Question

If $\hat{i},\hat{j},\hat{k}$ is an orthonormal system of vectors, $\vec{a}$ is a vector and $\vec{a}\times \hat{i}+2\vec{a}-5\hat{j}=\bar{0}$ then $\vec{a}=$

& A.2\hat{j}+\hat{k} \\\ & B.2\hat{i}-\hat{k} \\\ & C.2\hat{i}-\hat{j} \\\ & D.\text{None of these} \\\ \end{aligned}$$

Answer 1

Here, the orthogonal system of vectors means all 3 vectors are mutually perpendicular to each other. Whenever there is some unknown vector given in any such question, then assume it and represent it in any arbitrary variables with the 3 coordinate axes. Now, put this unknown vector in the given condition of the question. By doing some comparisons we will get the unknown variables and the required vector.

Complete step-by-step solution:
Now, let us assume:
$\vec{a}={{a}_{1}}\hat{i}+{{a}_{2}}\hat{j}+{{a}_{3}}\hat{k}$
And the given condition is:
$\vec{a}\times \hat{i}+2\vec{a}-5\hat{j}=\bar{0}\text{ }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. (i)}$
Now, compute:

& \vec{a}\times \hat{i}=\left| \begin{matrix} {\hat{i}} & {\hat{j}} & {\hat{k}} \\\ {{a}_{1}} & {{a}_{2}} & {{a}_{3}} \\\ 1 & 0 & 0 \\\ \end{matrix} \right| \\\ & \vec{a}\times \hat{i}=\left( {{a}_{2}}\times 0-0\times {{a}_{3}} \right)\hat{i}-\left( {{a}_{1}}\times 0-1\times {{a}_{3}} \right)\hat{j}+\left( {{a}_{1}}\times 0-1\times {{a}_{2}} \right) \\\ & \vec{a}\times \hat{i}={{a}_{3}}\hat{j}-{{a}_{2}}\hat{k} \\\ \end{aligned}$$ Now from expression (i) we have $$\begin{aligned} & \vec{a}\times \hat{i}+2\vec{a}-5\hat{j}=\vec{0} \\\ & \left( {{a}_{3}}\hat{j}-{{a}_{2}}\hat{k} \right)+2\left( {{a}_{1}}\hat{i}+{{a}_{2}}\hat{j}+{{a}_{3}}\hat{k} \right)-5\hat{j}=\vec{0} \\\ & \left( 2{{a}_{1}} \right)\hat{i}+\left( {{a}_{3}}+2{{a}_{2}}-5 \right)\hat{j}+\left( 2{{a}_{3}}-{{a}_{2}} \right)\hat{k}=0\hat{i}+0\hat{j}+0\hat{k} \\\ \end{aligned}$$ By comparing LHS and RHS, we have: $$\begin{aligned} & 2{{a}_{1}}=0\text{ }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. (ii)} \\\ &\Rightarrow {{a}_{3}}+2{{a}_{2}}-5=0\text{ }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. (iii)} \\\ &\Rightarrow 2{{a}_{3}}-{{a}_{2}}=0\text{ }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. (iv)} \\\ \end{aligned}$$ From (ii) we have ${{a}_{1}}=0$ By multiplying 2 in equation (iv) we have: $$4{{a}_{3}}-2{{a}_{2}}=0\text{ }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. (v)}$$ Now, add equation (iii) and (v) we get: $$\left( {{a}_{3}}+2{{a}_{2}}-5 \right)+\left( 4{{a}_{3}}-2{{a}_{2}} \right)=0$$ Now, cancelling $2{{a}_{2}}$from the equation=n, we get: $$\begin{aligned} & {{a}_{3}}+4{{a}_{3}}=5 \\\ &\Rightarrow 5{{a}_{3}}=5 \\\ &\Rightarrow {{a}_{3}}=1 \\\ \end{aligned}$$ Hence, from equation (iv) put value of ${{a}_{3}}$ $$\begin{aligned} & \therefore 2\times 1-{{a}_{2}}=0 \\\ & \Rightarrow {{a}_{2}}=2 \\\ \end{aligned}$$ Hence, we get ${{a}_{1}}=0,{{a}_{2}}=2,{{a}_{3}}=1$ $$\begin{aligned} & \vec{a}={{a}_{1}}\hat{i}+{{a}_{2}}\hat{j}+{{a}_{3}}\hat{k} \\\ & \Rightarrow 0+2\hat{j}+1\hat{k} \\\ & \Rightarrow 2\hat{j}+\hat{k} \\\ \end{aligned}$$ **Therefore, option A is the correct answer.** **Note:** Don’t compare the given expression directly with the $\vec{0}$ first assume the value of $\vec{a}$ and then resolve into the form which contains the all 3 coordinate axis. After that, do comparison. The overall question is simple but focus should be more on the calculation part. As if there is some minor mistake in case of writing $\hat{i},\hat{j},\hat{k}$ then, we can mark the wrong option. (Given options are quite similar to each other).