In the following cases, find the distance of each of the given points from the corresponding given plane.
Point Plane
(a) (0,0,0) 3x-4y+12z=3
(b) (3,-2,1) 2x-y+2z+3=0
(c) (2,3,-5) x+2y-2z=9
(d) (-6,0,0) 2x-3y+6z-2=0
It is known that the distance between a point P(x1,y1,z1)and a plane Ax-+By+Cz=D, is given by,
d=\(\begin{vmatrix}\frac{AX_1+By_1+Cz_1-D}{\sqrt{A^2+B^2+C^2}} \end{vmatrix}\)|...(1)
(a) The given point is (0,0,0) and the plane is 3x-4y+12z=3
∴d=\(\begin{vmatrix}\frac{3*0-4*0+12*0-3}{\sqrt{(3)^2+(-4)^2+(12)^2}} \end{vmatrix}\)
=\(\frac{3}{\sqrt{169}}\)=\(\frac{3}{13}\)
(b)The given point is (3,-2,1) and the plane is 2x-y+2z+3=0
∴d=\(\begin{vmatrix}\frac{2*3-(-2)+2*1+3}{\sqrt{(2)^2+(-1)^2+(2)^2}} \end{vmatrix}\)
=|\(\frac{13}{3}\)| =\(\frac{13}{3}\)
(c)The given point is (2,3,-5) and the plane is x+2y-2z=9
∴d=\(\begin{vmatrix}\frac{2+2*3-2(-5)-9}{\sqrt{(1)^2+(2)^2+(-2)^2}} \end{vmatrix}\)
=\(\frac{9}{3}\)
=3
(d)The given point is (-6,0,0) and the plane is 2x-3y+6z-2=0
∴d=\(\begin{vmatrix}\frac{2(-6)-3*0+6*0-2}{\sqrt{(2)^2+(-3)^2+(6)^2}} \end{vmatrix}\)
=\(\begin{vmatrix}\frac{-14}{\sqrt{49}} \end{vmatrix}\)
=\(\frac{14}{7}\)
=2
The vector equations of two lines are given as:
Line 1: \[ \vec{r}_1 = \hat{i} + 2\hat{j} - 4\hat{k} + \lambda(4\hat{i} + 6\hat{j} + 12\hat{k}) \]
Line 2: \[ \vec{r}_2 = 3\hat{i} + 3\hat{j} - 5\hat{k} + \mu(6\hat{i} + 9\hat{j} + 18\hat{k}) \]
Determine whether the lines are parallel, intersecting, skew, or coincident. If they are not coincident, find the shortest distance between them.
Show that the following lines intersect. Also, find their point of intersection:
Line 1: \[ \frac{x - 1}{2} = \frac{y - 2}{3} = \frac{z - 3}{4} \]
Line 2: \[ \frac{x - 4}{5} = \frac{y - 1}{2} = z \]
Determine the vector equation of the line that passes through the point \( (1, 2, -3) \) and is perpendicular to both of the following lines:
\[ \frac{x - 8}{3} = \frac{y + 16}{7} = \frac{z - 10}{-16} \quad \text{and} \quad \frac{x - 15}{3} = \frac{y - 29}{-8} = \frac{z - 5}{-5} \]
The shortest perpendicular distance from the point to the given plane is the distance between point and plane. In simple terms, the shortest distance from a point to a plane is the length of the perpendicular parallel to the normal vector dropped from the particular point to the particular plane. Let's see the formula for the distance between point and plane.
Read More: Distance Between Two Points