Q78 / 150 · MHT CET 2026 (19 April Shift 2) · Mathematics
If $\alpha, \beta, \gamma$ are the direction angles of the line $x = 4z + 3$ and $y = 2 - 3z$, then the value of $\cos\alpha + \cos\beta + \cos\gamma$ is...
- A$\dfrac{8}{\sqrt{26}}$
- B$\dfrac{6}{\sqrt{26}}$
- C$\dfrac{4}{\sqrt{26}}$
- Correct answerD$\dfrac{2}{\sqrt{26}}$
Solution
The equations of the line are given as $x = 4z + 3$ and $y = 2 - 3z$.
Rewriting these equations in symmetric form:
$\dfrac{x - 3}{4} = z$
$\dfrac{y - 2}{-3} = z$
Combining them, we get:
$\dfrac{x - 3}{4} = \dfrac{y - 2}{-3} = \dfrac{z - 0}{1}$
The direction ratios of the line are $4, -3, 1$.
The direction cosines $\cos\alpha, \cos\beta, \cos\gamma$ are given by dividing the direction ratios by their magnitude $\sqrt{4^2 + (-3)^2 + 1^2} = \sqrt{16 + 9 + 1} = \sqrt{26}$.
$\cos\alpha = \dfrac{4}{\sqrt{26}}$
$\cos\beta = \dfrac{-3}{\sqrt{26}}$
$\cos\gamma = \dfrac{1}{\sqrt{26}}$
Summing these values:
$\cos\alpha + \cos\beta + \cos\gamma = \dfrac{4 - 3 + 1}{\sqrt{26}} = \dfrac{2}{\sqrt{26}}$
Answer: $\dfrac{2}{\sqrt{26}}$
Rewriting these equations in symmetric form:
$\dfrac{x - 3}{4} = z$
$\dfrac{y - 2}{-3} = z$
Combining them, we get:
$\dfrac{x - 3}{4} = \dfrac{y - 2}{-3} = \dfrac{z - 0}{1}$
The direction ratios of the line are $4, -3, 1$.
The direction cosines $\cos\alpha, \cos\beta, \cos\gamma$ are given by dividing the direction ratios by their magnitude $\sqrt{4^2 + (-3)^2 + 1^2} = \sqrt{16 + 9 + 1} = \sqrt{26}$.
$\cos\alpha = \dfrac{4}{\sqrt{26}}$
$\cos\beta = \dfrac{-3}{\sqrt{26}}$
$\cos\gamma = \dfrac{1}{\sqrt{26}}$
Summing these values:
$\cos\alpha + \cos\beta + \cos\gamma = \dfrac{4 - 3 + 1}{\sqrt{26}} = \dfrac{2}{\sqrt{26}}$
Answer: $\dfrac{2}{\sqrt{26}}$
Hence, the correct answer is option: 'D'.