If $\vec{a}=(2 x+y) \hat{i}+3 \hat{j}+9 \hat{k}$ and $\vec{b}=2 \hat{i}+\hat{j}-(x-y) \hat{k}$ are two collinear vectors, then $x^3+27 y^3=$

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Aicharya · Accepted Answer

Given: $\vec{a}=(2 x+y) \hat{i}+3 \hat{j}+9 \hat{k}$ $\vec{b}=2 \hat{i}+\hat{j}-(x-y) \hat{k}$ $\because \quad \vec{a}$ and $\vec{b}$ are co-linear vectors $	herefore \quad \frac{2 x+y}{2}=\frac{3}{1}=\frac{9}{-(x-y)}$ $\Rightarrow \quad \frac{2 x+y}{2}=\frac{3}{1} \Rightarrow 2 x+y=6$ ...(i) and $\frac{3}{1}=\frac{9}{y-x} \Rightarrow y-x=3$ ...(ii) Solving eqns. (i) and (ii), we get $x=1$ and $y=4$ Now, $x^3+27 y^3=1+27 	imes 64=1729$.

If $\vec{a}=(2 x+y) \hat{i}+3 \hat{j}+9 \hat{k}$ and $\vec{b}=2 \hat{i}+\hat{j}-(x-y) \hat{k}$ are two…

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