Let the plane P contain the line 2 x + y - z - 3 = 0 = 5 x - 3 y + 4 z + 9 and be parallel to the line x + 2 2 = 3 - y - 4 = z - 7 5 . Then the distance of the point A ( 8 , - 1 , - 19 ) from the plane P measured parallel to the line x - 3 = y - 5

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Solution

Q.
Let the plane P contain the line $2 x + y - z - 3 = 0 = 5 x - 3 y + 4 z + 9$ 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 _______ . [2023]

Ans.
(26)

$Equation of the plane passing through the line of intersection of the given planes is$

$(2 x + y - z - 3) + λ (5 x - 3 y + 4 z + 9) = 0$

$\Rightarrow x (2 + 5 λ) + y (1 - 3 λ) + z (4 λ - 1) + 9 λ - 3 = 0 \dots (i)$

$The plane is parallel to the line$

$\frac{x + 2}{2} = \frac{3 - y}{- 4} = \frac{z - 7}{5}$    $i.e.,$    $\frac{x + 2}{2} = \frac{y - 3}{4} = \frac{z - 7}{5}$

$∴$ $2 (2 + 5 λ) + 4 (1 - 3 λ) + 5 (4 λ - 1) = 0$

$\Rightarrow 18 λ + 3 = 0 \Rightarrow λ = - \frac{1}{6}$

$Putting the value λ = - \frac{1}{6} in equation (i), we get$

$x (2 - \frac{5}{6}) + y (1 + \frac{3}{6}) + z (\frac{- 4}{6} - 1) - \frac{9}{6} - 3 = 0$

$\Rightarrow 7 x + 9 y - 10 z - 27 = 0$

$This is the required equation of plane.$

$Now, the given point is (8, - 1, - 19) and the equation of parallel line is \frac{x}{- 3} = \frac{y - 5}{4} = \frac{z - 2}{12}$

$∴$ $Equation of line passing through the given point and parallel to the line is$

   $\frac{x - 8}{- 3} = \frac{y + 1}{4} = \frac{z + 19}{12} = k$

$\Rightarrow x = - 3 k + 8, y = 4 k - 1, z = 12 k - 19$

$This will satisfy the plane: 7 x + 9 y - 10 z - 27 = 0$

   $- 21 k + 56 + 36 k - 9 - 120 k + 190 - 27 = 0$

$\Rightarrow - 105 k + 210 = 0 \Rightarrow k = 2$

$∴$    $x = 2, y = 7, z = 5$

$∴$ $Required distance = \sqrt{{(8 - 2)}^{2} + {(- 1 - 7)}^{2} + {(- 19 - 5)}^{2}}$

   $= \sqrt{36 + 64 + 576} = \sqrt{676} = 26 units$

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