Question

If $\left( 2,-1,2 \right)$ and $\left( k,3,5 \right)$ are the triads of the direction ratios of two lines and the angle between them is ${{45}^{\circ }}$ , then a value of $k$ is
A. 2
B. 3
C. 4
D. 6

Answer 1

Hint: We can use the formula for finding the angle between two lines whose direction ratios are given. If direction ratios are ${{a}_{1}},{{b}_{1}},{{c}_{1}}$ and ${{a}_{2}},{{b}_{2}},{{c}_{2}}$, then the formulas for $\theta $ is given by
$\cos \theta =\left| \dfrac{{{a}_{1}}{{a}_{2}}+{{b}_{1}}{{b}_{2}}+{{c}_{1}}{{c}_{2}}}{\sqrt{{{a}_{1}}^{2}+{{b}_{1}}^{2}+{{c}_{1}}^{2}}\sqrt{{{a}_{2}}^{2}+{{b}_{2}}^{2}+{{c}_{2}}^{2}}} \right|$.

Complete step by step answer:
Before proceeding with the question, we must know the formula for finding the angle between lines using the direction ratios. The formula is given by $\cos \theta =\left| \dfrac{{{a}_{1}}{{a}_{2}}+{{b}_{1}}{{b}_{2}}+{{c}_{1}}{{c}_{2}}}{\sqrt{{{a}_{1}}^{2}+{{b}_{1}}^{2}+{{c}_{1}}^{2}}\sqrt{{{a}_{2}}^{2}+{{b}_{2}}^{2}+{{c}_{2}}^{2}}} \right|$, where ${{a}_{1}},{{b}_{1}},{{c}_{1}}$ and ${{a}_{2}},{{b}_{2}},{{c}_{2}}$ are the direction ratios of the lines.

In this question, we have been given that $\left( 2,-1,2 \right)$ and $\left( k,3,5 \right)$ are the triads of the direction ratios of two lines and the angle between them is ${{45}^{\circ }}$. We have been asked to find the value of $k$.

We will first find the value of $\cos \theta $ using the formula that we have discussed above. So, we can substitute the values of ${{a}_{1}},{{b}_{1}},{{c}_{1}}$ as $\left( 2,-1,2 \right)$ and ${{a}_{2}},{{b}_{2}},{{c}_{2}}$ as $\left( k,3,5 \right)$ in the formula as shown below,

$\therefore \cos \theta =\left| \dfrac{2\times k+\left( -1\times 3 \right)+2\times 5}{\sqrt{{{2}^{2}}+{{\left( -1 \right)}^{2}}+{{2}^{2}}}\sqrt{{{k}^{2}}+{{3}^{2}}+{{5}^{2}}}} \right|$

$\Rightarrow \cos \theta =\left| \dfrac{2k-3+10}{\sqrt{4+1+4}\sqrt{{{k}^{2}}+9+25}} \right|$

$\Rightarrow \cos \theta =\left| \dfrac{2k+7}{\sqrt{9}\sqrt{{{k}^{2}}+34}} \right|$

$\Rightarrow \cos \theta =\left| \dfrac{2k+7}{3\sqrt{{{k}^{2}}+34}} \right|$

Now, we can substitute the value of $\theta ={{45}^{\circ }}$ in the above equation. So, we will get,

$\Rightarrow \cos {{45}^{\circ }}=\left| \dfrac{2k+7}{3\sqrt{{{k}^{2}}+34}} \right|$

$\Rightarrow \dfrac{1}{\sqrt{2}}=\left| \dfrac{2k+7}{3\sqrt{{{k}^{2}}+34}} \right|$

Squaring both the sides we get,

$\Rightarrow \dfrac{1}{2}=\dfrac{{{\left( 2k+7 \right)}^{2}}}{9\left( {{k}^{2}}+34 \right)}$

By cross-multiplication we get,

$\Rightarrow 9\left( {{k}^{2}}+34 \right)=2{{\left( 2k+7 \right)}^{2}}$

$\Rightarrow 9\left( {{k}^{2}}+34 \right)=2\left( 4{{k}^{2}}+49+28k \right)$

Opening the brackets we get:

$\Rightarrow 9{{k}^{2}}+306=8{{k}^{2}}+98+56k$

$\Rightarrow {{k}^{2}}-56k+208=0$

Using the formula $\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$ we can find the value of $k$.

$\Rightarrow k=\dfrac{-\left( -56 \right)\pm \sqrt{{{\left( -56 \right)}^{2}}-4\times 1\times 208}}{2\times 1}$

$\Rightarrow k=\dfrac{56\pm \sqrt{3136-832}}{2}$

$\Rightarrow k=\dfrac{56\pm \sqrt{2304}}{2}$

$\Rightarrow k=\dfrac{56\pm 48}{2}$

Taking the positive and negative values separately, we get the value of $k$as

$k=\dfrac{56+48}{2}$

$\therefore k=52$

And

$k=\dfrac{56-48}{2}$

$\therefore k=4$

Therefore, we get the value of $k$ is $52,4$.

Looking at the options, we can see that the value $4$ is option C and 52 is not an option.
Hence the correct answer is option C.

Note: Be careful about the values of $\cos \theta $ because it is enclosed in modulus, therefore, the negative values of $\cos \theta $ cannot be accepted. But, here we get two positive options, so we must be careful in choosing the correct options as there might be more than one correct answer.