Question

Find the length and foot of the perpendicular from point P ( 7, 14, 5 ) to planes 2x + 4y - z = 2 . Also find the image of the point P in the plane.

Answer 1

Hope u can understand and bro u r in class 12th n u know very well how this que is solve

Answer 2

Answer:

The length of perpendicular  from point P ( 7, 14, 5 ) to planes 2x + 4y - z = 2   is  $\sqrt{189}$

Image of Point P is (-5,-10,11).

Step-by-step explanation:

Refer the attached figure .

Let x,y,z be the coordinates of point M.

Equation of line PM

$\frac{x-7}{2} = \frac{y-14}{4}=\frac{z-5}{-1}= \lambda$

$x=2\lambda +7 , y = 4\lambda+14 , z = -\lambda+5$

Let x ,y ,z be the points that satisfies the equation of plane

$2(2\lambda +7)+4(4\lambda+14 )-(-\lambda+5)=2$

$4\lambda +14+16\lambda+56+\lambda-5=2$

$21\lambda +65=2$

$\lambda=\frac{-63}{21}=-3$

So, $x=2(-3) +7 , y = 4(-3)+14 , z = -(-3)+5$

$x=1 , y = 2, z = 8$

So, M =(1,2,8)

Length of PM = $\sqrt{(x_2-x_1)^2+(y_2-y_2)^2+(z_2-z_1)^2}$

Length of PM = $\sqrt{(7-1)^2+(14-2)^2+(5-8)^2}$

Length of PM = $\sqrt{189}$

Thus the length of perpendicular  from point P ( 7, 14, 5 ) to planes 2x + 4y - z = 2   is  $\sqrt{189}$

Now To Find P' i.e. Image of Point P

So, M is the mid point of PP'

So, (1,2,8)=$(\frac{7+\alpha}{2} , \frac{14+\beta}{2},\frac{5+\gamma}{2})$

So, $\alpha = -5, \beta = -10 , \gamma = 11$

Thus Image of Point P is (-5,-10,11).