Calculate the values of a and b if vectors ah | vedclass

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(D) Two vectors are perpendicular if their dot product is zero.Let $\vec{A} = a\hat{i} + b\hat{j}$ and $\vec{B} = \hat{i} + \hat{j}$.Since $\vec{A} \c

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$1/\sqrt{2}, -1/\sqrt{2}$

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(D) Two vectors are perpendicular if their dot product is zero.Let $\vec{A} = a\hat{i} + b\hat{j}$ and $\vec{B} = \hat{i} + \hat{j}$.Since $\vec{A} \cdot \vec{B} = 0$, we have $(a\hat{i} + b\hat{j}) \cdot (\hat{i} + \hat{j}) = 0$, which implies $a + b = 0$, so $b = -a$.Given that $\vec{A} = \hat{n}$ is a unit vector, its magnitude is $1$.$|\vec{A}| = \sqrt{a^2 + b^2} = 1$.Substituting $b = -a$, we get $\sqrt{a^2 + (-a)^2} = 1$, which simplifies to $\sqrt{2a^2} = 1$.Thus, $2a^2 = 1$, so $a^2 = 1/2$, which gives $a = \pm 1/\sqrt{2}$.If $a = 1/\sqrt{2}$, then $b = -1/\sqrt{2}$.If $a = -1/\sqrt{2}$, then $b = 1/\sqrt{2}$.Comparing with the options, option $D$ matches the condition.

Calculate the values of $a$ and $b$ if vectors $a\hat{i} + b\hat{j} = \hat{n}$ and $(\hat{i} + \hat{j})$ are perpendicular to each other. Note: $\hat{n}$ is a unit vector.

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