Q217 / 270 · NDA 2021 (Phase 1) · Mathematics

If $l, m, n$ are the direction cosines of the line $x-1=2(y+$ $3)=1-z$, then what is $l^4+m^4+n^4$ equal to?

  • A
    $1$
  • Correct answer
    B
    $\frac{11}{27}$
  • C
    $\frac{13}{27}$
  • D
    $4$

Solution

Let $x-1=(y-3)=1-z-k$ (Say) $\therefore l=k, m=\frac{k}{2}$ and $n=k$ $\begin{aligned} & \text { From } l^2+m^2+n^2=1 \\ & k^2+\left(\frac{k}{2}\right)^2+k^2=1 \Rightarrow k=\frac{2}{3}\end{aligned}$ $\begin{aligned} & \therefore l=\frac{2}{3}, m=\frac{1}{3} \text { and } n=\frac{2}{3} \\ & \therefore l^4+m^4+n^4 \\ & =\left(\frac{2}{3}\right)^4+\left(\frac{1}{3}\right)^4+\left(\frac{2}{3}\right)^4=\frac{11}{27}\end{aligned}$

Hence, the correct answer is option: 'B'.

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