Vectors $\to A = a^i + a^j + 3^k$ and $\to B = a^i - 2^j -^k$ are perpendicular to each other, then the positive value of $a$ is

3

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Solution

The correct option is A $3$
We know that,
If $\to A ⊥ \to B, \to A . \to B = 0$

$a^{2} - 2 a - 3 = 0$
$(a + 1) (a - 3) = 0$
$a = - 1$ or $3$
Hence the positive value of $a = 3$

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\to P = a^i + a^j + 3^k and \to Q = a^i - 2^j -^k

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\to A = a^i + a^j + 3^k

\to B = a^i - 2^j -^k

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\to P = a^i + a^j + 3^k

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