Question:
Let the plane P contain the line 2x+y-z-3=0=5x-3y+4z+9 and be parallel to the line \(\frac{x+2}{2}=\frac{3-y}{-4}=\frac{z-7}{5}\). Then the distance of the point
A(8, -1, -19) from the plane P measured parallel to the line \(\frac{x}{-3}=\frac{y-5}{4}=\frac{2-z}{-12}\) is equal to ___________.
Let the plane P contain the line 2x+y-z-3=0=5x-3y+4z+9 and be parallel to the line \(\frac{x+2}{2}=\frac{3-y}{-4}=\frac{z-7}{5}\). Then the distance of the point
A(8, -1, -19) from the plane P measured parallel to the line \(\frac{x}{-3}=\frac{y-5}{4}=\frac{2-z}{-12}\) is equal to ___________.
A(8, -1, -19) from the plane P measured parallel to the line \(\frac{x}{-3}=\frac{y-5}{4}=\frac{2-z}{-12}\) is equal to ___________.
Updated On: Mar 21, 2025
Hide Solution
Verified By Collegedunia
Correct Answer: 26
Solution and Explanation
Was this answer helpful?
0
0
Top Questions on Distance of a Point from a Plane
- The square of the distance of the point \(\left( \frac{15}{7}, \frac{32}{7}, 7 \right)\) from the line \(\frac{x+1}{3} = \frac{y+3}{5} = \frac{z+5}{7}\) in the direction of the vector \(\mathbf{i} + 4\mathbf{j} + 7\mathbf{k}\) is:
- JEE Main - 2025
- Mathematics
- Distance of a Point from a Plane
- An equation of a plane parallel to the plane \(x-2y+2z-5=0\) and at a unit distance from the origin is?
- JEE Main - 2024
- Mathematics
- Distance of a Point from a Plane
For \(a, b \in \mathbb{Z}\) and \(|a - b| \leq 10\), let the angle between the plane \(P: ax + y - z = b\) and the line \(L: x - 1 = a - y = z + 1\) be \(\cos^{-1}\left(\frac{1}{3}\right)\). If the distance of the point \((6, -6, 4)\) from the plane \(P\) is \(3\sqrt{6}\), then \(a^4 + b^2\) is equal to:
- JEE Main - 2023
- Mathematics
- Distance of a Point from a Plane
Let P₁ be the plane 3x-y-7z = 11 and P₂ be the plane passing through the points (2,-1,0), (2,0,-1), and (5,1,1). If the foot of the perpendicular drawn from the point (7,4,-1) on the line of intersection of the planes P₁ and P₂ is (α, β, γ), then a + ẞ+ y is equal to
- JEE Main - 2023
- Mathematics
- Distance of a Point from a Plane
- Let \(\lambda_1, \lambda_2\) be the values of \(\lambda\) for which the points \((1, \lambda, \frac{1}{2})\) and \((-2, 0, 1)\) are at equal distance from the plane \(2x + 3y - 6z + 7 = 0\). If \(\lambda_1 > \lambda_2\), then the distance of the point \((1 - \lambda_2, \lambda_2, \lambda_1)\) from the line \(\frac{x-5}{1} = \frac{y-1}{2} = \frac{z+7}{2}\) is:
- JEE Main - 2023
- Mathematics
- Distance of a Point from a Plane
View More Questions
Questions Asked in JEE Main exam
- If \( \sin x + \sin^2 x = 1 \), \( x \in \left(0, \frac{\pi}{2} \right) \), then the expression \[ (\cos^2 x + \tan^2 x) + 3(\cos^4 x + \tan^4 x + \cos^4 x + \tan^4 x) + (\cos^6 x + \tan^6 x) \] is equal to:
- JEE Main - 2025
- Trigonometric Identities
- The sum of the squares of all the roots of the equation \( x^2 + [2x - 3] - 4 = 0 \) is:
- JEE Main - 2025
- Quadratic Equations
- Let ABCD be a trapezium whose vertices lie on the parabola \( y^2 = 4x \). Let the sides AD and BC of the trapezium be parallel to the y-axis. If the diagonal AC is of length \( \frac{25}{4} \) and it passes through the point \( (1, 0) \), then the area of ABCD is:
- JEE Main - 2025
- Coordinate Geometry
- A molecule (P) on treatment with acid undergoes rearrangement and gives (Q). (Q) on ozonolysis followed by reflux under alkaline condition gives (R). The structure of (R) is given below. The structure of (P) is:
- JEE Main - 2025
- Organic Chemistry
- The product B formed in the following reaction sequence is: \[ {C}_6{H}_5{CN} \xrightarrow{{HCl}} (A) \xrightarrow{{AgCN}} (B) \]
- JEE Main - 2025
- Organic Chemistry
View More Questions