If $\vec{a}, \vec{b}, \vec{c}$ are vectors such that $\vec{a} + \vec{b} + \vec{c} = \vec{0}$ and $|\vec{a}| = 1, |\vec{b}| = 2, |\vec{c}| = 5$, then the expression $\vec{a}\cdot\vec{b} + \vec{b}\cdot\vec{c} + \vec{c}\cdot\vec{a}$ equals

If $\vec{a}, \vec{b}$ and $\vec{c}$ are three unit vectors such that $\vec{a} + \vec{b} + \vec{c} = 0$, then the value of $\vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{c} + \vec{c} \cdot \vec{a}$ is

For any vector $\vec{a}$, the value of $|\vec{a} \times \hat{i}|^2 + |\vec{a} \times \hat{j}|^2 + |\vec{a} \times \hat{k}|^2$ is equal to:

Let $\vec{a} = \hat{i} + 4\hat{j}$, $\vec{b} = 4\hat{j} + \hat{k}$ and $\vec{c} = \hat{i} - 2\hat{k}$. If $\vec{d}$ is a vector perpendicular to both $\vec{a}$ and $\vec{b}$ such that $\vec{c} \cdot \vec{d} = 16$, then $|\vec{d}|$ is equal to

The value of $\lambda$, for which the two vectors $2\hat{i} - \hat{j} + 2\hat{k}$ and $3\vec{i} + \lambda\vec{j} + \hat{k}$ are perpendicular, is:

In $\triangle ABC$, Which of the following are True?

A. $\overline{AB}+\overline{BC}+\overline{CA}=\vec{0}$

B. $\overline{AB}+\overline{BC}-\overline{AC}=\vec{0}$

C. $\overline{AB}+\overline{BC}-\overline{CA}=\vec{0}$

D. $\overline{AB}-\overline{CB}+\overline{CA}=\vec{0}$

Choose the correct answer from the options given below:

If $\lvert\vec{a}\rvert = \lvert\vec{b}\rvert = \lvert\vec{a}+\vec{b}\rvert = 1$, then

Match List-I with List-II

Choose the correct answer from the options given below: