Direction cosines and direction ratios - class-XII

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

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.

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

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.

We have

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 $-55=-11K$
$\Rightarrow a=-100+60=-40$

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

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

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

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

For three points to be collinear, the vectors connecting them must be proportional. The vectors are P1(x, y, z), P2(1, 2, 0), and P3(-1, -1, 0). The vector P2P3 is (-2, -3, 0). The vector P1P2 is (1-x, 2-y, -z). For these to be parallel, the ratios of components must be equal, implying z=0 and a specific linear relationship between x and y that is not listed in A, B, or C.

Since A(1, 2, 3), B(2, 10, 1), and Q(Qx, Qy, Qz) are collinear, the vector AB = (1, 8, -2) must be proportional to AQ = (Qx-1, Qy-2, Qz-3). Given Qx = -1, the x-component of AQ is -2. Since -2 is -2 times the x-component of AB, the other components must follow the same ratio: Qy-2 = -2(8) = -16 (Qy = -14) and Qz-3 = -2(-2) = 4 (Qz = 7).

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 $q \neq 1$

Three vectors are parallel if they are scalar multiples of each other. Here, the vectors are v1 = 2a - 4b + 4c, v2 = a - 2b + 4c, and v3 = -a + 2b + 4c. Since the coefficients of a and b are not proportional to the constant c across all three vectors, they cannot be parallel.

Let the points be $A(-1,3,2) B(-4,2,-2) C(5,5,\lambda)$
Direction ratio of $AB{=}(-3,-1,-4)$
If $A,B,C$ are collinear points then direction ratio of $AB$ and $BC$ must be proportional.
Direction ratio of $BC{=}(9,3,\lambda+2)$
$\therefore 9{=}\alpha (-3)$
$\therefore 3{=}\alpha(-1)$
$\therefore \lambda+2{=}\alpha(-4)$ and $\alpha{=} -3$
$\therefore \dfrac{\lambda+2}{-3}{=}-4$
$\therefore \lambda{=} 10$

Let the points be $A(8,-7,a), B(5,2,4), C(6,-1,2)$
Direction ratio of $BC =(1,-3,-2)$
If $A,B,C$ are collinear points then direction ratio of $BC$ and $AB$ must be proportional.
Direction ratio of $AB=(-3,9.4-a)$
$\therefore -3=\lambda1$
$\therefore 9=\lambda(-3)$
$\therefore 4-a=\lambda(-2)$ and $\lambda=-3$
$\therefore \dfrac{4-a}{-3}=-2$

Given:- $A(1,-1,0)$; $B(3,1,2)$;$C(2,-2,4)$ and $D(-1,1,-1)$ 

D.R of AB are $ 2 -1, 6-2,2-(-1) i.e. 1,4,3$

$A(1,-1,0) , B(3,1,2), C(2,-2,4), D(-1,1,-1)\ \vec { AB } = <3-1, 1-(-1), 2-0>$ 

D.R. of AB are $2−1,6−2,2−(−1)i.e.,1,4,3.$

When points $x, y, z$ are collinear, we have $\alpha x + \beta y = (\alpha + \beta)z$
Similarly, $x(2\vec{a} - \vec{b} + 3\vec{c}) + y(\vec{a} - 2\vec{b} + \lambda \vec{c}) = (x + y)(\mu \vec{a} - 5\vec{b})$
$\Rightarrow$ comparing the coefficients of $\vec{a} \rightarrow 2x + y = x\mu + y\mu $
$\vec{b} \rightarrow - x - 2y = - 5x - 5y$
$\vec{c} \rightarrow 3x + \lambda y = 0$
$\Rightarrow 4x = -3y$ and so $\lambda = \dfrac{9}{4}$
Also, $\mu = -2$

We have

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 $\displaystyle \left ( 0,\frac{17}{2},\frac{-13}2{} \right )$
$\displaystyle \therefore \frac{0-5}{-2}=\frac{17/2-1}{b-1}=\frac{-13/2-a}{1-a}=\lambda  :$

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 $\lambda = 3$

Vectors starting from the same point (1,1) are collinear if their components are proportional. The vectors are v1 = (2, 3), v2 = (5, 6), and v3 = (8, lambda). The vector v2 - v1 = (3, 3). The vector v3 - v2 = (3, lambda - 6). For these to be collinear, the slopes must be equal, so (lambda - 6) / 3 = 3 / 3, which gives lambda - 6 = 3, so lambda = 9.

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 $

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 $

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

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| $

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 points with position vectors a, b, c are collinear, then (b-a) is parallel to (c-b). This implies (b-a) x (c-b) = 0. Expanding this cross product gives b x c - b x b - a x c + a x b = 0. Since b x b = 0, we get b x c + a x b - a x c = 0, which rearranges to a x b + b x c + c x a = 0.

$\overrightarrow { a } ,\overrightarrow { b } ,\overrightarrow { c } $ are collinear

Three points with position vectors p1, p2, p3 are collinear if (p2-p1) is a multiple of (p3-p2). Here, p2-p1 = -2b and p3-p2 = (lambda-1)b. Since both vectors are multiples of b, they are parallel for any real value of lambda.

Line passing through $A(2,4,5)$ and $B(-7,-2,8)$ is
$L: \dfrac{x-2}{9} = \dfrac{y-4}{6} = \dfrac{z-5}{-3}$
Lets suppose $O$ is collinear with $A$ and $B$, then $L = k$
$\therefore$ Coordinates of $O$ are $(2+9k,4+6k,5-3k)$
Looking at options, only option C satisfies the coordinates of $O$ for $k = -\dfrac{1}{3}$

The equation of line passing through $A$ and $B$ is $\displaystyle\frac { x-1 }{ 2-1 } =\frac { y-2 }{ 10-2 } =\frac { z-3 }{ 1-3 } \Rightarrow \frac { x-1 }{ 1 } =\frac { y-2 }{ 8 } =\frac { z-3 }{ -2 } =r$

$Positive, , vector, , are, , , 2i+j+k,6\hat { i } +\hat { j } +2\hat { k }  \ and, , 4i-5j+pk, , are, , collinear, , then, , P=2 \ if, , three, , position, , vectors, , are, , collinear, , then,, , its, , { { determinantsis } }\left( 0 \right)  \ \Rightarrow \left| \begin{matrix} 2, , , , , , , 1, , , , , , , 1 \ 6, , , , -1, , , , , , 2 \ 14, , -5, , , , P \  \end{matrix} \right| =0 \ \Rightarrow 2\left( { -P+10 } \right) -1\left( { 6P-28 } \right) +\left( { -30+14 } \right) =0 \ \Rightarrow -2P+20-6P+28-16=0 \ \Rightarrow -8P+32=0 \ \therefore P=4, $

If $\vec{a}$, $\vec{b}$ and $\vec{c}$ are collinear vectors, therefore they lie on the same line. Hence they are parallel. Hence there cross products will be 0.
Therefore
$(\vec{a}\times\vec{b})=(\vec{c}\times\vec{b})=(\vec{a}\times\vec{c})=0$
And application of the section formula gives us
$\vec{c}=\dfrac{\lambda \vec{a}+\mu\vec{b}}{\lambda+\mu}$
Or
$\vec{c}(\lambda+\mu)=\lambda \vec{a}+\mu\vec{b}$

Since, $\vec{a}+2\vec{b}$ is collinear with $\vec{c}$.
$\therefore \vec{a}+2\vec{b}=x\vec{c}, x\epsilon R$ and $\vec{b}+3\vec{c}$ is collinear with $\vec{a}$.
$\therefore \vec{b}+3\vec{c}=y\vec{a}, y\epsilon R$
$\Rightarrow \vec{a}+2\vec{b}+6\vec{c}=(1+2y)\vec{a}$
Also, $\vec{a}+2\vec{b}+6\vec{c}=(x+6)\vec{c}$
$\therefore (x+6)\vec{c}=(1+2y)\vec{a}$
$\Rightarrow x+6=0$
and $1+2y=0$
$\Rightarrow x=-6$ and $y=-1/2$
$\therefore \vec{a}+2\vec{b}+6\vec{c}=\vec{0}$

Given points are $A(i+8j-5k)$ , $B(6i-2j)$ and $C(9i+4j-3k)$

Consider the given points $A(\bar{a}),B(\bar{b}),C(\bar{c})$.
These three points determine two vectors $\vec{AB}$ and $\vec{AC}$

We know that, "two vectors $a,b$ are collinear if and only if $\vec{a} \times \vec{b}=0$".

Therefore two vectors $\vec{AB}$ and $\vec{AC}$ are collinear if and only if $\vec{AB} \times \vec{AC}=0$

$\vec{AB}=\vec{OB}-\vec{OA}=\bar{b}-\bar{a}$ and $\vec{AC}=\vec{OC}-\vec{OA}=\bar{c}-\bar{a}$

We have, two vectors $\vec{AB}$ and $\vec{AC}$ are collinear if and only if $\vec{AB} \times \vec{AC}=0$

$ \Rightarrow$ two vectors $\vec{AB}$ and $\vec{AC}$ are collinear if and only if $(\bar{b}-\bar{a}) \times (\bar{c}-\bar{a})=0$

Since the three points determine two vectors $\vec{AB}$ and $\vec{AC}$, we conclude that

Three points $A(\bar{a}),B(\bar{b}),C(\bar{c})$ are collinear if and only if $(\bar{b}-\bar{a}) \times (\bar{c}-\bar{a})=0$.