If vec{P} = b hat{i} + 6 hat{j} + hat{k} and | vedclass

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(B) Two vectors $\vec{P}$ and $\vec{Q}$ are perpendicular if their dot product is zero,i.e.,$\vec{P} \cdot \vec{Q} = 0$. Given $\vec{P} = b \hat{i} +

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$a = 1, b = 2$

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(B) Two vectors $\vec{P}$ and $\vec{Q}$ are perpendicular if their dot product is zero,i.e.,$\vec{P} \cdot \vec{Q} = 0$. Given $\vec{P} = b \hat{i} + 6 \hat{j} + \hat{k}$ and $\vec{Q} = \hat{i} - a \hat{j} + 4 \hat{k}$. Calculating the dot product: $(b)(1) + (6)(-a) + (1)(4) = 0$. This simplifies to $b - 6a + 4 = 0$,or $b - 6a = -4$. We are also given the equation $3b - a = 5$,which implies $a = 3b - 5$. Substituting $a$ into the first equation: $b - 6(3b - 5) = -4$. $b - 18b + 30 = -4$. $-17b = -34$,so $b = 2$. Now,find $a$: $a = 3(2) - 5 = 6 - 5 = 1$. Thus,$a = 1$ and $b = 2$.

If $\vec{P} = b \hat{i} + 6 \hat{j} + \hat{k}$ and $\vec{Q} = \hat{i} - a \hat{j} + 4 \hat{k}$ are perpendicular to each other,and $3b - a = 5$,find the values of $a$ and $b$.

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