The direction cosines of the line 6x - 2 = 3y | vedclass

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Question

(B) The given equation of the line is $6x - 2 = 3y + 1 = 2z - 2$.Rewrite the equation in the standard form $\frac{x - x_1}{a} = \frac{y - y_1}{b} = \f

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$\frac{1}{\sqrt{14}}, \frac{2}{\sqrt{14}}, \frac{3}{\sqrt{14}}$

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(B) The given equation of the line is $6x - 2 = 3y + 1 = 2z - 2$.Rewrite the equation in the standard form $\frac{x - x_1}{a} = \frac{y - y_1}{b} = \frac{z - z_1}{c}$:$6(x - \frac{1}{3}) = 3(y + \frac{1}{3}) = 2(z - 1)$Dividing by the least common multiple of $6, 3, 2$,which is $6$:$\frac{x - 1/3}{1/6} = \frac{y + 1/3}{1/3} = \frac{z - 1}{1/2}$Multiplying the denominators by $6$ to get direction ratios $(a, b, c)$:$a = 1, b = 2, c = 3$The magnitude of the direction vector is $\sqrt{1^2 + 2^2 + 3^2} = \sqrt{1 + 4 + 9} = \sqrt{14}$.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,the direction cosines are $\frac{1}{\sqrt{14}}, \frac{2}{\sqrt{14}}, \frac{3}{\sqrt{14}}$.

The direction cosines of the line $6x - 2 = 3y + 1 = 2z - 2$ are:

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