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

Suppose $\vec{a}+\vec{b}+\vec{c}=0$,$|\vec{a}|=3$,$|\vec{b}|=5$,$|\vec{c}|=7$. Then the angle between $\vec{a}$ and $\vec{b}$ is:

If the vectors $\hat{i}+3 \hat{j}+4 \hat{k}$ and $\lambda \hat{i}-4 \hat{j}+\hat{k}$ are orthogonal to each other,then $\lambda$ is equal to

Let $\vec{a} = 5\hat{i} - \hat{j} - 3\hat{k}$ and $\vec{b} = \hat{i} + 3\hat{j} + 5\hat{k}$ be two vectors. Then which one of the following statements is $TRUE$?

For what values of $x$ is the angle between the vectors $\vec{a} = x\hat{i} - 3\hat{j} - \hat{k}$ and $\vec{b} = 2x\hat{i} + x\hat{j} - \hat{k}$ acute,and the angle between the vector $\vec{b}$ and the $y$-axis obtuse?

Let $\vec{c}$ and $\vec{d}$ be vectors such that $|\vec{c}+\vec{d}|=\sqrt{29}$ and $\vec{c}\times(2\hat{i}+3\hat{j}+4\hat{k})=(2\hat{i}+3\hat{j}+4\hat{k})\times\vec{d}$. If $\lambda_1, \lambda_2$ $(\lambda_1 > \lambda_2)$ are the possible values of $(\vec{c}+\vec{d}) \cdot (-7\hat{i}+2\hat{j}+3\hat{k})$, then the equation $K^{2}x^{2}+(K^{2}-5K+\lambda_{1})xy+(3K+\frac{\lambda_{2}}{2})y^{2}-8x+12y+\lambda_{2}=0$ represents a circle, for $K$ equal to: