If $\mathbf{a}=2 \mathbf{u}+3 \mathbf{v}+7 \mathbf{w}, b=\mathbf{u}+\mathbf{v}-2 \mathbf{w}$ and $\mathbf{c}=-\mathbf{u}-2 \mathbf{v}-3 \mathbf{w}$ then $\left|\frac{[\mathbf{u} \mathbf{v} \mathbf{w}]}{[\mathbf{a} \mathbf{b} \mathbf{c}]}\right|(\mathbf{a}+\mathbf{b}+\mathbf{c})=$
Let $\mathbf{V}=2 \hat{\mathbf{i}}+\hat{\mathbf{j}}-\hat{\mathbf{k}}$ and $\mathbf{W}=\hat{\mathbf{i}}+3 \hat{\mathbf{k}}$. If $\mathbf{U}$ is a unit vector, then the maximum value of $[\mathbf{U} \mathbf{V} \mathbf{W}]$ is
$L_1$ is a line passing through the points with position vectors $\hat{\mathbf{i}}-2 \hat{\mathbf{j}}-\hat{\mathbf{k}}$ and $4 \hat{\mathbf{i}}-3 \hat{\mathbf{k}} . L_2$ is a line passing through the points with position vectors $\hat{\mathbf{i}}+2 \hat{\mathbf{j}}-\hat{\mathbf{k}}$ and $2 \hat{\mathbf{i}}-4 \hat{\mathbf{j}}-5 \hat{\mathbf{k}}$. Then the distance between $L_1$ and $L_2$ is
The mean and standard deviation of 100 observations $x_1, x_2, \ldots, x_{100}$ were calculated as 40 and 5.1 respectively by a student who took by mistake 50 instead of 40 for one observation. Then the correct value of $\sum_{i=1}^{100} x_i^2=$
TS EAMCET Papers
All year-wise previous year question papers