Question

If \(α,β,γ\) are the direction angles of the line \(x = 4z+3\) and \(y = 2-3z\), then the value of \(cosα+cosβ+cosγ\) is...

• A. \(\frac{8}{\sqrt{26}}\)
• B. \(\frac{6}{\sqrt{26}}\)
• C. \(\frac{4}{\sqrt{26}}\)
• D. \(\frac{2}{\sqrt{26}}\)

Hint:

Write the line in symmetric form to read off direction ratios, then divide by the magnitude.

Answer 1

The Correct Option is D

Step 1: Parametrise
Take $z = t$. Then $x = 4t + 3$ and $y = 2 - 3t$, so the direction vector is $(4, -3, 1)$.

Step 2: Normalise
Divide by $\sqrt{4^2 + (-3)^2 + 1^2} = \sqrt{26}$ to get the unit vector, whose components are the direction cosines.

Step 3: Add
$\dfrac{4}{\sqrt{26}} - \dfrac{3}{\sqrt{26}} + \dfrac{1}{\sqrt{26}} = \dfrac{2}{\sqrt{26}}$.

Step 4: Check
Squares of the cosines add to $\frac{16 + 9 + 1}{26} = 1$, as they must.

Final Answer:
The sum is 2/sqrt(26). This is option (D). \[ \boxed{\text{(D) }\frac{2}{\sqrt{26}}} \]