If the angle between the lines whose direction ratios are $4,-3,5$ and $3,4, k$ is $\frac{\pi}{3}$, then $k=$
If the angle between the lines whose direction ratios are $4,-3,5$ and $3,4, k$
is $\frac{\pi}{3}$, then $k=$
- $\pm$ 7
- $\pm$ 10
- $\pm$ 5
- $\pm$ 6
Solution
$\begin{aligned}
\cos \frac{\pi}{3} &=\left|\frac{4(3)+(-3)(4)+5 \mathrm{k}}{\sqrt{4^{2}+(-3)^{2}+5^{2}} \sqrt{3^{2}+4^{2}+\mathrm{k}^{2}}}\right| \\
\therefore \quad \frac{1}{2} &=\left|\frac{5 \mathrm{k}}{5 \sqrt{2} \sqrt{25+\mathrm{k}^{2}}}\right|
\end{aligned}$
On squaring both side we get
$\begin{aligned}
& \frac{1}{4}=\frac{25 \mathrm{k}^{2}}{50 \times\left(25+\mathrm{k}^{2}\right)} \\
\therefore \quad & 100 \mathrm{k}^{2}=50\left(25+\mathrm{k}^{2}\right) \Rightarrow \mathrm{k}^{2}=25 \Rightarrow \mathrm{k}=\pm 5
\end{aligned}$
Asked in: MHT CET 2020 (14 Oct Shift 2)
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