What are the direction cosines of the line jo | vedclass

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(B) Let the points be $P(4, 3, -5)$ and $Q(-2, 1, -8)$.First,we find the direction ratios $(a, b, c)$ of the line $PQ$:$a = x_2 - x_1 = -2 - 4 = -6$$b

Vedclass · Accepted Answer

$\frac{6}{7}, \frac{2}{7}, \frac{3}{7}$

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(B) Let the points be $P(4, 3, -5)$ and $Q(-2, 1, -8)$.First,we find the direction ratios $(a, b, c)$ of the line $PQ$:$a = x_2 - x_1 = -2 - 4 = -6$$b = y_2 - y_1 = 1 - 3 = -2$$c = z_2 - z_1 = -8 - (-5) = -3$Alternatively,taking the vector $QP$:$a = 4 - (-2) = 6, b = 3 - 1 = 2, c = -5 - (-8) = 3$.The magnitude of the vector is $\sqrt{a^2 + b^2 + c^2} = \sqrt{6^2 + 2^2 + 3^2} = \sqrt{36 + 4 + 9} = \sqrt{49} = 7$.The direction cosines $(l, m, n)$ are given by $\frac{a}{\sqrt{a^2+b^2+c^2}}, \frac{b}{\sqrt{a^2+b^2+c^2}}, \frac{c}{\sqrt{a^2+b^2+c^2}}$.Thus,$(l, m, n) = (\frac{6}{7}, \frac{2}{7}, \frac{3}{7})$.

What are the direction cosines of the line joining the points $(4, 3, -5)$ and $(-2, 1, -8)$?

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