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Find the value of a for which the vectors $3\hat{i} + 2\hat{j} + 9\hat{k}$ and $\hat{i} + \hat{aj} + 3\hat{k}$ are perpendicular
1 Answer
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Let $ \overrightarrow a$ = 3$ \hat i $+2$\hat j $+9$\hat k $ and $ \overrightarrow b$ =$ \hat i $+$\hat aj$+3$\hat k$.

For two vectors to be perpendicular, $ \overrightarrow a$ ⊥ $ \overrightarrow b$ = $ \overrightarrow a$.$ \overrightarrow b$=0

(3$ \hat i $+2$ \hat j $+9$\hat k$). ($\hat i$+a$\hat j$+3$\hat k$) =0

3+2a+27=0

30+2a=0

2a = -30

a= -15

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