Vector overrightarrow P = ahat{i} + ahat{j} + 3hat{k} and overrightarrow Q = ahat{i} - 2hat{j} - hat

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3-1.Vectors

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The vector $\overrightarrow P = a\hat i + a\hat j + 3\hat k$ and $\overrightarrow Q = a\hat i - 2\hat j - \hat k$ are perpendicular to each other. The positive value of $a$ is

A

$3$

B

$4$

C

$9$

D

$13$

(AIIMS-2002)

Solution

(a) $\bar P.\bar Q = 0$

$⇒$ ${a^2} – 2a – 3 = 0$ $⇒$ $a = 3$

Standard 11

Physics

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