MCQ
If $a\,.i\,=a\,.\,(i+j)=a\,.\,(i+j+k)$ , then $a = $
- ✓$i$
- B$k$
- C$j$
- D$i + j + k$
Then $a\,.\,i = (xi + yj + zk)\,.\,i = x$ and $a\,.\,(i + j) = x + y$ and $a\,.\,(i + j + k) = x + y + z$
Given that $x = x + y = x + y + z$
Now $x = x + y\,\,\, \Rightarrow y = 0$ and $x + y = x + y + z\,\, \Rightarrow \,\,z = 0$
Hence $x = 1$; $\therefore \,\,a = i$.
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