If n^=ai^+bj^ is perpendicular to the vector (i^+j^), then the value of a and b will be

If $\hat{n} = a \hat{i} + b \hat{j}$ is perpendicular to the vector $(\hat{i} + \hat{j}),$ then the value of $a$ and $b$ will be

A
$1, 0$
B
$- 2, 0$
C
$3, 0$
D
$\frac{1}{\sqrt{2}}, - \frac{1}{\sqrt{2}}$

Solution:

The dot product of two perpendicular vectors is zero $I f \vec{A} ⊥ \vec{B} t h e n \vec{A} . \vec{B} = 0$

$∴ a = \frac{1}{\sqrt{2}}$ and $b = - \frac{1}{\sqrt{2}}$

$(\frac{\hat{i}}{\sqrt{2}} - \frac{\hat{j}}{\sqrt{2}}) . (\hat{i} + \hat{j})$

$= \frac{1}{\sqrt{2}} - \frac{1}{\sqrt{2}} = 0$

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