Question 452 of 521MediumMCQ
The d.r. of normal to the plane through (1,0,0),(0,1,0) which makes an angle \[\pi /4\] with plane x+y=3 are : Solutions · 1
Official solution
Equation of plane through \[\left( {1,0,0} \right)\] is
\[a\left( {x - 1} \right) + by + cz = 0\,\,\,\,\,\,\,\,\,\,...\left( i \right)\]
(i) passes through \[\left( {0,1,0} \right).\]
\[ - a + b = 0 \Rightarrow b = a;\]
Also, \[\cos {45^ \circ }\]
\[ = {{a + a} \over {\sqrt {2\left( {2{a^2} + {c^2}} \right)} }} \Rightarrow 2a = \sqrt {2{a^2} + {c^2}} \]
\[ \Rightarrow 2{a^2} = {c^2}\]
\[ \Rightarrow c = \sqrt 2 a.\]
So d.r of normal area a, \[a\sqrt {2a} \] i.e. 1, \[1,\sqrt 2 .\] Answer the question to unlock