If the equation of the plane passing through the point $(1,1,2)$ and perpendicular to the line $x-3 y+$ $2 z-1=0=4 x-y+z$ is $A t+ B y+ C z=1$, then $140( C - B + A )$ is equal to______
Correct Answer: 15
Approach Solution - 1
We are given the equations of two planes:
\[ x - 3y + 2z - 1 = 0 \]
\[ 4x - y + z = 0 \]
Find the Direction Ratios of the Normal to the Plane
The direction ratios of the normals to the planes are:
\[ \vec{n}_1 = \langle 1, -3, 2 \rangle, \quad \vec{n}_2 = \langle 4, -1, 1 \rangle \]
The cross product \(\vec{n}_1 \times \vec{n}_2\) gives the direction ratios of the normal to the required plane:
\[ \vec{n}_1 \times \vec{n}_2 = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & -3 & 2 \\ 4 & -1 & 1 \end{vmatrix} = -\hat{i} + 7\hat{j} + 11\hat{k} \]
Thus, the direction ratios of the normal to the plane are:
\[ \langle -1, 7, 11 \rangle \]
Find the Equation of the Plane
The equation of the plane passing through the point \( (1, 1, 2) \) and having normal direction ratios \( -1, 7, 11 \) is:
\[ -1(x - 1) + 7(y - 1) + 11(z - 2) = 0 \]
Simplifying:
\[ -x + 7y + 11z = 28 \]
Normalize the Equation
Divide through by 28 to express the equation in the form \( Ax + By + Cz = 1 \):
\[ \frac{-1}{28}x + \frac{7}{28}y + \frac{11}{28}z = 1 \]
Here:
\[ A = \frac{-1}{28}, \quad B = \frac{7}{28}, \quad C = \frac{11}{28} \]
Verify the Given Expression
We are asked to compute:
\[ 140(C - B + A) \]
Substitute the values of \( A \), \( B \), and \( C \):
\[ 140 \left( \frac{11}{28} - \frac{7}{28} - \frac{1}{28} \right) = 140 \times \frac{3}{28} = 15 \]
Final Answer: 15
Approach Solution -2
Dr of normal to the plane is
Equation of plane :
Top Questions on Plane
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Questions Asked in JEE Main exam
- Density of 3 M NaCl solution is 1.25 g/mL. The molality of the solution is:
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In the given circuit the sliding contact is pulled outwards such that the electric current in the circuit changes at the rate of 8 A/s. At an instant when R is 12 Ω, the value of the current in the circuit will be A.- JEE Main - 2025
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Concepts Used:
Plane
A surface comprising all the straight lines that join any two points lying on it is called a plane in geometry. A plane is defined through any of the following uniquely:
- Using three non-collinear points
- Using a point and a line not on that line
- Using two distinct intersecting lines
- Using two separate parallel lines
Properties of a Plane:
- In a three-dimensional space, if there are two different planes than they are either parallel to each other or intersecting in a line.
- A line could be parallel to a plane, intersects the plane at a single point or is existing in the plane.
- If there are two different lines that are perpendicular to the same plane then they must be parallel to each other.
- If there are two separate planes which are perpendicular to the same line then they must be parallel to each other.