MHTCET 15 Apr 2026 Shift 2 Paper

Concept:When three vectors add to zero, the sum of any two equals the negative of the third, allowing their magnitudes to be related using the dot product.Explanation:Given $\vec{a}+\vec{b}+\vec{c}=\overrightarrow{0}$, we can write $\vec{a}=-(\vec{b}+\vec{c})$.Taking magnitudes on both sides gives $|\vec{a}|=|\vec{b}+\vec{c}|$.Since $|\vec{a}|=|\vec{b}|=|\vec{c}|=3$, we get $|\vec{b}+\vec{c}|=3$.Use the formula $|\vec{b}+\vec{c}|^{2}=|\vec{b}|^{2}+|\vec{c}|^{2}+2|\vec{b}||\vec{c}|\cos\theta$.Substitute the magnitudes: $3^{2}=3^{2}+3^{2}+2(3)(3)\cos\theta$.This simplifies to $9=18+18\cos\theta$, so $18\cos\theta=-9$.Thus $\cos\theta=-\frac{1}{2}$, which means $\theta=120^\circ$.Now $\tan 120^\circ=-\sqrt{3}$, so $\tan^{2}\theta=3$.Also $\cot^{2}\theta=\frac{1}{\tan^{2}\theta}=\frac{1}{3}$.Therefore $\tan^{2}\theta+\cot^{2}\theta=3+\frac{1}{3}=\frac{10}{3}$.Answer:$\frac{10}{3}$, which is option D.