The distance between the planes 2 x -3 y -4 z -4 and 4 x -6 y -8 z -12 isA- 2 unitsB- 4 unitsC- 8 unitsD- 2-29 units

The correct option is D 2√(29) units2x+3y+4z=4 4x+6y+8z=12 ⇒ 2x+3y+4z=6 nbsp; Both the planes are parallel. Required distance=|4 6|√(2^2+3^2+4^2)=2√(29) nb ...

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Standard XII

Mathematics

Condition for Two Lines to be on the Same Plane

The distance ...

Question

The distance between the planes $2 x + 3 y + 4 z = 4$ and $4 x + 6 y + 8 z = 12$ is

2 units

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4 units

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8 units

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\frac{2}{\sqrt{29}} units

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Solution

The correct option is D $\frac{2}{\sqrt{29}} units$
$2 x + 3 y + 4 z = 4$
$4 x + 6 y + 8 z = 12 \Rightarrow 2 x + 3 y + 4 z = 6$
Both the planes are parallel.
$Required distance = \frac{| 4 - 6 |}{\sqrt{2^{2} + 3^{2} + 4^{2}}} = \frac{2}{\sqrt{29}}$

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Similar questions

Q. The distance between the planes

2 x + 3 y + 4 z = 4

and

4 x + 6 y + 8 z = 12

(a) \frac{2}{\sqrt{29}} units (b) \frac{10}{\sqrt{29}} units

(c) 4 units (d) 2 units

Choose the correct answer in Question
Distance between the two planes 2x + 3y + 4z = 4 and 4x + 6y+ 8z =12 is
(a) 2 units
(b) 4 units
(c) 8 units
(d) $\frac{2}{\sqrt{29}} u n i t s$

Distance between the two planes: and is

(A)2 units (B)4 units (C)8 units

(D)

Q. Distance between the two planes :

2 x + 3 y + 4 z = 4

and

4 x + 6 y + 8 z = 12

The distance between two parallel planes 2x + 3y + 4z = 4 and 4x + 6y + 8z = 12 is

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The distance between the planes 2x+3y+4z=4 and 4x+6y+8z=12 is

The correct option is D 2√29 units2x+3y+4z=4 4x+6y+8z=12⇒2x+3y+4z=6 Both the planes are parallel. Required distance=|4−6|√22+32+42=2√29

The distance between the planes $2 x + 3 y + 4 z = 4$ and $4 x + 6 y + 8 z = 12$ is

The correct option is D $\frac{2}{\sqrt{29}} units$
$2 x + 3 y + 4 z = 4$
$4 x + 6 y + 8 z = 12 \Rightarrow 2 x + 3 y + 4 z = 6$
Both the planes are parallel.
$Required distance = \frac{| 4 - 6 |}{\sqrt{2^{2} + 3^{2} + 4^{2}}} = \frac{2}{\sqrt{29}}$