Respuesta :
Answer:
Distance of the point from its image = 8.56 units
Step-by-step explanation:
Given,
Co-ordinates of point is (-2, 3,-4)
Let's say
[tex]x_1\ =\ -2[/tex]
[tex]y_1\ =\ 3[/tex]
[tex]z_1\ =\ -4[/tex]
Distance is measure across the line
[tex]\dfrac{x+2}{3}\ =\ \dfrac{2y+3}{4}\ =\ \dfrac{3z+4}{5}[/tex]
So, we can write
[tex]\dfrac{x-x_1+2}{3}\ =\ \dfrac{2(y-y_1)+3}{4}\ =\ \dfrac{3(z-z_1)+4}{5}\ =\ k[/tex]
[tex]=>\ \dfrac{x-(-2)+2}{3}\ =\ \dfrac{2(y-3)+3}{4}\ =\ \dfrac{3(z-(-4))+4}{5}\ =\ k[/tex]
[tex]=>\ \dfrac{x+4}{3}\ =\ \dfrac{2y-3}{4}\ =\ \dfrac{3z+16}{5}\ =\ k[/tex]
[tex]=>\ x\ =\ 3k-4,\ y\ =\ \dfrac{4k+3}{2},\ z\ =\ \dfrac{5k-16}{3}[/tex]
Since, the equation of plane is given by
x+y+z=3
The point which intersect the point will satisfy the equation of plane.
So, we can write
[tex]3k-4+\dfrac{4k+3}{2}+\dfrac{5k-16}{3}\ =\ 3[/tex]
[tex]=>6(3k-4)+3(4k+3)+2(5k-16)\ =\ 18[/tex]
[tex]=>18k-24+12k+9+10k-32\ =\ 18[/tex]
[tex]=>\ k\ =\dfrac{13}{8}[/tex]
So,
[tex]x\ =\ 3k-4[/tex]
[tex]=\ 3\times \dfrac{13}{8}-4[/tex]
[tex]=\ \dfrac{7}{4}[/tex]
[tex]y\ =\ \dfrac{4k+3}{2}[/tex]
[tex]=\ \dfrac{4\times \dfrac{13}{8}+3}{2}[/tex]
[tex]=\ \dfrac{19}{4}[/tex]
[tex]z\ =\ \dfrac{5k-16}{3}[/tex]
[tex]=\ \dfrac{5\times \dfrac{13}{8}-16}{3}[/tex]
[tex]=\ \dfrac{-21}{8}[/tex]
Now, the distance of point from the plane is given by,
[tex]d\ =\ \sqrt{(x-x_1)^2+(y-y_1)^2+(z-z_1)^2}[/tex]
[tex]=\ \sqrt{(-2-\dfrac{7}{4})^2+(3-\dfrac{19}{4})^2+(-4+\dfrac{21}{8})^2}[/tex]
[tex]=\ \sqrt{(\dfrac{-15}{4})^2+(\dfrac{-7}{4})^2+(\dfrac{9}{8})^2}[/tex]
[tex]=\ \sqrt{\dfrac{225}{16}+\dfrac{49}{16}+\dfrac{81}{64}}[/tex]
[tex]=\ \sqrt{\dfrac{1177}{64}}[/tex]
[tex]=\ 4.28[/tex]
So, the distance of the point from its image can be given by,
D = 2d = 2 x 4.28
= 8.56 unit
So, the distance of a point from it's image is 8.56 units.
