Question

Direction cosines of line $x=y=1-z$ are

• A. $1,1,1$
• B. $\dfrac{1}{\sqrt{3}},\dfrac{1}{\sqrt{3}},-\dfrac{1}{\sqrt{3}}$
• C. $0,0,1$
• D. $\dfrac{1}{\sqrt{3}},\dfrac{1}{\sqrt{3}},\dfrac{1}{\sqrt{3}}$

Hint:

To find direction cosines of a line, first determine the direction ratios and then divide each ratio by the magnitude of the vector.

Answer 1

The Correct Option is B

Step 1: Express the line in parametric form.
Given \[ x=y=1-z \] Let \[ x=y=t \] Then \[ t=1-z \] Thus \[ z=1-t \] So the parametric form becomes \[ x=t \] \[ y=t \] \[ z=1-t \] Step 2: Find direction ratios.
Direction ratios are coefficients of \(t\).
\[ x=t \] \[ y=t \] \[ z=-t \] Thus direction ratios are \[ (1,1,-1) \] Step 3: Convert to direction cosines.
Magnitude of vector \[ \sqrt{1^2+1^2+(-1)^2} \] \[ =\sqrt{3} \] Therefore direction cosines are \[ \frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}},-\frac{1}{\sqrt{3}} \] Conclusion:
Hence the direction cosines are \[ \left(\frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}},-\frac{1}{\sqrt{3}}\right) \] Final Answer: $\boxed{\dfrac{1}{\sqrt{3}},\dfrac{1}{\sqrt{3}},-\dfrac{1}{\sqrt{3}}}$