A line passes through the points A(6, -7, -1) | vedclass

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(A) The direction ratios of the line passing through $A(6, -7, -1)$ and $B(2, -3, 1)$ are given by $(x_2 - x_1, y_2 - y_1, z_2 - z_1) = (2 - 6, -3 - (

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$\frac{2}{3}, -\frac{2}{3}, -\frac{1}{3}$

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(A) The direction ratios of the line passing through $A(6, -7, -1)$ and $B(2, -3, 1)$ are given by $(x_2 - x_1, y_2 - y_1, z_2 - z_1) = (2 - 6, -3 - (-7), 1 - (-1)) = (-4, 4, 2)$.The magnitude of the vector is $\sqrt{(-4)^2 + 4^2 + 2^2} = \sqrt{16 + 16 + 4} = \sqrt{36} = 6$.The direction cosines $(l, m, n)$ are $\left( \frac{-4}{6}, \frac{4}{6}, \frac{2}{6} \right) = \left( -\frac{2}{3}, \frac{2}{3}, \frac{1}{3} \right)$ or $\left( \frac{2}{3}, -\frac{2}{3}, -\frac{1}{3} \right)$.Let $\alpha$ be the angle made by the line with the positive $x$-axis. Then $\cos \alpha = l$. For $\alpha$ to be an acute angle,$\cos \alpha > 0$,which implies $l > 0$.Comparing the two sets of direction cosines,the set with $l > 0$ is $\left( \frac{2}{3}, -\frac{2}{3}, -\frac{1}{3} \right)$.

$A$ line passes through the points $A(6, -7, -1)$ and $B(2, -3, 1)$. Find the direction cosines of the line such that the angle made by the line with the positive direction of the $x$-axis is acute.

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