  \right) \hat { i } +2\hat { j }  \right) \left( 2\hat { i } +4\hat { j }  \right) \times \left( \left( 4-\alpha  \right) \hat { i } +6\hat { j }  \right) \\ \left( \left( 2-\alpha  \right) \hat { i } +2\hat { j }  \right) .\left[ 12\hat { k } -16\hat { k } +4\alpha \hat { k }  \right] =0$
$4\alpha =4$
 $\alpha =1$
Also the direction vector will be proportion
$\left( 2-\alpha,2 \right)=\lambda\left( 4-2.5-1\right)$
$\left( 2-\alpha,2 \right)=\lambda\left( 2,4\right)$
$\lambda=\dfrac{1}{2}$ as $2=4\lambda$
$2-\alpha=1$
$\therefore \alpha=1$

If the points $\bar a + \bar b,\bar a - \bar b,\bar a + k\bar b$ are collinear, then  

direction cosines and direction ratios three dimensional geometry maths

  1. $k$ has only one real value

  2. $k$ has two real value

  3. $k$ has no real values

  4. $k$ has infinite number of real values

Show answer
Correct Option: D
Explanation:
As the $3$ point should be collinear area of triangle termed by then should be zero

considering $2$ direction to be $(\bar{a}+\bar{b})-(\bar{a}-\bar{b})=2\bar{b}$
& $(\bar{a}+\bar{b})-(\bar{a}+k\bar{b})=(1-k)\bar{b}$

$(2\bar{b})\times (1-k)\bar{b}=0$

this will be for any value of $k$ as cross produced of $2$ linear vector $=0$


If $A = (1,2,3) , B  = (2,10,1), Q$ are collinear points and $Q _{x}=-1$ then $Q _{z}$ is

direction cosines and direction ratios three dimensional geometry maths

  1. $-3$

  2. $7$

  3. $-14$

  4. $-7$

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Correct Option: B

If points $\hat i + \hat j, \hat i - \hat j$ and $p \hat i + q \hat j + r \hat k$ are collinear, then

direction cosines and direction ratios three dimensional geometry maths

  1. $p = 1$

  2. $r = 0$

  3. $q \in R$

  4. $q \neq 1$

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Correct Option: A,B,D
Explanation:

Points $A(\hat i + \hat j), B (\hat i - \hat j)$ and $C(p \hat i + q \hat j + r \hat k)$ are collinear
Now $\vec{AB} = - 2 \hat j$ and $\vec{BC} = (p -1) \hat i + (q - 1) \hat j + r k$
Vectors $\vec{AB}$ and $\vec{BC}$ must be collinear
$\Rightarrow p = 1,  r= 0 $ and $q \neq 1$

If  $\bar { a }, \bar { b }, \bar { c }$ are non-coplaner vector , then the vectors $2\bar { a }- 4\bar { b }+ 4\bar { c }, \bar { a }- 2\bar { b }+ 4\bar { c }$ and $-\bar { a }+ 2\bar { b }+ 4\bar { c }$ are parellel.

direction cosines and direction ratios three dimensional geometry maths

  1. True

  2. False

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Correct Option: B