Determine the vector equation of the straight line passing through the two points, $P (3, - 1, 5)$ and $Q (- 1, - 4, 2)$ .

\to r = 3 i + j + 5 k - t (4 i - 3 j + 3 k)

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\to r = 4 i - 3 j + 3 k - t (3 i + j + 5 k)

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\to r = 3 i - j + 5 k - t (4 i + 3 j + 3 k)

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\to r = 4 i - 3 j + 3 k - t (3 i - j - 5 k)

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Solution

The correct option is C $\to r = 3 i - j + 5 k - t (4 i + 3 j + 3 k)$
given line is passing through P(3,-1,5) and Q(-1,-4,2)
position vector of line is P
$\to P = 3^i -^j + 5^k$
normal vector can be obtained by
$P Q = Q - P = -^i - 4^j + 2^k - (3^i -^j + 5^k)$
$P Q = - 4^i - 3^j - 3^k$
$\to n = P Q = - (4^i + 3^j + 3^k)$
vector eq of line
$\to r =_{0} + t \to n$
$\to r = 3^i -^j + 5^k - t (4^i + 3^j + 3^k)$

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