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
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What will be the equilibrium constant of the given reaction carried out in a \(5 \,L\) vessel and having equilibrium amounts of \(A_2\) and \(A\) as \(0.5\) mole and \(2 \times 10^{-6}\) mole respectively?
The reaction : \(A_2 \rightleftharpoons 2A\)
A black body is at a temperature of 2880 K. The energy of radiation emitted by this body with wavelength between 499 nm and 500 nm is U1, between 999 nm and 1000 nm is U2 and between 1499 nm and 1500 nm is U3. The Wien's constant, b = 2.88×106 nm-K. Then,
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