Let vec{a} = 2hat{i} + lambda_{1}hat{j} + 3ha | vedclass

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(B) Given $\vec{b} = 2\vec{a}$,we have $4\hat{i} + (3 - \lambda_{2})\hat{j} + 6\hat{k} = 2(2\hat{i} + \lambda_{1}\hat{j} + 3\hat{k}) = 4\hat{i} + 2\la

Vedclass · Accepted Answer

$(-\frac{1}{2}, 4, 0)$

Vedclass · Answer

(B) Given $\vec{b} = 2\vec{a}$,we have $4\hat{i} + (3 - \lambda_{2})\hat{j} + 6\hat{k} = 2(2\hat{i} + \lambda_{1}\hat{j} + 3\hat{k}) = 4\hat{i} + 2\lambda_{1}\hat{j} + 6\hat{k}$.Comparing the coefficients of $\hat{j}$,we get $3 - \lambda_{2} = 2\lambda_{1}$,which implies $\lambda_{2} = 3 - 2\lambda_{1}$ ... $(i)$.Since $\vec{a} \perp \vec{c}$,their dot product is zero: $\vec{a} \cdot \vec{c} = 0$.$(2\hat{i} + \lambda_{1}\hat{j} + 3\hat{k}) \cdot (3\hat{i} + 6\hat{j} + (\lambda_{3} - 1)\hat{k}) = 0$.$2(3) + \lambda_{1}(6) + 3(\lambda_{3} - 1) = 0$.$6 + 6\lambda_{1} + 3\lambda_{3} - 3 = 0$.$3\lambda_{3} = -3 - 6\lambda_{1} \Rightarrow \lambda_{3} = -1 - 2\lambda_{1}$ ... $(ii)$.Thus,the triplet is $(\lambda_{1}, 3 - 2\lambda_{1}, -1 - 2\lambda_{1})$.For $\lambda_{1} = -\frac{1}{2}$,we get $\lambda_{2} = 3 - 2(-\frac{1}{2}) = 4$ and $\lambda_{3} = -1 - 2(-\frac{1}{2}) = 0$.Therefore,the possible value is $(-\frac{1}{2}, 4, 0)$.

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