The points $A(a), B(b), C(c)$ will be collinear if

Area of a rectangle having vertices $A, B, C$ and $D$ with position vectors $-\hat{i}+\frac{1}{2} \hat{j}+4 \hat{k}, \hat{i}+\frac{1}{2} \hat{j}+4 \hat{k}, \hat{i}-\frac{1}{2} \hat{j}+4 \hat{k}$ and $-\hat{i}-\frac{1}{2} \hat{j}+4 \hat{k}$ respectively is

If $\overline{a} = \hat{i} + \hat{j} + \hat{k}$ and $\overline{b} = \hat{j} - \hat{k}$,then the vector $\overline{r}$ satisfying $\overline{a} \times \overline{r} = \overline{b}$ and $\overline{a} \cdot \overline{r} = 3$ is

Let $\vec{a} = \vec{j} - \vec{k}$ and $\vec{c} = \vec{i} - \vec{j} - \vec{k}$. Find the vector $\vec{b}$ satisfying $\vec{a} \times \vec{b} + \vec{c} = 0$ and $\vec{a} \cdot \vec{b} = 3$.

Let $\vec{a}=\hat{i}+\hat{j}-\hat{k}$ and $\vec{c}=2 \hat{i}-3 \hat{j}+2 \hat{k}$. Then the number of vectors $\vec{b}$ such that $\vec{b} \times \vec{c}=\vec{a}$ and $|\vec{b}| \in\{1, 2, \ldots, 10\}$ is

Find a vector of magnitude $6,$ which is perpendicular to both the vectors $\vec{a} = 2 \hat{i}-\hat{j}+2 \hat{k}$ and $\vec{b} = 4 \hat{i}-\hat{j}+3 \hat{k}$.