MHT CET · Maths · Three Dimensional Geometry
If the angle between the lines whose direction ratios are \(4,-3,5\) and \(3,4, k\)
is \(\frac{\pi}{3}\), then \(k=\)
- A \(\pm\) 7
- B \(\pm\) 10
- C \(\pm\) 5
- D \(\pm\) 6
Answer & Solution
Correct Answer
(C) \(\pm\) 5
Step-by-step Solution
Detailed explanation
\(\cos \frac{\pi}{3} =\left|\frac{4(3)+(-3)(4)+5 \mathrm{k}}{\sqrt{4^{2}+(-3)^{2}+5^{2}} \sqrt{3^{2}+4^{2}+\mathrm{k}^{2}}}\right| \)
\( \therefore \frac{1}{2} =\left|\frac{5 \mathrm{k}}{5 \sqrt{2} \sqrt{25+\mathrm{k}^{2}}}\right|\)
On squaring both side we get
\(
\begin{aligned}
& \frac{1}{4}=\frac{25 \mathrm{k}^{2}}{50 \times\left(25+\mathrm{k}^{2}\right)} \\
\therefore & 100 \mathrm{k}^{2}=50\left(25+\mathrm{k}^{2}\right) \Rightarrow \mathrm{k}^{2}=25 \Rightarrow \mathrm{k}=\pm 5
\end{aligned}
\)
\( \therefore \frac{1}{2} =\left|\frac{5 \mathrm{k}}{5 \sqrt{2} \sqrt{25+\mathrm{k}^{2}}}\right|\)
On squaring both side we get
\(
\begin{aligned}
& \frac{1}{4}=\frac{25 \mathrm{k}^{2}}{50 \times\left(25+\mathrm{k}^{2}\right)} \\
\therefore & 100 \mathrm{k}^{2}=50\left(25+\mathrm{k}^{2}\right) \Rightarrow \mathrm{k}^{2}=25 \Rightarrow \mathrm{k}=\pm 5
\end{aligned}
\)
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