In a $\triangle ABC$,$|CB|=a$,$|CA|=b$,$|AB|=c$ and $CD$ is the median through the vertex $C$. Then,$CA \cdot CD=$

If $\vec{a} = 2\hat{i} - \hat{j} + \hat{k}$,$\vec{b} = \hat{i} + \hat{j} - 2\hat{k}$,and $\vec{c} = \hat{i} + 3\hat{j} - (\lambda^2 + 3\lambda)\hat{k}$ (where $\lambda$ is a constant) and $\vec{a}$ is perpendicular to $\vec{c} - \lambda\vec{b}$,then the sum of different values of $\lambda$ is:

Let $\vec{AB} = 2 \hat{i} + 4 \hat{j} - 5 \hat{k}$ and $\vec{AD} = \hat{i} + 2 \hat{j} + \lambda \hat{k}$, $\lambda \in R$. Let the projection of the vector $\vec{v} = \hat{i} + \hat{j} + \hat{k}$ on the diagonal $\vec{AC}$ of the parallelogram $ABCD$ be of length $1$ unit. If $\alpha, \beta$, where $\alpha > \beta$, are the roots of the equation $\lambda^2 x^2 - 6 \lambda x + 5 = 0$, then $2 \alpha - \beta$ is equal to

The points $O, A, B, C, D$ are such that $\overrightarrow{OA} = \vec{a}$,$\overrightarrow{OB} = \vec{b}$,$\overrightarrow{OC} = 2\vec{a} + 3\vec{b}$,and $\overrightarrow{OD} = \vec{a} - 2\vec{b}$. If $|\vec{a}| = 3|\vec{b}|$,then the angle between $\overrightarrow{BD}$ and $\overrightarrow{AC}$ is:

If $a \neq 0, b \neq 0$ and $|a + b| = |a - b|,$ then the vectors $a$ and $b$ are

Let $\vec{r}$ be a vector in the plane of $\hat{i} - 2\hat{j} + \hat{k}$ and $\hat{i} - \hat{j} - \hat{k}$ such that $\vec{r} \cdot (\hat{i} + \hat{j}) + 2 = 0$ and the length of the projection of $\vec{r}$ on $\hat{i} - \hat{j}$ is $4\sqrt{2}$. Then,the magnitude of vector $\vec{r}$ is: