UPSC NDA I 2021 Mathematics Solved Paper

Concept: If $a, b, c$ are the direction ration ratios of a line passing through the point $(x_{1}, y_{1}, z_{1})$, then the equation of line is given by: $\frac{x-x_{1}}{a}=\frac{y-y_{1}}{b}=\frac{z-z_{1}}{c}$ If $a, b, c$ are the direction ration ratiosof a line then the direction cosine of the line is given by: $1=\frac{a}{\sqrt{a^{2}+b^{2}+c^{2}}} m=\frac{b}{\sqrt{a^{2}+b^{2}+c^{2}}}, n=\frac{c}{\sqrt{a^{2}+b^{2}+c^{2}}}$ Calculation Given: $1, m, n$ are the direction cosinesof the line $x-1=2(y+3)=1-z$ The given equation of lines can be re-written as $\frac{x-1}{1}=\frac{y+3}{\frac{1}{2}}=\frac{z-1}{-1}$ So, by comparing the equation $\frac{x-1}{1}=\frac{y+3}{\frac{1}{2}}=\frac{z-1}{-1}$ with $\frac{x-x_{1}}{a}=\frac{y-y_{1}}{b}=\frac{z-z_{1}}{c}$we get $\Rightarrow a=1, b=\frac{1}{2} \text{ and } c=-1$ As we know that, if $a, b, c$ are the direction ration ratios of a line then the direction cosine of the line is given by: $1=\frac{a}{\sqrt{a^{2}+b^{2}+c^{2}}} m=\frac{b}{\sqrt{a^{2}+b^{2}+c^{2}}}, n=\frac{c}{\sqrt{a^{2}+b^{2}+c^{2}}}$ The direction cosine of the given line is: $\Rightarrow 1=\frac{2}{3}, m=\frac{1}{3}, n=\frac{-2}{3}$ $\Rightarrow 1^{4}+m^{4}+n^{4}=\frac{16}{81}+\frac{1}{81}+\frac{16}{81}=\frac{33}{81}=\frac{11}{27}$ Hence, the correct option is 2 .