If the foot of the perpendicular from the origin to a plane is $(1, 2, 3)$ , then the equation of the plane is ___.

3 x + 2 y + z = 14

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

x + 2 y + 3 z = 14

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

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

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

x + 2 y + 3 z = 6

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

Open in App

Solution

The correct option is A $x + 2 y + 3 z = 14$
The vector normal to plane is
$\to n =^i + 2^j + 3^k$
Hence, vector equation of plane is
$(\to r -_{o}) . \to n = 0$ where $\to r = x^i + y^j + z^k$
And, $_{o} =^i + 2^j + 3^k$
Hence, $x + 2 y + 3 z = 14$

Suggest Corrections

Similar questions

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

x - 1 = 2 - y = z - 3

\frac{x - 1}{3} = \frac{y - 5}{2} = \frac{z + 3}{- 4}

(1, - 2, 3)

(- 1, 2, 0)

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

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

The equation of the plane passing through the origin and perpendicular to the line x=2 y=3 z is

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

\frac{x}{3} = \frac{y}{4} = \frac{z}{2} and \frac{x}{4} = \frac{y}{2} = \frac{z}{3}

Join BYJU'S Learning Program