Tag: scalars and vectors

$\begin{matrix} then\, a=? \ The\, equation\, lineAB\, is\,  \ \Rightarrow \dfrac { { x-8 } }{ { 5-8 } } =\dfrac { { y+7 } }{ { 2+7 } } =\dfrac { { z-a } }{ { 4-a } } ....\left( 1 \right)  \ po{ int }\, c\, lies\, in\, the\, line\, AB \ po{ int }\left( { 6,-1,2 } \right) satisfy\, eq\left( 1 \right)  \ \Rightarrow \dfrac { { -2 } }{ { -3 } } =\dfrac { { -1+7 } }{ 9 } =\dfrac { { 2-a } }{ { 4-a } }  \ \Rightarrow \dfrac { 2 }{ 3 } =\dfrac { { 2-a } }{ { 4-a } }  \ \Rightarrow 8-2a=6-3a \ \Rightarrow 3a-2a=6-8 \ a=-2 \  \end{matrix}$

Eliminating t from the given equation, we get the equation of the path $\dfrac{x}{2}=\dfrac{y}{-4}=\dfrac{z}{4}=t $

AB${=}$ $\sqrt{{(7-5)}^{2}+{(-4+1)}^{2}+{(7-1)}^{2}}$
AB${=}$ $\sqrt{{(2)}^{2}+{(-3)}^{2}+{(6)}^{2}}$
AB${=}$ $\sqrt{49}$
AB${=}$ 7
Similarly you find that BC${=}$ $\sqrt{49}$  CD${=}$ 7  and DA${=}$7
Hence all sides of quadrilateral are equal, Now we check the diagonals
AC${=}$ $\sqrt{{(1-5)}^{2}+{(-6+1)}^{2}+{(10-1)}^{2}}$
AC${=}$ $\sqrt{122}$
similarly BD${=}$ $\sqrt{74}$ 
Diagonals are not equal
direction ratio of line passing through AC is (-4,-5,9)
direction ratio of line passing through  BD is (-8,1,-3), As the dot product dr of AC and BD are equal to 0 which means AC is perpendicular to BD,
All sides are equal and diagonal are not equal but bisect each other at right angle
hence it is rhombus

AB${=}$ $\sqrt{{(-1-1)}^{2}+{(-2-2)}^{2}+{(-1-3)}^{2}}$
AB${=}$ $\sqrt{{(-2)}^{2}+{(-4)}^{2}+{(-4)}^{2}}$
AB${=}$ $\sqrt{36}$
AB${=}$ 6
Similarly you find that BC${=}$ $\sqrt{43}$  CD${=}$ 6  and DA${=}$ $\sqrt{43}$
Hence opposite sides of quadrilateral are equal, Now we check the diagonals
AC${=}$ $\sqrt{{(2-1)}^{2}+{(3-2)}^{2}+{(2-3)}^{2}}$
AC${=}$ $\sqrt{3}$
similarly BD${=}$ $\sqrt{155}$ 
Diagonals are not equal
direction ratio of line passing through AB is (-2,-4,-4)
direction ratio of line passing through  CD is (2,4,4), As the dr of AB and CD are proportional which means AB is parallel to CD,
Similarly check for BC and DA then you will find that they are also parallel
hence it is parallelogram

Given points are, $A= \left ( 0,0,2 \right ),B= \left (\sqrt{2},\sqrt{2},2 \right ),C=

\left ( \sqrt{2},\sqrt{2},0 \right )$ and $D= \left ( \displaystyle

\frac{8\sqrt{2}-20}{17},\frac{12\sqrt{2}+4}{17},\frac{20-8\sqrt{2}}{17}

\right )$

Length $AB = \sqrt{\sqrt{2}^2 + \sqrt{2}^2 + (2-2)^2} = 2$
Length $BC = \sqrt{(\sqrt{2}-\sqrt{2})^2 + (\sqrt{2}-\sqrt{2})^2 + 2^2} = 2$
Length $CD = \sqrt{\left(

\dfrac{8\sqrt{2}-20}{17}-\sqrt{2}\right)^2 + \left(\dfrac{12\sqrt{2}+4}{17}-\sqrt{2}\right)^2 + \left(\dfrac{20-8\sqrt{2}}{17}\right)^2} = 2$
Length $AD = \sqrt{\left(

\dfrac{8\sqrt{2}-20}{17}\right)^2 + \left(\dfrac{12\sqrt{2}+4}{17}\right)^2 + \left(\dfrac{20-8\sqrt{2}}{17} - 2\right)^2} = 2$

Angle between two vectors $\overline{AB} = p _{1}\hat{i}+q _{1}\hat{j}+r _{1}\hat{k} = \sqrt{2}\hat{i} + \sqrt{2}\hat{j} + 0\hat{k}$ and $\overline{BC} = p _{2}\hat{i}+q _{2}\hat{j}+r _{2}\hat{k} = 0\hat{i} + 0\hat{j} + 2 \hat{k}$ is $cos\theta = 0 \Rightarrow \theta = 90^o$

All sides are equal and angle between $AB$ and $BC$ is $90^o$, Hence, its a square.

AB${=}$ $\sqrt{{(2-1)}^{2}+{(5-2)}^{2}+{(-2+1)}^{2}}$
AB${=}$ $\sqrt{{(1)}^{2}+{(3)}^{2}+{(-1)}^{2}}$
AB${=}$ $\sqrt{11}$
Similarly you find that BC${=}$ $\sqrt{6}$  CD${=}$ $\sqrt{11}$  and DA${=}$$\sqrt{6}$
Hence opposite sides of quadrilateral are equal, Now we check the diagonals
AC${=}$ $\sqrt{{(4-1)}^{2}+{(4-2)}^{2}+{(-3+1)}^{2}}$
AC${=}$ $\sqrt{17}$
similarly BD${=}$ $\sqrt{17}$ 
Diagonals are not equal
direction ratio of line passing through AB is (1,3,-1)
direction ratio of line passing through  BC is (2,-1,-1), As the dot product dr of AB and BC are equal to 0 which means AB is perpendicular to BC,similarly check for others sides too
opposite sides are equal and diagonal are equal
hence it is rectangle

The plane forming the parallelipiped are

Let $P(x,y,z)$ be the required point.