If alpha, beta, gamma are the direction angle | vedclass

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Question

(D) The given equations of the line are $x = 4z + 3$ and $y = -3z + 2$.Rewrite these in terms of $z$: $\frac{x - 3}{4} = z$ and $\frac{y - 2}{-3} = z$

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$\frac{2}{\sqrt{26}}$

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(D) The given equations of the line are $x = 4z + 3$ and $y = -3z + 2$.Rewrite these in terms of $z$: $\frac{x - 3}{4} = z$ and $\frac{y - 2}{-3} = z$.Thus, the symmetric form of the line is $\frac{x - 3}{4} = \frac{y - 2}{-3} = \frac{z - 0}{1}$.The direction ratios of the line are $(a, b, c) = (4, -3, 1)$.The magnitude of the direction vector is $\sqrt{4^2 + (-3)^2 + 1^2} = \sqrt{16 + 9 + 1} = \sqrt{26}$.The direction cosines are $\cos \alpha = \frac{4}{\sqrt{26}}$, $\cos \beta = \frac{-3}{\sqrt{26}}$, and $\cos \gamma = \frac{1}{\sqrt{26}}$.Summing these values: $\cos \alpha + \cos \beta + \cos \gamma = \frac{4 - 3 + 1}{\sqrt{26}} = \frac{2}{\sqrt{26}}$.

If $\alpha, \beta, \gamma$ are the direction angles of the line $x = 4z + 3, y = 2 - 3z$, then the value of $\cos \alpha + \cos \beta + \cos \gamma$ is...

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