if i,j,k is an orthonormal system of vectors, abar is a vector and a cross i+2a-5j=0 then a=

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$$\begin{gathered} \mathrm{Let} \overrightarrow{\mathrm{a}}=\mathrm{x}\hat{\mathrm{i}}+\mathrm{y}\hat{\mathrm{j}}+\mathrm{z}\hat{\mathrm{k}} \\ \mathrm{According} \mathrm{to} \mathrm{question} \\ \overrightarrow{\mathrm{a}}\times \hat{\mathrm{i}}+2\overrightarrow{\mathrm{a}}-5\hat{\mathrm{j}}=\overrightarrow{0} \\ \Rightarrow \left(\mathrm{x}\hat{\mathrm{i}}+\mathrm{y}\hat{\mathrm{j}}+\mathrm{z}\hat{\mathrm{k}}\right)\times \hat{\mathrm{i}}+2\left(\mathrm{x}\hat{\mathrm{i}}+\mathrm{y}\hat{\mathrm{j}}+\mathrm{z}\hat{\mathrm{k}}\right)-5\hat{\mathrm{j}}=\overrightarrow{0} \\ \Rightarrow \mathrm{x}\left(\hat{\mathrm{i}}\times \hat{\mathrm{i}}\right)+\mathrm{y}\left(\hat{\mathrm{j}}\times \hat{\mathrm{i}}\right)+\mathrm{z}\left(\hat{\mathrm{k}}\times \hat{\mathrm{i}}\right)+2 \left(\mathrm{x}\hat{\mathrm{i}}+\mathrm{y}\hat{\mathrm{j}}+\mathrm{z}\hat{\mathrm{k}}\right)-5\hat{\mathrm{j}}=\overrightarrow{0} \\ \Rightarrow \mathrm{x}\left(0\right)+\mathrm{y}\left(-\hat{\mathrm{k}}\right)+\mathrm{z}\left(\hat{\mathrm{j}}\right)+2\mathrm{x}\hat{\mathrm{i}}+2\mathrm{y}\hat{\mathrm{j}}+2\mathrm{z}\hat{\mathrm{k}}-5\hat{\mathrm{j}}=\overrightarrow{0} \\ \Rightarrow -\mathrm{y}\hat{\mathrm{k}}+\mathrm{z}\hat{\mathrm{j}}+2\mathrm{x}\hat{\mathrm{i}}+2\mathrm{y}\hat{\mathrm{j}}+2\mathrm{z}\hat{\mathrm{k}}-5\hat{\mathrm{j}}=\overrightarrow{0} \\ \Rightarrow 2\mathrm{x}\hat{\mathrm{i}}+\left(\mathrm{z}+2\mathrm{y}-5\right)\hat{\mathrm{j}}+\left(-\mathrm{y}+2\mathrm{z}\right)\hat{\mathrm{k}}=\overrightarrow{0} \\ \mathrm{On} \mathrm{comparing} \mathrm{with} \mathrm{R}.\mathrm{H}.\mathrm{S}. \mathrm{we} \mathrm{get}: \\ 2\mathrm{x}=0 \\ \Rightarrow \mathrm{x}=0 \\ \mathrm{z}+2\mathrm{y}-5=0 \\ \Rightarrow \mathrm{z}+2\mathrm{y}=5...\left(\mathrm{i}\right) \\ -\mathrm{y}+2\mathrm{z}=0 \\ \Rightarrow \mathrm{y}=2\mathrm{z} \\ \mathrm{Put} \mathrm{this} \mathrm{in} \left(\mathrm{i}\right) \\ \mathrm{z}+4\mathrm{z}=5 \\ \Rightarrow 5\mathrm{z}=\mathrm{z} \\ \Rightarrow \mathrm{z}=1 \\ \mathrm{y}=2\mathrm{z}=2\times 1=2 \\ \mathrm{x}=0, \mathrm{y}=2, \mathrm{z}=1 \\ \mathbf{\Rightarrow }\overrightarrow{\mathbf{a}}\mathbf{=}\mathbf{2}\hat{\mathbf{j}}\mathbf{+}\hat{\mathbf{k}} \end{gathered}$$

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