If $\overline{\mathrm{a}}=3 \hat{\mathrm{i}}-5 \hat{\mathrm{j}}, \overline{\mathrm{b}}=6 \hat{\mathrm{i}}-3…

If $\overline{\mathrm{a}}=3 \hat{\mathrm{i}}-5 \hat{\mathrm{j}}, \overline{\mathrm{b}}=6 \hat{\mathrm{i}}-3 \hat{\mathrm{j}}$ are two vectors and $\overline{\mathrm{c}}$ is vector such that $\bar{c}=\bar{a} \times \bar{b}$, then $a: b: c$ is
  1. $\sqrt{34}: \sqrt{45}: \sqrt{39}$
  2. $\sqrt{34}: \sqrt{45}: 39$
  3. $34: 39: 45$
  4. $39: 35: 34$

Solution

$\begin{aligned} & \overline{\mathrm{c}}=\overline{\mathrm{a}} \times \overline{\mathrm{b}}=\left|\begin{array}{ccc} \hat{\mathrm{i}} & \hat{\mathrm{j}} & \hat{\mathrm{k}} \\ 3 & -5 & 0 \\ 6 & 3 & 0 \end{array}\right|=\hat{\mathrm{i}}(0)-\hat{\mathrm{j}}(0)+\hat{\mathrm{k}}(9+30)=39 \hat{\mathrm{k}} \\ & |\overline{\mathrm{a}}|=\sqrt{(3)^2+(-5)^2}=\sqrt{34} \text { and }|\overline{\mathrm{b}}|=\sqrt{(6)^2+(3)^2}=\sqrt{45} \\ & |\overline{\mathrm{c}}|=\sqrt{39^2}=39 \\ & \therefore \mathrm{a}: \mathrm{b}: \mathrm{c}=\sqrt{34}: \sqrt{45}: 39 \end{aligned}$

Asked in: MHT CET 2021 (21 Sep Shift 1)

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