If $a$ and $b$ are two unit vectors such that $a+b$ is also a unit vector,then $|a-b|^2=$

Let $\vec{a}$ and $\vec{b}$ be two vectors such that $|2 \vec{a}+3 \vec{b}|=|3 \vec{a}+\vec{b}|$ and the angle between $\vec{a}$ and $\vec{b}$ is $60^{\circ}$. If $\frac{1}{8} \vec{a}$ is a unit vector,then $|\vec{b}|$ is equal to :

If $|\vec{a}| = 3, |\vec{b}| = 4, |\vec{c}| = 5$ such that each vector is perpendicular to the sum of the other two, then $|\vec{a} + \vec{b} + \vec{c}|$ is equal to

If the vectors $\vec{AB} = p \hat{i} + q \hat{j} + r \hat{k}$, $\vec{AC} = s \hat{i} + 3 \hat{j} + 4 \hat{k}$, and $\vec{CB} = 3 \hat{i} + \hat{j} - 2 \hat{k}$ form a $\triangle ABC$, then the values of $p, q, r$ and $s$ such that the area of that $\triangle ABC$ is $5 \sqrt{6}$ are:

If $|a| = 3$ and $|b| = 4$,for what value of $\lambda$ is the vector $(a + \lambda b)$ perpendicular to $(a - \lambda b)$?

The position vectors of the vertices of a triangle $ABC$ are $4i - 2j$,$i + 4j - 3k$,and $-i + 5j + k$ respectively. Then $\angle ABC = $