#### . Find the length and the foot of the perpendicular from the point (1, 3/2, 2) to the plane 2x - 2y + 4z + 5 = 0.

Given, point P (1, 3/2, 2)

The plane is 2x - 2y + 4z + 5 = 0

We must find the foot of the perpendicular from the point P to the equation of the given plane.

Also, we must find the distance from the point P to the plane.

Let us consider the foot of the perpendicular from point P to be Q.

Let Q be Q (x1 , y1 , z1)

So, the direction ratio of PQ is given by

(x1 - 1, y1 - 3/2, z1 - 2)

Now, let us consider the normal to the plane 2x - 2y + 4z + 5 = 0:

It is obviously parallel to PQ, since a normal is a line or vector that is perpendicular to a given object. The direction ratio simply states the number of units to move along each axis.

For any plane, ax + by + cz = d, where, a, b, and c are normal vectors to the plane.

Hence, the direction ratios are (a, b, c).

Therefore, the direction ratio of the normal = (2, -2, 4) for plane 2x - 2y  + 4z + 5 = 0.

The Cartesian equation of the line PQ, where P(1, 3/2, 2) and Q (x1 , y1 , z1)   is:

$\frac{x_{1}-1}{2}=\frac{y_{1}-\frac{3}{2}}{-2}=\frac{z_{1}-2}{4}=\lambda(say)$

To find any point on this line,

$\frac{x_{1}-1}{2}=\lambda,\frac{y_{1}-\frac{3}{2}}{-2}=\lambda,\frac{z_{1}-2}{4}=\lambda$

$from \frac{x_{1}-1}{2}=\lambda\\\Rightarrow x_{1}-1=2\lambda\\\Rightarrow x_{1}=2\lambda+1\\ \\ \\ from \frac{y_{1}-\frac{3}{2}}{-2}=\lambda\\ \Rightarrow y_{1}-\frac{3}{2}=-2\lambda\\\Rightarrow y_{1}=\frac{3}{2}-2\lambda\\ \\ \\ from \frac{z_{1}-2}{4}=\lambda\\\Rightarrow z_{1}-2=4\lambda \\\Rightarrow z_{1}=4\lambda+2$

Any point on the line is $(2\lambda+ 1, \frac{3}{2} - 2\lambda, 4\lambda + 2).$

This point is Q.

$Q\left ( x_{1},y_{1},z_{1} \right )=Q(2\lambda+ 1, \frac{3}{2} - 2\lambda, 4\lambda + 2)....(i)$

And, it was assumed that is lies on the given plane. Substituting x1, y1, and z1 in the plane equation, we get:

2x1 - 2y1 + 4z1 + 5 = 0

$\Rightarrow 2\left (2\lambda+ 1 \right )-2\left ( \frac{3}{2} - 2\lambda \right )+4\left (4\lambda + 2 \right )+5=0$

Simplifying to find the value of $\lambda$

$\Rightarrow 4\lambda + 2 - 3 + 4\lambda + 16\lambda + 8 + 5 = 0\\ \Rightarrow 4\lambda + 4\lambda + 16\lambda + 2 - 3 + 8 + 5 = 0\\ \Rightarrow 24\lambda + 12 = 0\\ \Rightarrow 24\lambda = -12\\ \Rightarrow \lambda =-\frac{12}{24}\\ \Rightarrow \lambda =-\frac{1}{2}$

Since Q is the foot of the perpendicular from the point P,

We substitute the value of $\lambda$ in equation (i) to get:

$Q\left ( x_{1},y_{1},z_{1} \right )=Q(2(-\frac{1}{2})+ 1, \frac{3}{2} - 2(-\frac{1}{2}), 4(-\frac{1}{2}) + 2)\\ \Rightarrow Q\left ( x_{1},y_{1},z_{1} \right )=Q\left ( -1+1,\frac{3}{2}+1,-2+2 \right )\\ \Rightarrow Q\left ( x_{1},y_{1},z_{1} \right )=Q\left ( 0,\frac{5}{2},0 \right )$

Then, to find $\left | \vec{PQ} \right |$

Where, P = (1, 3/2, 2) and Q = (0, 5/2, 0)

$\left | \vec{PQ} \right |=\sqrt{\left (0-1 \right )^{2}+\left ( \frac{5}{2}-\frac{3}{2} \right )^{2}+\left ( 0-1 \right )^{2}}\\ \Rightarrow \left | \vec{PQ} \right |=\sqrt{(-1)^{2}+(1)^{2}+(-2)^{2}}\\ \Rightarrow \left | \vec{PQ} \right |=\sqrt{1+1+4}\\ \Rightarrow \left | \vec{PQ} \right |=\sqrt{6}$

Thus, the foot of the perpendicular from the given point to the plane is   (0, 5/2, 0) and the distance is $\sqrt{6}$ units.