In a $\triangle A B C$, if $\frac{r}{r_1}=\frac{1}{2}$, then $4 \tan \frac{A}{2}\left(\tan \frac{B}{2}+\tan \frac{C}{2}\right)=$
If $2 \hat{\mathbf{i}}-\hat{\mathbf{j}}+\hat{\mathbf{k}}, \hat{\mathbf{i}}-3 \hat{\mathbf{j}}-5 \hat{\mathbf{k}}$ are the position vectors of the points $\mathbf{A}$ and $\mathbf{B}$ respectively, $\mathbf{C}$ divides $\mathbf{A B}$ in the ratio $2: 3$ and $\mathbf{M}$ is the mid-point of $A B$, then 5 (position vector of $\mathbf{C})-2($ position vector of $\mathbf{M})=$
- If $\mathbf{a}, \mathbf{b}$ and $\mathbf{c}$ are the non-coplanar vectors and $\mathbf{a}-2 \mathbf{b}+3 \mathbf{c},-4 \mathbf{a}+5 \mathbf{b}-6 \mathbf{c}, x \mathbf{a}-9 \mathbf{b}+z \mathbf{c}$ are collinear points, then $2 x-z=$
The point which lies on the plane passing through the point $\hat{\mathbf{i}}-2 \hat{\mathbf{j}}-3 \hat{\mathbf{k}}, 3 \hat{\mathbf{i}}-\hat{\mathbf{j}}+4 \hat{\mathbf{k}},-3 \hat{\mathbf{i}}+2 \hat{\mathbf{j}}-5 \hat{\mathbf{k}}$ is
TS EAMCET Papers
All year-wise previous year question papers