Let $\vec{a}, \vec{b}$ and $\vec{c}$ be three vectors such that $|\vec{a}|=3, |\vec{b}|=4, |\vec{c}|=5$ and each one of them is perpendicular to the sum of the other two. Find $|\vec{a}+\vec{b}+\vec{c}|$.

If $\vec{a}=5 \hat{i}-\hat{j}-3 \hat{k}$ and $\vec{b}=\hat{i}+3 \hat{j}-5 \hat{k},$ then show that the vectors $\vec{a}+\vec{b}$ and $\vec{a}-\vec{b}$ are perpendicular.

If $\vec{a}$ and $\vec{b}$ are perpendicular unit vectors and vector $\vec{c}$ is such that $\vec{c} = \vec{a} + \vec{b}$,then $(\vec{a} \times \vec{b}) \cdot (\vec{b} \times \vec{c}) + (\vec{b} \times \vec{c}) \cdot (\vec{c} \times \vec{a}) + (\vec{c} \times \vec{a}) \cdot (\vec{a} \times \vec{b})$ is

If vectors $\vec{a}$ and $\vec{b}$ are not perpendicular,and $\vec{c}$ and $\vec{d}$ are two vectors satisfying $\vec{b} \times \vec{c} = \vec{b} \times \vec{d}$ and $\vec{a} \cdot \vec{d} = 0$,then the vector $\vec{d}$ is equal to:

Let $\vec{a} = \hat{i} + \hat{j} + \sqrt{2}\hat{k}$,$\vec{b} = b_{1}\hat{i} + b_{2}\hat{j} + \sqrt{2}\hat{k}$,and $\vec{c} = 5\hat{i} + \hat{j} + \sqrt{2}\hat{k}$ be three vectors such that the projection vector of $\vec{b}$ on $\vec{a}$ is $\vec{a}$. If $\vec{a} + \vec{b}$ is perpendicular to $\vec{c}$,then $|\vec{b}|$ is equal to

The altitude through vertex $A$ of $\triangle ABC$ with position vectors of points $A, B, C$ as $\bar{a}, \bar{b}, \bar{c}$ respectively is