MHT CET · Maths · Three Dimensional Geometry
The direction cosines \(\ell, \mathrm{m}, \mathrm{n}\) of the line \(\frac{\mathrm{x}+2}{2}=\frac{2 \mathrm{y}-4}{3} ; \mathrm{z}=-1\)
- A \(\ell= \pm \frac{1}{\sqrt{5}}, \mathrm{~m}=0,= \pm \frac{2}{\sqrt{5}}\)
- B \(\ell= \pm \frac{3}{5}, \mathrm{~m}= \pm \frac{4}{5}, \mathrm{n}=0\)
- C \(\ell= \pm \frac{4}{5}, \mathrm{~m}= \pm \frac{3}{5}, \mathrm{n}=0\)
- D \(\ell= \pm \frac{1}{\sqrt{3}}, \mathrm{~m}= \pm \frac{1}{\sqrt{3}}, \mathrm{n}= \pm \frac{1}{\sqrt{3}}\)
Answer & Solution
Correct Answer
(C) \(\ell= \pm \frac{4}{5}, \mathrm{~m}= \pm \frac{3}{5}, \mathrm{n}=0\)
Step-by-step Solution
Detailed explanation
We have line \(\frac{x+2}{2}=\frac{2 y-5}{3}, z=-1\)
i.e. \(\frac{x-(-2)}{2}=\frac{2\left(y-\frac{5}{2}\right)}{3}, z=-1\) i.e. \(\frac{x-(-2)}{2}=\frac{\left(y-\frac{5}{2}\right)}{\left(\frac{3}{2}\right)}, z=-1\)
Here direction ratios are \(2, \frac{3}{2}\)
Also \(\sqrt{(2)^2+\left(\frac{3}{2}\right)^2+0}= \pm \frac{5}{2}\)
Here required direction cosines are
\(\frac{2}{\left( \pm \frac{5}{2}\right)}, \frac{\left(\frac{3}{2}\right)}{\left( \pm \frac{5}{2}\right)}, \frac{0}{\left( \pm \frac{5}{2}\right)} \text { i.e. } \pm \frac{4}{5}, \pm \frac{3}{5}, 0\)
i.e. \(\frac{x-(-2)}{2}=\frac{2\left(y-\frac{5}{2}\right)}{3}, z=-1\) i.e. \(\frac{x-(-2)}{2}=\frac{\left(y-\frac{5}{2}\right)}{\left(\frac{3}{2}\right)}, z=-1\)
Here direction ratios are \(2, \frac{3}{2}\)
Also \(\sqrt{(2)^2+\left(\frac{3}{2}\right)^2+0}= \pm \frac{5}{2}\)
Here required direction cosines are
\(\frac{2}{\left( \pm \frac{5}{2}\right)}, \frac{\left(\frac{3}{2}\right)}{\left( \pm \frac{5}{2}\right)}, \frac{0}{\left( \pm \frac{5}{2}\right)} \text { i.e. } \pm \frac{4}{5}, \pm \frac{3}{5}, 0\)
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