MHT CET · Maths · Vector Algebra
Let \(\vec{v}=2 \hat{i}+2 \hat{j}-\hat{k}\) and \(\bar{w}=\hat{i}+3 \hat{k}\). If \(\bar{u}\) is a unit vector, then the maximum value of the scalar triple product \([\overline{\mathrm{u}} \overline{\mathrm{v}} \overline{\mathrm{W}}]\) is
- A \(\sqrt{6}\)
- B \(\sqrt{10}\)
- C \(\sqrt{13}\)
- D \(\sqrt{89}\)
Answer & Solution
Correct Answer
(D) \(\sqrt{89}\)
Step-by-step Solution
Detailed explanation
\(\overline{ v } \times \overline{ w }=\left|\begin{array}{ccc}\hat{ i } & \hat{ j } & \hat{ k } \\ 2 & 2 & -1 \\ 1 & 0 & 3\end{array}\right|=\hat{ i }(6)-\hat{ j }(7)+\hat{ k }(-2)=6 \hat{ i }-7 \hat{ j }-2 \hat{ k }\)
\(\therefore|\overline{ v } \times \overline{ w }|=\sqrt{(6)^2+(7)^2+(2)^2}=\sqrt{89}\)
Also \(|\overline{ u }|=1\)
\(\therefore[\overline{ uvw }]=\sqrt{89}\)
\(\therefore|\overline{ v } \times \overline{ w }|=\sqrt{(6)^2+(7)^2+(2)^2}=\sqrt{89}\)
Also \(|\overline{ u }|=1\)
\(\therefore[\overline{ uvw }]=\sqrt{89}\)
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