BITSAT 2013
- Exam
- BITSAT
- Year
- 2013
- Subject
- Mathematics
- Chapter
- Three Dimensional Geometry
BITSAT 2013 · Mathematics · Three Dimensional Geometry
Find the angle between the two planes 2 x+y-2 z=5 and 3 x-6 y-2 z=7
Find the angle between the two planes and
Options
- A
- B
- C
- D
Correct answer
A.
Step-by-step solution
The angle between two planes is the angle between their normals. From the equation of the planes, the normal vectors are \overrightarrow{\mathrm{N}_{1}=2 \hat{\mathrm{i}}+\hat{\mathrm{j}}-2 \hat{\mathrm{k}} and \overrightarrow{\mathrm{N}_{2}=3 \hat{\mathrm{i}}-6 \hat{\mathrm{j}}-2 \hat{\mathrm{k}} Therefore, \cos \theta=\left|\frac{\overrightarrow{\mathrm{N}_{1} \cdot \overrightarrow{\mathrm{N}_{2}}{\left|\overrightarrow{\mathrm{N}_{1}\right|\left|\overrightarrow{\mathrm{N}_{2}\right|}\right|=\left|\frac{(2 \hat{i}+\hat{j}-2 \hat{k}) \cdot(3 \hat{i}-6 \hat{j}-2 \hat{k})}{\sqrt{4+1+4} \sqrt{9+36+4}}\right|= \left(\frac{4}{21}\right)Hence, \theta=\cos ^{-1} \left(\frac{4}{21}\right)