Vectors ahat{i} + bhat{j} + hat{k} and 2hat{i | vedclass

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(B) Two vectors $\vec{A} = a\hat{i} + b\hat{j} + \hat{k}$ and $\vec{B} = 2\hat{i} - 3\hat{j} + 4\hat{k}$ are perpendicular if their dot product is zer

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

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 that $3a + 2b = 7$, and the ratio of $a$ to $b$ is $x/2$, find the value of $x$.

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