Equation of the plane through origin and perpendicular through the line $\frac{x - 1}{5} = \frac{y - 2}{4} = \frac{z - 3}{5}$ is

x + y - z = 0

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x - 5 y + 3 z = 0

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x + 5 y - 5 z = 0

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2 x + 5 y - 6 z = 0

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Solution

The correct option is B $x - 5 y + 3 z = 0$
$\frac{x - 1}{5} = \frac{y - 2}{4} = \frac{2 - 3}{5}$ …….. $(1)$

DRs of the given line $(5, 4, 5)$

plane passes through $(1, 2, 3)$ from $(1)$

also through $(0, 0, 0)$

DRs of line perpendicular to the plane

$= ∣ ∣ ∣ ∣ ∣ \begin{matrix} ^i & ^j & ^k 1 & 2 & 3 5 & 4 & 5 \end{matrix} ∣ ∣ ∣ ∣ ∣$

$=^i (10 - 12) -^j (5 - 15) +^k (4 - 10)$

$= - 2^i + 10^j - 6^k$

Direction ratios $= (- 2, 10, - 6)$

$∴$ Equation of plane

$- 2 (x - 0) + 10 (y - 0) - 6 (z - 0) = 0$

$\Rightarrow - 2 x + 10 y - 6 z = 0$

(or) $x - 5 y + 3 z = 0$ .

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Similar questions

\frac{x}{1} = \frac{y}{2} = \frac{z}{3}

Find the value of $x, y$ and $z$ from the given linear equations: $x + y + z = 0, 2 x + 5 y + 7 z = 0, 2 x - 5 y + 3 z = 0$

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x + y + z = 6

2 x + 3 y + 4 z + 5 = 0

4 x + 5 y - 3 z - 8 = 0

2 x + 3 y - 4 z + 5 = 0

x - 5 y + z + 6 = 0

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