The position vector of a point in which a line through the origin perpendicular to the plane $2 x - y - z = 4$ meets the plane $\to r \cdot (3^i - 5^j + 2^k = 6$ , is

(1, - 1, - 1)

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(- 1, - 1, 2)

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(4, 2, 2)

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(\frac{4}{3}, \frac{- 2}{3}, \frac{- 2}{3})

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Solution

The correct option is D $(\frac{4}{3}, \frac{- 2}{3}, \frac{- 2}{3})$

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Similar questions

2 x - y - z = 4,

\to r \cdot (3^i - 5^j + 2^k) = 6

The point where the line $\frac{x - 1}{2} = \frac{y - 2}{- 3} = \frac{z + 3}{4}$ meets the plane 2x + 4y - z = 1, is [DSSE 1981]

(4, - 1, 2)

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

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

(3, - 1, 2)

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

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

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