If P, Q, R, S are the points (4, 5, 3) (6, 3, 4), (2, 4, -1), (0, 5, 1) the length of projection RS and PQ is

\frac{4}{3}

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\frac{2}{3}

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4

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6

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Solution

The correct option is A $\frac{4}{3}$
Projection of RS on PQ
$= | (2 - 0) \frac{2}{3} + (4 - 5) (- \frac{2}{3}) + (- 1 - 1) \frac{1}{3} | = | \frac{4}{3} + \frac{2}{3} - \frac{2}{3} | = | \frac{4}{3} | = \frac{4}{3}$

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Similar questions

P = (3, 4, 5)

Q = (4, 6, 3)

R = (- 1, 2, 4)

S = (1, 0, 5)

R S

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P = (3, 4, 5), Q = (4, 6, 3), R = (- 1, 2, 4)

S = (1, 0, 5)

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If the points P , Q, R and S are collinear points and PQ=RS then prove that PR=QS using Euclid’s postulate.

P (1, 0, 1), Q (2, 0, - 3), R (- 1, 2, 0)

S (3, - 2, - 1)

- - \to P Q

- - \to R S

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