If $\vec{a}$ is a nonzero vector of magnitude $a$ and $\lambda$ is a nonzero scalar,then $\lambda \vec{a}$ is a unit vector if

If $\vec{a}=\vec{b}+\vec{c},$ then is it true that $|\vec{a}|=|\vec{b}|+|\vec{c}|$ ? Justify your answer.

Let $\bar{a}$ and $\bar{b}$ be the position vectors of points $A$ and $B$ respectively. $C$ and $D$ are points on the line $AB$ such that $\overline{AC} = 3 \overline{AB}$ and $\overline{BD} = 2 \overline{BA}$. Find the vector $\overline{CD}$.

For any vector $\vec{a} = a_1 \hat{i} + a_2 \hat{j} + a_3 \hat{k}$,with $10|a_i| < 1$,$i = 1, 2, 3$,consider the following statements:
$(A): \max \{|a_1|, |a_2|, |a_3|\} \leq |\vec{a}|$
$(B): |\vec{a}| \leq 3 \max \{|a_1|, |a_2|, |a_3|\}$

If $\vec{a} = \hat{i} + \hat{j} - 2 \hat{k}$,$\vec{b} = 2 \hat{i} - \hat{j} + \hat{k}$ and $\vec{c} = 3 \hat{i} - \hat{k}$. If $\vec{c} = m \vec{a} + n \vec{b}$ then $m + n = $

The position vectors of two points $A$ and $B$ are $\bar{i}+2\bar{j}+3\bar{k}$ and $7\bar{i}-\bar{k}$ respectively. The point $P$ with position vector $-2\bar{i}+3\bar{j}+5\bar{k}$ is on the line $AB$. If the point $Q$ is the harmonic conjugate of $P$ with respect to $A$ and $B$, then the sum of the scalar components of the position vector of $Q$ is