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

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Answer 1

Since
$(a \hat{i} + b \hat{j}) ⊥ (\hat{i} + \hat{j})$

Let $\hat{i} + \hat{j} = B$
Also we know dot product of two vectors =0
Thus $(a \hat{i} + b \hat{j}) . (\hat{i} + \hat{j}) = 0$
$a + b = 0$
$a = - b$ -----------(1)
As we know n=1 as n is a unit vector
$∣ n ∣ = a^{2} + b^{2} = 1$
$a^{2} + b^{2} = 1$
$(- b)^{2} + b^{2} = 1$
$b = \frac{1}{2}$ and $a = \frac{- 1}{2}$

Answer 2

Since

(a \hat{i} + b \hat{j}) ⊥ (\hat{i} + \hat{j})

Let

\hat{i} + \hat{j} = B

Also we know dot product of two vectors =0

Thus

(a \hat{i} + b \hat{j}) . (\hat{i} + \hat{j}) = 0

a + b = 0

a = - b

-----------(1)

As we know n=1 as n is a unit vector

∣ n ∣ = a^{2} + b^{2} = 1

a^{2} + b^{2} = 1

(- b)^{2} + b^{2} = 1

b = \frac{1}{2}

and

a = \frac{- 1}{2}

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

$1, 0$

$- 2, 0$

$3, 0$

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Solution

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