If bar{c} = 5bar{a} + 6bar{b} and 3bar{c} = b | vedclass

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(C) Given equations are: $(1)$ $\bar{c} = 5\bar{a} + 6\bar{b}$ $(2)$ $3\bar{c} = \bar{a} - 4\bar{b}$ Multiply equation $(1)$ by $3$: $3\bar{c} = 15\ba

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$\bar{a}, \bar{c}$ are in the same direction but $\bar{a}, \bar{b}$ are in the opposite direction

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(C) Given equations are: $(1)$ $\bar{c} = 5\bar{a} + 6\bar{b}$ $(2)$ $3\bar{c} = \bar{a} - 4\bar{b}$ Multiply equation $(1)$ by $3$: $3\bar{c} = 15\bar{a} + 18\bar{b}$ Now,equate this with equation $(2)$: $15\bar{a} + 18\bar{b} = \bar{a} - 4\bar{b}$ $14\bar{a} = -22\bar{b}$ $\bar{a} = -\frac{11}{7}\bar{b}$ Since $\bar{a}$ is a scalar multiple of $\bar{b}$ with a negative constant,$\bar{a}$ and $\bar{b}$ are collinear and in opposite directions. Substitute $\bar{a} = -\frac{11}{7}\bar{b}$ into equation $(1)$: $\bar{c} = 5(-\frac{11}{7}\bar{b}) + 6\bar{b} = -\frac{55}{7}\bar{b} + \frac{42}{7}\bar{b} = -\frac{13}{7}\bar{b}$ Since $\bar{c}$ is also a negative scalar multiple of $\bar{b}$,$\bar{c}$ and $\bar{b}$ are in opposite directions. Since $\bar{a} = -\frac{11}{7}\bar{b}$ and $\bar{c} = -\frac{13}{7}\bar{b}$,we have $\bar{a} = \frac{11}{13}\bar{c}$. Since $\bar{a}$ is a positive scalar multiple of $\bar{c}$,$\bar{a}$ and $\bar{c}$ are in the same direction. Thus,$\bar{a}$ and $\bar{c}$ are in the same direction,while $\bar{a}$ and $\bar{b}$ are in opposite directions.

If $\bar{c} = 5\bar{a} + 6\bar{b}$ and $3\bar{c} = \bar{a} - 4\bar{b}$,then:

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