The position vectors of the points $A, B, C$ are $(2\hat{i}+\hat{j}-\hat{k}), (3\hat{i}-2\hat{j}+\hat{k})$ and $(\hat{i}+4\hat{j}-3\hat{k})$ respectively. These points

Vector $\vec{x}$ is a vector in the direction of $(2, -2, 1)$ and has a magnitude of $6$ units. Vector $\vec{y}$ is a vector in the direction of $(1, 1, -1)$ and has a magnitude of $\sqrt{3}$ units. Then,$|\vec{x} + 2\vec{y}| = $ . . . . . . .

If $G$ is the centroid of the $\triangle ABC$,then $\vec{GA} + \vec{GB} + \vec{GC}$ is equal to

Let $\vec{a}, \vec{b}, \vec{c}$ be co-initial vectors and $\vec{a}=2 \hat{i}-\hat{j}+5 \hat{k}$ and $\vec{b}=3 \hat{i}+7 \hat{j}-\hat{k}$. Let $(\vec{a}, \vec{b})=\theta$ be an acute angle and $\vec{c}$ be the vector along the bisector of the angle $\theta$. If $\lambda, x, y \in R$,then $\vec{c}=$

If $a$ and $b$ are two vectors such that $a \cdot b = 0$ and $a \times b = 0$,then:

If $\bar{a}, \bar{b}, \bar{c}$ are three unit vectors such that $|\bar{a}+\bar{b}|^2+|\bar{a}+\bar{c}|^2=8$,then $|\bar{a}+3\bar{b}|^2+|\bar{a}+3\bar{c}|^2=$