The foot of the perpendicular drawn from the origin to a plane is $(1, 2, - 3)$ . Find the equation of the plane.

x - 2 y - 3 z = 14

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x - 2 y + 3 z = 14

No worries! We‘ve got your back. Try BYJU‘S free classes today!

x + 2 y - 3 z = 14

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x + 2 y + 3 z = 14

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Solution

The correct option is C $x + 2 y - 3 z = 14$
Given:

Let the equation of plane passing through $(1, 2, 3)$ and perpendicular to $O A$ .

$A =^i + 2^j - 3^k$

$\to n = - - \to O A =^i + 2^j - 3^k$
$a = |^i + 2^j - 3^k | = \sqrt{1 + 4 + 9} = \sqrt{14}$

$^n = \frac{^i + 2^j - 3^k}{\sqrt{14}}$
The equation of the required plane is $\to r \cdot \to n = a$

$\Rightarrow \to r \cdot (\frac{^i + 2^j - 3^k}{\sqrt{14}}) = \sqrt{14}$

$\Rightarrow \to r \cdot (^i + 2^j - 3^k) = 14$

$\Rightarrow x + 2 y - 3 z = 14$

Thus, the equation of the plane is $x + 2 y - 3 z = 14$ .

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L_{1} : \frac{x}{3} = \frac{y + 1}{0} = \frac{z - 2}{- 1}

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