Let $P$ be an arbitrary point having sum of the squares of the distance from the planes $x+y+z=0$, $\ell x-n z=0$ and $x-2 y+z=0$ equal to 9. If the locus of the point $P$ is $x^{2}+y^{2}+z^{2}=9$, then the value of $\ell-n$ is equal to ____.

JEE Main · 2021 · 17 Mar 2021 · Shift 2 · Three Dimensional Geometry

Let

P

be an arbitrary point having sum of the squares of the distance from the planes

x + y + z = 0

ℓ x - n z = 0

and

x - 2 y + z = 0

equal to 9. If the locus of the point

P

x^{2} + y^{2} + z^{2} = 9

, then the value of

ℓ - n

is equal to ____.

Answer: 0

Solution

For

P (x, y, z)

the three squared distances are

\frac{( x + y + z ) ^{2}}{3}, \frac{( ℓ x - n z ) ^{2}}{ℓ ^{2} + n ^{2}}, \frac{( x - 2 y + z ) ^{2}}{6}

and each is the square of the component of

O P

along the corresponding unit normal.
If three unit vectors are mutually perpendicular, the squares of the components of any vector along them add up to the square of its length. The locus being

x^{2} + y^{2} + z^{2} = 9

is exactly this statement, so the three normals

(1, 1, 1)

(ℓ, 0, - n)

and

(1, - 2, 1)

must be mutually perpendicular.
The first and third already satisfy

1 - 2 + 1 = 0

. Perpendicularity of

(ℓ, 0, - n)

with

(1, 1, 1)

gives

ℓ - n = 0

, and with

(1, - 2, 1)

gives the same condition.
Hence

ℓ - n = 0

JEE Main · 2021 · 17 Mar 2021 · Shift 2 · Integer answer

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JEE Main · 2021 · 17 Mar 2021 · Shift 2 · Three Dimensional Geometry

Solution

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