Tag: three dimensional geometry

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.