Question

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

Answer 1

Answer:

Length of the foot of perpendicular $3\sqrt{21} \:\:units$

Step-by-step explanation:

To find the length and foot of the perpendicular from the point (7, 14, 5) to the plane

2x + 4y - z = 2....eq1

the direction ratio's of the normal to this plane are 2,4,-1,because the standard equation of a plane in cartesian form is ax+by+cz+d=0,where a,b and c are direction ratio

equation of the line passing through the points (7,14,5) with direction ratio's 2,4,-1 are

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

so,general point on the line is $(2\lambda+7,4\lambda+14,-\lambda+5)$

for some value $\lambda$.

let the points Q$(2\lambda+7,4\lambda+14,-\lambda+5)$ lie on the plane of eq1,thus it satisfies the equation

$2(2\lambda+7)+4(4\lambda+14)-(-\lambda+5) = 2\\\\\lambda=-3$

the coordinates of the foot of the perpendicular PQ are

Q[(2(-3)+7,4(-3)+14,-(-3+5)]

Q(1,2,8)

Distance between two points P(7, 14, 5) and Q(1,2,8)

$PQ=\sqrt{(7-1)^{2}+(14-2)^{2}+(5-8)^{2}   } \\\\\\=\sqrt{36+144+9}\\ \\ \\=\sqrt{189}\\\\=3\sqrt{21} \:\:units$

Answer 2

Answer:

it's a simple question

also it is in sample paper