MathematicsHard202×since 2002Q5851Let a→=a1i^+a2j^+a3k^\overrightarrow{\mathrm{a}}=\mathrm{a}_1 \hat{i}+\mathrm{a}_2 \hat{j}+\mathrm{a}_3 \hat{k}a=a1i^+a2j^+a3k^ and b→=b1i^+b2j^+b3k^\overrightarrow{\mathrm{b}}=\mathrm{b}_1 \hat{i}+\mathrm{b}_2 \hat{j}+\mathrm{b}_3 \hat{k}b=b1i^+b2j^+b3k^ be two vectors such that ∣a→∣=1,a⃗⋅b⃗=2|\overrightarrow{\mathrm{a}}|=1, \vec{a} \cdot \vec{b}=2∣a∣=1,a⋅b=2 and ∣b⃗∣=4|\vec{b}|=4∣b∣=4. If c⃗=2(a⃗×b⃗)−3b⃗\vec{c}=2(\vec{a} \times \vec{b})-3 \vec{b}c=2(a×b)−3b, then the angle between b⃗\vec{b}b and c⃗\vec{c}c is equal to:Acos−1(−13)\cos ^{-1}\left(-\frac{1}{\sqrt{3}}\right)cos−1(−31)Bcos−1(23)\cos ^{-1}\left(\frac{2}{3}\right)cos−1(32)Ccos−1(23)\cos ^{-1}\left(\frac{2}{\sqrt{3}}\right)cos−1(32)Dcos−1(−32)\cos ^{-1}\left(-\frac{\sqrt{3}}{2}\right)cos−1(−23)Check answerSkip