1
TS EAMCET 2020 (Online) 14th September Morning Shift
MCQ (Single Correct Answer)
+1
-0

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 $\left|\frac{[\mathbf{u} \mathbf{v} \mathbf{w}]}{[\mathbf{a} \mathbf{b} \mathbf{c}]}\right|(\mathbf{a}+\mathbf{b}+\mathbf{c})=$

A

$12(\mathbf{u}+\mathbf{v}+\mathbf{w})$

B

$3(\mathbf{u}+\mathbf{v}+\mathbf{w})$

C

$\frac{2}{3}(\mathbf{u}+\mathbf{v}+\mathbf{w})$

D

$\frac{1}{3}(u+v+w)$

2
TS EAMCET 2020 (Online) 14th September Morning Shift
MCQ (Single Correct Answer)
+1
-0

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

A

-1

B

$\sqrt{10}+\sqrt{16}$

C

$\sqrt{59}$

D

$\sqrt{60}$

3
TS EAMCET 2020 (Online) 14th September Morning Shift
MCQ (Single Correct Answer)
+1
-0

$L_1$ 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

A

0

B

$\frac{3}{4}$

C

$\frac{4}{3}$

D

$\frac{2}{3}$

4
TS EAMCET 2020 (Online) 14th September Morning Shift
MCQ (Single Correct Answer)
+1
-0

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 $\sum_{i=1}^{100} x_i^2=$

A

3990

B

161701

C

162601

D

4000

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