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