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

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(D) The direction ratios of the line passing through $A(x_1, y_1, z_1)$ and $B(x_2, y_2, z_2)$ are proportional to $(x_2 - x_1, y_2 - y_1, z_2 - z_1)$

Vedclass · Accepted Answer

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

Vedclass · Answer

(D) The direction ratios of the line passing through $A(x_1, y_1, z_1)$ and $B(x_2, y_2, z_2)$ are proportional to $(x_2 - x_1, y_2 - y_1, z_2 - z_1)$.For points $A(6, -7, -1)$ and $B(2, -3, 1)$,the direction ratios are $(2 - 6, -3 - (-7), 1 - (-1)) = (-4, 4, 2)$.Dividing by $2$,we get the simplified direction ratios as $(-2, 2, 1)$.The magnitude of the vector is $\sqrt{(-2)^2 + 2^2 + 1^2} = \sqrt{4 + 4 + 1} = \sqrt{9} = 3$.The direction cosines are $\pm \frac{-2}{3}, \pm \frac{2}{3}, \pm \frac{1}{3}$.The angle $\alpha$ made with the $x$-axis is acute,which means $\cos \alpha > 0$. The direction cosine corresponding to the $x$-axis is $l = \pm \frac{-2}{3}$.To satisfy $\cos \alpha > 0$,we must choose the negative sign for the direction ratios so that $l = -(\frac{-2}{3}) = \frac{2}{3}$.Thus,the direction cosines are $(\frac{2}{3}, -\frac{2}{3}, -\frac{1}{3})$.

$A$ 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 $\alpha$ made with the $x$-axis is acute.

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