The cartesian from of equation a line passing through the point position vector $2^i -^j + 2^k$ and is in the direction of $- 2^i +^j +^k$ , is

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

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

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

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Solution

The correct option is A $\frac{x - 2}{- 2} = \frac{y + 1}{1} = \frac{z - 2}{1}$
Equation of a line passing through a point with position vector $\to a$ and parallel to $\to b$ is

$\to r = \to a + λ \to b$

Here $\to a = 2^i -^j + 2^k$ and $\to b = - 2^i +^j +^k$

So, $\to r = 2^i -^j + 2^k + λ (- 2^i +^j +^k)$

Equation of the line in vector form is $2^i -^j + 2^k + λ (- 2^i +^j +^k)$

Since the line passes through a point with position vector $2^i -^j + 2^k$

$∴ x_{1} = 2, y_{1} = - 1, z_{1} = 2$

Also, line is in the direction of $- 2^i +^j +^k$

Direction ratios : $a = - 2, b = 1, c = 1$

Equation of line in cartesian form is

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

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The equation of the line which passes through the point (1, 1, 1) and intersecting the lines $\frac{x - 1}{2} = \frac{y - 2}{3} = \frac{z - 3}{4}$ and $\frac{x + 2}{1} = \frac{y - 3}{2} = \frac{z + 1}{4}$ is

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x + 2 y - z - 5 = 0

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The cartesian from of equation a line passing through the point position vector 2^i−^j+2^k and is in the direction of −2^i+^j+^k, is

The cartesian from of equation a line passing through the point position vector $2^i -^j + 2^k$ and is in the direction of $- 2^i +^j +^k$ , is