Vectors — Maths STD 12 Science — Question

Then $\begin{aligned} \bar{a} \times \hat{i} & =(x \hat{i}+y \hat{j}+z \hat{k}) \times \hat{i} \\ & =x(\hat{i} \times \hat{i})+y(\hat{j} \times \hat{i})+z(\hat{k} \times \hat{i}) \\ & =z \hat{j}-y \hat{k} \ldots[\because \hat{i} \times \hat{i}=\overline{0}, \hat{j} \times \hat{i}=-\hat{k}, \hat{k} \times \hat{i}=\hat{j}]\end{aligned}$

It is given that

$\bar{a} \times \hat{i}+2 \bar{a}-5 \hat{j}=\overline{0}$

$\begin{aligned} & \therefore z \hat{j}-y \hat{k}+2(x \hat{i}+y \hat{j}+z \hat{k})-5 \hat{j}=0 \\ & \therefore z \hat{j}-y \hat{k}+2 x \hat{i}+2 y \hat{j}+2 z \hat{k}-5 \hat{j}=\overline{0} \\ & \therefore 2 x \hat{i}+(2 y+z-5) \hat{j}+(2 z-y) \hat{k}=\overline{0}\end{aligned}$

By equality of vectors

2x = 0 i.e. x = 0 2y + z – 5 = 0 … (1) 2z – y = 0 … (2) From (2), y = 2z Substituting y = 2z in (1), we get 4z + z = 5 ∴ z = 1 ∴ y = 2z = 2(1) = 2 ∴ x = 0, y = 2, z = 1

$\therefore \bar{a}=2 \hat{j}+\hat{k}$