\(L:\) \(\frac{x+2}{4}=\frac{y−1}{2}=\frac{z+1}{3}\)
Let
\(P=(4t−2,2t+1,3t−1)\)
\(∵ P\) is the foot of perpendicular of \((1, 2, 4)\)
\(∴ 4(4t – 3) + 2(2t – 1) + 3(3t – 5) = 0\)
\(⇒29t=29⇒t=1\)
\(∴ P = (2, 3, 2)\)
Now, distance of \(P\) from the plane
\(3x + 4y + 12z + 23 = 0\), is
\(\begin{vmatrix}\frac{6+12+24+23}{\sqrt{9+16+144}}\end{vmatrix}=\frac{65}{13}=5\)
The motion of an airplane is represented by the velocity-time graph as shown below. The distance covered by the airplane in the first 30.5 seconds is km.
The length of the perpendicular drawn from the point to the line is the distance of a point from a line. The shortest difference between a point and a line is the distance between them. To move a point on the line it measures the minimum distance or length required.
The following steps can be used to calculate the distance between two points using the given coordinates:
Note: If the two points are in a 3D plane, we can use the 3D distance formula, d = √(m2 - m1)2 + (n2 - n1)2 + (o2 - o1)2.
Read More: Distance Formula