Question

Let $\alpha$ be the angle between the lines whose direction cosines satisfy the equations $l+m–n=0$ and $l^2+m^2–n^2=0$. Then the value of ${\sin }^4\alpha +{\cos }^4\alpha$ is:

• A. $\frac{3}{4}$
• B. $\frac{1}{2}$
• C. $\frac{5}{8}$
• D. $\frac{3}{8}$

Answer 1

The correct option is C

$\frac{5}{8}$

Explanation for the correct option:

Finding the value of ${\sin }^4\alpha +{\cos }^4\alpha$:

Step 1: Finding the value of $m$ and $n$

The given equations are

$$\begin{gathered} l+m=n\dots .\left(i\right) \\ {l}^2+m^2–n^2=0\dots .\left(ii\right) \end{gathered}$$

Squaring the equation $\left(i\right)$

$$\begin{gathered} \Rightarrow {\left(l+m\right)}^2=n^2 \\ \Rightarrow l^2+m^2+2lm=n^2\dots .\left(iii\right) \end{gathered}$$

Subtracting the equation $\left(ii\right)$from $\left(iii\right)$

$$\begin{gathered} \Rightarrow l^2+m^2+2lm-\left(n^2{l}^2+m^2–n^2\right)=n^2-0 \\ \Rightarrow 2lm=0 \\ \Rightarrow l=0\text{or}m=0 \end{gathered}$$

Case $1$: When $l=0$

$\Rightarrow m=n\mathbf{[}\mathbf{}\mathbf{\text{from}}\mathbf{}\mathbf{(}\mathit{i}\mathbf{)}\mathbf{}\mathbf{]}$

Since we know

$l^2+m^2+n^2=1$

$$\begin{gathered} \Rightarrow 0+m^2+m^2=1\mathbf{}\left[\because m=n\right] \\ \Rightarrow 2m^2=1 \\ \Rightarrow m=n=\pm \frac{1}{\sqrt{2}} \end{gathered}$$

Therefore, $(l,m,n)=\left(0,\frac{1}{\sqrt{2}},\frac{1}{\sqrt{2}}\right)\text{or}\left(0,-\frac{1}{\sqrt{2}},-\frac{1}{\sqrt{2}}\right)$

So, let $\overrightarrow{a}=\left(0,\frac{1}{\sqrt{2}},\frac{1}{\sqrt{2}}\right)$

Case $2$: $m=0$

$$\begin{gathered} \Rightarrow l=n\mathbf{[}\mathbf{\text{from}}\mathbf{}\mathbf{(}\mathit{i}\mathbf{\left)}\mathbf{\right]} \\ \because l^2+m^2+n^2=1 \end{gathered}$$

$$\begin{gathered} \Rightarrow l^2+0+l^2=1\mathbf{}\left[\because m=n\right] \\ \Rightarrow 2l^2=1 \\ \Rightarrow l=n=\pm \frac{1}{\sqrt{2}} \end{gathered}$$

$(l,m,n)=\left(\frac{1}{\sqrt{2}},0,\frac{1}{\sqrt{2}}\right)\text{or}\left(-\frac{1}{\sqrt{2}},0,-\frac{1}{\sqrt{2}}\right)$

So, let $\overrightarrow{b}=\left(\frac{1}{\sqrt{2}},0,\frac{1}{\sqrt{2}}\right)$

Step 2: Finding the values of ${\sin }^4\alpha$and ${\cos }^4\alpha$ and equating ${\sin }^4\alpha +{\cos }^4\alpha$

$$\begin{gathered} \because \cos \alpha =\frac{\overrightarrow{a}·\overrightarrow{b}}{\left|\overrightarrow{a}\right|·\left|\overrightarrow{b}\right|} \\ =\frac{\left(0,\frac{1}{\sqrt{2}},\frac{1}{\sqrt{2}}\right)\left(\frac{1}{\sqrt{2}},0,\frac{1}{\sqrt{2}}\right)}{\left|\left(0,\frac{1}{\sqrt{2}},\frac{1}{\sqrt{2}}\right)\right|\left|\left(\frac{1}{\sqrt{2}},0,\frac{1}{\sqrt{2}}\right)\right|} \\ =\frac{\left(0+0+{\displaystyle \frac{1}{2}}\right)}{\left(\sqrt{0+{\left({\displaystyle \frac{1}{\sqrt{2}}}\right)}^2+{\left(\frac{1}{\sqrt{2}}\right)}^2}\right)\left(\sqrt{{\left(\frac{1}{\sqrt{2}}\right)}^2+0+{\left(\frac{1}{\sqrt{2}}\right)}^2}\right)} \\ =\frac{{\displaystyle \frac{1}{2}}}{\left(1\right)\left(1\right)} \\ \Rightarrow \cos \alpha =\frac{1}{2}...\left(iv\right) \\ \Rightarrow {\cos }^4\alpha ={\left(\frac{1}{2}\right)}^4\mathbf{[}\mathbf{\text{taking}}\mathbf{}\mathbf{4}\mathbf{\text{th power both sides}}\mathbf{]} \\ \Rightarrow {\cos }^4\alpha =\frac{1}{16}.....\left(v\right) \end{gathered}$$

$$\begin{gathered} \because {\sin }^2\alpha =1-{\cos }^2\alpha \\ =1-{\left(\frac{1}{2}\right)}^2\mathbf{[}\mathbf{}\mathbf{\text{from}}\mathbf{(}\mathit{i}\mathit{v}\mathbf{)}\mathbf{}\mathbf{]} \\ \Rightarrow {\sin }^2\alpha =\frac{3}{4} \\ \Rightarrow {\left({\sin }^2\alpha \right)}^2={\left(\frac{3}{4}\right)}^2\mathbf{}\left[\text{squaring both sides}\right] \\ \Rightarrow {\sin }^4\alpha =\frac{9}{16}....\left(vi\right) \end{gathered}$$

Adding equations $\left(v\right)\&\left(vi\right)$

$$\begin{gathered} \Rightarrow {\cos }^4\alpha +{\sin }^4\alpha =\frac{1}{16}+\frac{9}{16} \\ =\frac{10}{16} \\ =\frac{5}{8} \\ \Rightarrow {\cos }^4\alpha +{\sin }^4\alpha =\frac{5}{8} \end{gathered}$$

Therefore, option (C) is the correct answer.