A) \[0{}^\circ \]
B) \[60{}^\circ \]
C) \[30{}^\circ \]
D) \[90{}^\circ \]
E) \[45{}^\circ \]
Correct Answer: A
Solution :
Given equation of line \[r=(i+2j+3k)+\lambda (2i+3j+4k)\] and equation of plane \[r.(i+2j-2k)=0\] Now, angle between line and plane is \[\sin \theta =\frac{(2i+3j+4k).(i+2j-2k)}{\sqrt{4+9+16}.\sqrt{1+4+4}}\] \[\sin \theta =\frac{2+6-8}{\sqrt{29}\sqrt{9}}\] \[=0=\sin 0{}^\circ \] \[\Rightarrow \] \[\theta =0{}^\circ \]
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