The direction cosines of the line frac{3x + 1 | vedclass

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(D) To find the direction cosines,first rewrite the equation of the line in the standard form $\frac{x - x_1}{a} = \frac{y - y_1}{b} = \frac{z - z_1}{

Vedclass · Accepted Answer

$\left( -\frac{1}{\sqrt{6}}, \frac{2}{\sqrt{6}}, -\frac{1}{\sqrt{6}} \right)$

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(D) To find the direction cosines,first rewrite the equation of the line in the standard form $\frac{x - x_1}{a} = \frac{y - y_1}{b} = \frac{z - z_1}{c}$.Given: $\frac{3x + 1}{-3} = \frac{3y + 2}{6} = \frac{z}{-1}$.Divide the numerator and denominator of each term by the coefficient of the variable:$\frac{3(x + 1/3)}{-3} = \frac{3(y + 2/3)}{6} = \frac{z}{-1} \implies \frac{x + 1/3}{-1} = \frac{y + 2/3}{2} = \frac{z}{-1}$.The direction ratios are $(a, b, c) = (-1, 2, -1)$.The magnitude of the direction vector is $\sqrt{(-1)^2 + 2^2 + (-1)^2} = \sqrt{1 + 4 + 1} = \sqrt{6}$.The direction cosines $(l, m, n)$ are given by $\left( \frac{a}{\sqrt{a^2+b^2+c^2}}, \frac{b}{\sqrt{a^2+b^2+c^2}}, \frac{c}{\sqrt{a^2+b^2+c^2}} \right)$.Thus,the direction cosines are $\left( \frac{-1}{\sqrt{6}}, \frac{2}{\sqrt{6}}, \frac{-1}{\sqrt{6}} \right)$.

The direction cosines of the line $\frac{3x + 1}{-3} = \frac{3y + 2}{6} = \frac{z}{-1}$ are

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