If the distance between the planes x – $$2y$$ $$+$$ z $$=$$ d and the plane containing the lines x−$$12=y$$−$$23=z$$−$$34$$ and x−$$23=y$$−$$34=z$$−$$45$$ is $$6$$, then |d| is

Equation of the plane containing the given line is: x−$$1y$$−$$2z$$−$$3234345=0x-2y+z=0$$ Equation of given plane is x−$$2y+z=d$$ . Distance between the planes is $$6$$ . ∴|$$d-0$$|$$1+4+1=6$$⇒|d|$$=6$$

Planes in 3D

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If the distance between the planes x – 2y + z = d and the plane containing the lines $\frac{x - 1}{2} = \frac{y - 2}{3} = \frac{z - 3}{4} and \frac{x - 2}{3} = \frac{y - 3}{4} = \frac{z - 4}{5} is \sqrt{6}, then | d | is$

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Solution

Equation of the plane containing the given line is:
$|\begin{array}{ccc} x - 1 & y - 2 & z - 3 \\ 2 & 3 & 4 \\ 3 & 4 & 5 \end{array}| = 0$
$x - 2 y + z = 0$
$Equation of given plane is x - 2 y + z = d .$
$Distance between the planes is \sqrt{6} .$
$\begin{array}{l} ∴ \frac{| d - 0 |}{\sqrt{1 + 4 + 1}} = \sqrt{6} \\ \Rightarrow | d | = 6 \end{array}$

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Planes in 3D

Remember concepts with our Masterclasses.

If the distance between the planes x – 2y + z = d and the plane containing the lines x−12=y−23=z−34 and x−23=y−34=z−45 is 6, then |d| is

Equation of the plane containing the given line is: x−1y−2z−3234345=0x-2y+z=0 Equation of given plane is x−2y+z=d . Distance between the planes is 6 . ∴|d-0|1+4+1=6⇒|d|=6

Ready to Test Your Skills?

free study material

If the distance between the planes x – 2y + z = d and the plane containing the lines $\frac{x - 1}{2} = \frac{y - 2}{3} = \frac{z - 3}{4} and \frac{x - 2}{3} = \frac{y - 3}{4} = \frac{z - 4}{5} is \sqrt{6}, then | d | is$