The plane containing the line $\frac{x - 3}{2} = \frac{y + 2}{- 1} = \frac{z - 1}{3}$ and also containing its projection on the plane $2 x + 3 y - z = 5$ , contains which one of the following point?

(2, 0, - 2)

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(- 2, 2, 2)

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(0, - 2, 2)

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(2, 2, 0)

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Solution

The correct option is A $(2, 0, - 2)$
The normal vector of required plane

$= (2^i -^j + 3^k) \times (2^i + 3^j -^k)$

$= - 8^i + 8^j + 8^k$

So, direction ratio of normal is $(- 1, 1, 1)$

So required plane is

$- (x - 3) + (y + 2) + (z - 1) = 0$

$⟹ - x + y + z + 4 = 0$

Which is satisfied by $(2, 0, - 2)$

Suggest Corrections

Similar questions

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

2 x + 3 y - z = 5

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

2 x + 3 y - z = 5

(3, 2, 0)

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

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

π : x - 2 y + z = 6

π

L

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

\frac{x}{4} = \frac{y - 2}{- 2} = \frac{z + 1}{6}

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

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