Question

Find the equation of the plane passing through the intersection of the planes $x - 2 y + z = 1$ and $2 x + y + z = 8$ , and parallel to the line with direction ratios $1, 2, 1$ . Also, find the perpendicular distance of $(1, 1, 1)$ from the plane.

Accepted Answer

Equation of plane through the line of intersection of planes in Cartesian form is

$A_{1} x + B_{1} y + C_{1} z + D_{1} + λ (A_{2} x + B_{2} y + C_{2} z + D_{2}) = 0$ ........(1)

For the standard equation of planes,

$A_{1} x + B_{1} y + C_{1} z + D_{1} = 0$ and $A_{2} x + B_{2} y + C_{2} z + D_{2} = 0$

So, putting $x - 2 y + z = 1$ and $2 x + y + z = 8$ in equation (1), we have

$x - 2 y + z - 1 + λ (2 x + y + z - 8) = 0$

$\Rightarrow (1 + 2 λ) x + (- 2 + 1) y + (1 + λ) z - 1 - 8 λ = 0$ .......(2)

If $A_{1} x + B_{1} y + C_{1} z + D_{1} = 0$ and $A_{2} x + B_{2} y + C_{2} z + D_{2} = 0$ are perpenidicular then $A_{1} A_{2} + B_{1} B_{2} + C_{1} C_{2} = 0$

$∴ (1 + 2 λ) \cdot 1 + (- 2 + 1) \cdot 2 + (1 + 1) \cdot 1 = 0$

$\Rightarrow 1 + 2 λ - 4 + 2 λ + 1 + λ = 0$

$\Rightarrow - 2 + 5 λ = 0$

$\Rightarrow λ = \frac{2}{5}$

Putting the value of $λ$ in equation (2)

$\Rightarrow (1 + 2 \cdot (\frac{2}{5})) \cdot x + (- 2 + \frac{2}{5}) \cdot y + (1 + \frac{2}{5}) \cdot z - 1 - 8 \cdot (\frac{2}{5}) = 0$

$\Rightarrow (\frac{5 + 4}{5}) x + (\frac{- 10 + 2}{5}) y + (\frac{5 + 2}{5}) z + \frac{- 5 - 16}{5} = 0$

$\Rightarrow 9 x - 8 y + 7 z - 21 = 0$

$\Rightarrow 9 x - 8 y + 7 z = 21$

Distance from a point $(x_{1}, y_{1}, z_{1})$ to a plane $A x + B y + C z + D = 0$ is $d = \frac{|A x_{1} + B y_{1} + C z_{1} + D|}{\sqrt{A^{2} + B^{2} + C^{2}}}$

Distance from $(1, 1, 1)$ to $9 x - 8 y + 7 z = 21$ is $|\frac{9.1 - 8.1 + 7.1 - 21}{\sqrt{(9)^{2} + (- 8)^{2} + (7)^{2}})}| = |\frac{9 - 8 + 7 - 21}{\sqrt{81} + 64 + 49}| = |\frac{13}{\sqrt{194}}|$

So, the required equation of the plane is $9 x - 8 y + 7 z = 21$ , and distance of the plane from $(1, 1, 1)$ is $d = \frac{13}{\sqrt{194}}$

R S Aggarwal Solutions for Exercise 5: EXERCISE

R S Aggarwal Mathematics Solutions for Exercise - R S Aggarwal Solutions for Exercise 5: EXERCISE

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