Vectors a hat{i} + b hat{j} + hat{k} and 2 ha | vedclass

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(A) Two vectors are perpendicular if their dot product is zero.$(a \hat{i} + b \hat{j} + \hat{k}) \cdot (2 \hat{i} - 3 \hat{j} + 4 \hat{k}) = 0$$2a -

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(A) Two vectors are perpendicular if their dot product is zero.$(a \hat{i} + b \hat{j} + \hat{k}) \cdot (2 \hat{i} - 3 \hat{j} + 4 \hat{k}) = 0$$2a - 3b + 4 = 0 \Rightarrow 2a - 3b = -4$We are given the second equation: $3a + 2b = 7$.To solve for $a$ and $b$,multiply the first equation by $2$ and the second by $3$:$4a - 6b = -8$$9a + 6b = 21$Adding these equations: $13a = 13 \Rightarrow a = 1$.Substituting $a = 1$ into $3a + 2b = 7$: $3(1) + 2b = 7 \Rightarrow 2b = 4 \Rightarrow b = 2$.The ratio $\frac{a}{b} = \frac{1}{2}$.Given $\frac{a}{b} = \frac{x}{2}$,we have $\frac{1}{2} = \frac{x}{2}$,which implies $x = 1$.

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 $\frac{x}{2}$,find the value of $x$.

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