Foot of the Perpendicular from a Point on a Plane

Q. Let $P(a,b,c)$ be any point on the plane $3x+2y+z=7.$ Then the least value of $2(a^2+b^2+c^2)$ is

Q. Find the coordinates of the foot of the perpendicular drawn from the origin to the plane 2x – 3y + 4z – 6 = 0.

1. $\left(12,-18,24\right)$
2. $\left(\frac{12}{29},-\frac{18}{29},\frac{24}{29}\right)$
3. $\left(\frac{12}{\sqrt{29}},-\frac{18}{\sqrt{29}},\frac{24}{\sqrt{29}}\right)$
4. $(29,-29,29)$

Q. Find the coordinates of the foot of the perpendicular drawn from the origin to the plane 2x – 3y + 4z – 6 = 0.

1. $\left(12,-18,24\right)$
2. $\left(\frac{12}{29},-\frac{18}{29},\frac{24}{29}\right)$
3. $\left(\frac{12}{\sqrt{29}},-\frac{18}{\sqrt{29}},\frac{24}{\sqrt{29}}\right)$
4. $(29,-29,29)$

Q.

Perpendicular are drawn from points on the line $\frac{x+2}{2}=\frac{y+1}{-1}=\frac{z}{3}$ to the plane x + y + z = 3. The feet of perpendiculars lie on the line

1. $\frac{x}{5}=\frac{y-1}{8}=\frac{z-2}{-13}$

2. $\frac{x}{2}=\frac{y-1}{3}=\frac{z-2}{-5}$

3. $\frac{x}{4}=\frac{y-1}{3}=\frac{z-2}{-7}$

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

Q. The length and foot of the perpendicular from the point (7, 14, 5) to the plane 2x + 4y - z = 2, are

1. $\sqrt{21},(1,2,8)$
2. $3\sqrt{21},(3,2,8)$
3. $21\sqrt{3},(1,2,8)$
4. $3\sqrt{21},(1,2,8)$