Let the foot of the perpendicular from the point \((1, 2, 4)\) on the line \(\frac{x+2}{4}=\frac{y−1}{2}=\frac{z+1}{3}\) be \(P\), Then the distance of \(P\) from the plane \(3x+4y+12z+23=0\) is
- \(5\)
- \(\frac{50}{13}\)
- \(4\)
- \(\frac{63}{13}\)
The Correct Option is A
Solution and Explanation
\(L:\) \(\frac{x+2}{4}=\frac{y−1}{2}=\frac{z+1}{3}\)
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
\(\begin{vmatrix}\frac{6+12+24+23}{\sqrt{9+16+144}}\end{vmatrix}=\frac{65}{13}=5\)
Top Questions on Distance of a Point From a Line
- If the image of the point \( P(1, 2, a) \) in the line \[ \frac{x - 6}{3} = \frac{y - 7}{2} = \frac{7 - z}{2} \] is \( Q(5, b, c) \), then \( a^2 + b^2 + c^2 \) is equal to
- JEE Main - 2026
- Mathematics
- Distance of a Point From a Line
- Find the point on the line \( \frac{x-1}{3} = \frac{y+1}{2} = \frac{z-4}{3} \) at a distance of \( \sqrt{2} \) units from the point \( (-1, -1, 2) \).
- CBSE CLASS XII - 2025
- Mathematics
- Distance of a Point From a Line
- Find the foot of the perpendicular drawn from the point \( (1, 1, 4) \) on the line \( \frac{x+2}{5} = \frac{y+1}{2} = \frac{z-4}{-3} \).
- CBSE CLASS XII - 2025
- Mathematics
- Distance of a Point From a Line
- Find the point on the line \( \frac{x-1}{3} = \frac{y+1}{2} = \frac{z-4}{3} \) at a distance of \( \sqrt{2} \) units from the point \( (-1, -1, 2) \).
- CBSE CLASS XII - 2025
- Mathematics
- Distance of a Point From a Line
- The distance of the point \( P(-3,4,5) \) from the yz-plane is:
- KCET - 2025
- Mathematics
- Distance of a Point From a Line
Questions Asked in JEE Main exam
- Identify the correct statements: The presence of –NO\(_2\) group in benzene ring A. activates the ring towards electrophilic substitutions. B. deactivates the ring towards electrophilic substitutions. C. activates the ring towards nucleophilic substitutions. D. deactivates the ring towards nucleophilic substitutions. Choose the correct answer from the options given below:
- JEE Main - 2026
- General Chemistry
- If the distances of the point \( (1,2,a) \) from the line \[ \frac{x-1}{1}=\frac{y}{2}=\frac{z-1}{1} \] along the lines \[ L_1:\ \frac{x-1}{3}=\frac{y-2}{4}=\frac{z-a}{b} \quad \text{and} \quad L_2:\ \frac{x-1}{1}=\frac{y-2}{4}=\frac{z-a}{c} \] are equal, then \( a+b+c \) is equal to:
- JEE Main - 2026
- Three Dimensional Geometry
Two circular discs of radius \(10\) cm each are joined at their centres by a rod, as shown in the figure. The length of the rod is \(30\) cm and its mass is \(600\) g. The mass of each disc is also \(600\) g. If the applied torque between the two discs is \(43\times10^{-7}\) dyne·cm, then the angular acceleration of the system about the given axis \(AB\) is ________ rad s\(^{-2}\).
- JEE Main - 2026
- Rotational motion
- In an experiment, a set of readings are obtained as follows: \[ 1.24~\text{mm},\ 1.25~\text{mm},\ 1.23~\text{mm},\ 1.21~\text{mm}. \] The expected least count of the instrument used in recording these readings is _______ mm.
- JEE Main - 2026
- General Physics
Method used for separation of mixture of products (B and C) obtained in the following reaction is:
- JEE Main - 2026
- p -Block Elements
Concepts Used:
Distance of a Point From a Line
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.
To Find the Distance Between two points:
The following steps can be used to calculate the distance between two points using the given coordinates:
- A(m1,n1) and B(m2,n2) are the coordinates of the two given points in the coordinate plane.
- The distance formula for the calculation of the distance between the two points is, d = √(m2 - m1)2 + (n2 - n1)2
- Finally, the given solution will be expressed in proper units.
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