KCET · Maths · Linear Programming
The sine of the angle between the straight line \(\frac{x-2}{3}=\frac{3-y}{-4}=\frac{z-4}{5}\) and the plane \(2 x-2 y+z=5\) is
- A \(\frac{3}{\sqrt{30}}\)
- B \(\frac{3}{50}\)
- C \(\frac{4}{5 \sqrt{2}}\)
- D \(\frac{\sqrt{2}}{10}\)
Answer & Solution
Correct Answer
(D) \(\frac{\sqrt{2}}{10}\)
Step-by-step Solution
Detailed explanation
Given straight line \(\frac{x-2}{3}=\frac{y-3}{4}=\frac{z-4}{5}\) and plane \(2 x-2 y+z=5\)
Let \(\theta\) be the angle between line and plane then
\(\begin{aligned}
\sin \theta &=\left|\frac{3 \times 2+(4)(-2)+5 \times 1}{\sqrt{(3)^{2}+(4)^{2}+(5)^{2}} \sqrt{(2)^{2}+(-2)^{2}+(1)^{2}}}\right| \\
&=\left|\frac{6-8+5}{\sqrt{9+16+25} \sqrt{4+4+1}}\right| \\
&=\left|\frac{3}{\sqrt{50} \sqrt{9}}\right|=\frac{1}{5 \sqrt{2}}=\frac{\sqrt{2}}{10}
\end{aligned}\)
Let \(\theta\) be the angle between line and plane then
\(\begin{aligned}
\sin \theta &=\left|\frac{3 \times 2+(4)(-2)+5 \times 1}{\sqrt{(3)^{2}+(4)^{2}+(5)^{2}} \sqrt{(2)^{2}+(-2)^{2}+(1)^{2}}}\right| \\
&=\left|\frac{6-8+5}{\sqrt{9+16+25} \sqrt{4+4+1}}\right| \\
&=\left|\frac{3}{\sqrt{50} \sqrt{9}}\right|=\frac{1}{5 \sqrt{2}}=\frac{\sqrt{2}}{10}
\end{aligned}\)
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