The line x-2-3-y-3-4-z-4-5is parallel to the plane

Explanation for the correct option:Given: line x 2/3=y 3/4=z 4/5let x 2/3=y 3/4=z 4/5=λ3î+4ĵ+5k̂=a will be perpendicular to normal vector of a plane.Checking o ...

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Byju's Answer

Standard IX

Mathematics

If Two Sides of a Triangle Are Unequal, Then the Angle Opposite to the Greater Side Is Greater Than Angle Opposite to the Smaller Side

The line x-2/...

Question

The line $\frac{x - 2}{3} = \frac{y - 3}{4} = \frac{z - 4}{5}$ is parallel to the plane

$3 x + 4 y + 5 z = 7$

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$2 x + 3 y + 4 z = 0$

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$x + y - z = 2$

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$2 x + y - 2 z = 0$

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Solution

The correct option is D
$2 x + y - 2 z = 0$

Explanation for the correct option:
Given: line $\frac{x - 2}{3} = \frac{y - 3}{4} = \frac{z - 4}{5}$
let $\frac{x - 2}{3} = \frac{y - 3}{4} = \frac{z - 4}{5} = λ$
$3 \hat{i} + 4 \hat{j} + 5 \hat{k} = \vec{a}$ will be perpendicular to normal vector of a plane.
Checking option D
So,
$= 3 (2) + 4 (1) + 5 (- 2) = 6 + 4 - 10 = 0$
Hence, $2 x + y - 2 z = 0$ is parallel to $\frac{x - 2}{3} = \frac{y - 3}{4} = \frac{z - 4}{5}$ .
Explanation for the in-correct options:
Option A $3 x + 4 y + 5 z = 7$
$3 \hat{i} + 4 \hat{j} + 5 \hat{k} = \vec{a}$ and normal vecotor to the plane is $3 \hat{i} + 4 \hat{j} + 5 \hat{k} = \vec{n}$
Both vector are parallel so option (A) is incorrect.
Option(B) $2 x + 3 y + 4 z = 0$
$2 \hat{i} + 3 \hat{j} + 4 \hat{k} = \vec{n}$ and $3 \hat{i} + 4 \hat{j} + 5 \hat{k} = \vec{a}$
$\vec{a} . \vec{n} = 6 + 12 + 20 \neq 0$
So, option B is also incorrect.
Option(C) $x + y - z = 2$
$\hat{i} + \hat{j} - \hat{k} = \vec{n}$ and $3 \hat{i} + 4 \hat{j} + 5 \hat{k} = \vec{a}$
$\vec{a} . \vec{n} = 3 + 4 - 5 \neq 0$
So, option C is also incorrect.
Hence,option D is correct.

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Similar questions

Q. Find the equation of the plane through the intersection of the planes 3x − 4y + 5z = 10 and 2x + 2y − 3z = 4 and parallel to the line x = 2y = 3z.

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If the distance between the plane, $23 x - 10 y - 2 z + 48 = 0$ and the plane containing the lines $\frac{x + 1}{2} = \frac{y - 3}{4} = \frac{z + 1}{3}$ and $\frac{x + 3}{2} = \frac{y + 2}{6} = \frac{z - 1}{λ}$ $(λ \in ℝ)$ is equal to $\frac{k}{\sqrt{633}}$ then $k$ is equal to

Q. The line

\frac{x - 2}{3} = \frac{y - 3}{4} = \frac{z - 4}{5}

is parallel to the plane.

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The line $\frac{x - 2}{3} = \frac{y - 3}{4} = \frac{z - 4}{5}$ is parallel to the plane