MHT CET · Maths · Vector Algebra
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
- A \(\sqrt{34}: \sqrt{45}: \sqrt{39}\)
- B \(\sqrt{34}: \sqrt{45}: 39\)
- C \(34: 39: 45\)
- D \(39: 35: 34\)
Answer & Solution
Correct Answer
(B) \(\sqrt{34}: \sqrt{45}: 39\)
Step-by-step Solution
Detailed explanation
\(\overline{\text c }=\overline{\text a } \times \overline{\text b }=\left|\begin{array}{ccc}\hat{\text i } & \hat{\text j } & \hat{\text k } \\ 3 & -5 & 0 \\ 6 & 3 & 0\end{array}\right|=\hat{\text i }(0)-\hat{\text j }(0)+\hat{\text k }(9+30)\) \(=39 \hat{\text k }\)
\(|\overline{\mathrm{a}}|=\sqrt{(3)^2+(-5)^2}=\sqrt{34} \text { and }|\overline{\mathrm{b}}|=\sqrt{(6)^2+(3)^2}=\) \(\sqrt{45} \)
\( |\overline{\mathrm{c}}|=\sqrt{39^2}=39 \)
\( \therefore \mathrm{a}: \mathrm{b}: \mathrm{c}=\sqrt{34}: \sqrt{45}: 39\)
\(|\overline{\mathrm{a}}|=\sqrt{(3)^2+(-5)^2}=\sqrt{34} \text { and }|\overline{\mathrm{b}}|=\sqrt{(6)^2+(3)^2}=\) \(\sqrt{45} \)
\( |\overline{\mathrm{c}}|=\sqrt{39^2}=39 \)
\( \therefore \mathrm{a}: \mathrm{b}: \mathrm{c}=\sqrt{34}: \sqrt{45}: 39\)
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