If the angle $\theta$ between the vectors $\overrightarrow{a}=2 x^2 \hat{i}+4 x \hat{j}+\hat{k}$ and $\overrightarrow{b}=7 \hat{i}-2 \hat{j}+x \hat{k}$ is such that $90^{\circ} < \theta < 180^{\circ}$,then $x$ lies in the interval

Let $\overrightarrow{a} = \hat{i} + 2\hat{j} + 3\hat{k}$,$\overrightarrow{b} = \hat{i} - \hat{j} + 2\hat{k}$,and $\overrightarrow{c} = 5\hat{i} - 3\hat{j} + 3\hat{k}$ be three vectors. If $\overrightarrow{r}$ is a vector such that $\overrightarrow{r} \times \overrightarrow{b} = \overrightarrow{c} \times \overrightarrow{b}$ and $\overrightarrow{r} \cdot \overrightarrow{a} = 0$,then $25|\overrightarrow{r}|^2$ is equal to:

Let $\vec{a}=2\hat{i}-\hat{j}-\hat{k}$, $\vec{b}=\hat{i}+3\hat{j}-\hat{k}$ and $\vec{c}=2\hat{i}+\hat{j}+3\hat{k}$. Let $\vec{v}$ be a vector in the plane of the vectors $\vec{a}$ and $\vec{b}$, such that the length of its projection on the vector $\vec{c}$ is equal to $\frac{1}{\sqrt{14}}$. Then $|\vec{v}|$ is equal to:

If the vectors $\vec{a}=2 \hat{i}+2 \hat{j}+3 \hat{k}$,$\vec{b}=-\hat{i}+2 \hat{j}+\hat{k}$ and $\vec{c}=3 \hat{i}+\hat{j}$ are such that $(\vec{a}+\lambda \vec{b})$ is perpendicular to $\vec{c}$,then the value of $\lambda$ is

If the position vectors of three points $A, B, C$ are $\hat{i}+2\hat{j}+\hat{k}$, $2\hat{i}-\hat{j}+2\hat{k}$ and $\hat{i}+\hat{j}+2\hat{k}$ respectively, then the perpendicular distance of the point $C$ from the line $AB$ is

The angle between two consecutive sides $\vec{a}$ and $\vec{b}$ of a parallelogram is $\frac{\pi}{6}$. Given $\vec{a} = (2, -2, 1)$ and $\vec{b} = 2|\vec{a}|$,the area of the parallelogram is . . . . . . sq. units.