Two vectors a hat{i} + b hat{j} + hat{k} and | vedclass

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(C) Two vectors are perpendicular if their dot product is zero.Let $\vec{A} = a \hat{i} + b \hat{j} + \hat{k}$ and $\vec{B} = 2 \hat{i} - 3 \hat{j} +

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$1$

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(C) Two vectors are perpendicular if their dot product is zero.Let $\vec{A} = a \hat{i} + b \hat{j} + \hat{k}$ and $\vec{B} = 2 \hat{i} - 3 \hat{j} + 4 \hat{k}$.$\vec{A} \cdot \vec{B} = (a)(2) + (b)(-3) + (1)(4) = 0$.$2a - 3b + 4 = 0$,which implies $2a - 3b = -4$.We are given the system of equations:$1) 2a - 3b = -4$$2) 3a + 2b = 7$Multiply equation $(1)$ by $2$ and equation $(2)$ by $3$:$4a - 6b = -8$$9a + 6b = 21$Adding these equations: $13a = 13$,so $a = 1$.Substitute $a = 1$ into $3a + 2b = 7$: $3(1) + 2b = 7 \implies 2b = 4 \implies b = 2$.The ratio $a/b = 1/2$.We are given $a/b = x/2$,so $1/2 = x/2$,which means $x = 1$.

Two vectors $a \hat{i} + b \hat{j} + \hat{k}$ and $2 \hat{i} - 3 \hat{j} + 4 \hat{k}$ are perpendicular to each other. Given $3a + 2b = 7$,the ratio of $a$ to $b$ is $\frac{x}{2}$. The value of $x$ is:

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