Determine the equation of straight line passing through the point with position vector $i - 3 j + k$ and parallel to the vector, $2 i + 3 j - 4 k$ .

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

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

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

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

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Solution

The correct option is C $\frac{x - 1}{2} = \frac{y + 3}{3} = \frac{z - 1}{- 4}$
Given position vector
$\to a =^i - 3^j +^k$
parallel to vector so also normal vector of new line
$\to b = 2^i + 3^j - 4^k$
eq of line
$\to r =^i - 3^j +^k + t (2^i + 3^j - 4^k)$
cartesian form
$\frac{x - 1}{2} = \frac{y + 3}{3} = \frac{z - 1}{- 4}$

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(- 4, 3, 1)

x + 2 y - z - 5 = 0

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

(2, 3, 1)

x - 2 y - z + 5 = 0 and x + y + 3 z = 6

(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}

(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}

(1, 1, 1)

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

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

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