1
TS EAMCET 2020 (Online) 11th September Morning Shift
MCQ (Single Correct Answer)
+1
-0

If $\mathbf{a}, \mathbf{b}, \mathbf{c}$ are the position vectors of the points $A, B, C$ respectively, then match the items of list-I with those of list-II.

$$
\text { List-I }
$$
$$
\text { List-II }
$$
A. $$
\text { } \begin{aligned}
\mathbf{a} & =2 \hat{\mathbf{i}}+3 \hat{\mathbf{j}}+4 \hat{\mathbf{k}}, \\
\mathbf{b} & =3 \hat{\mathbf{i}}+4 \hat{\mathbf{j}}+2 \hat{\mathbf{k}}, \\
\mathbf{c} & =4 \hat{\mathbf{i}}+2 \hat{\mathbf{j}}+3 \hat{\mathbf{k}}
\end{aligned}
$$
I. $A, B, C$ are collinear
B. $$
\text { } \begin{aligned}
\mathbf{a} & =\hat{\mathbf{i}}+2 \hat{\mathbf{j}}+3 \hat{\mathbf{k}}, \\
\mathbf{b} & =3 \hat{\mathbf{i}}+4 \hat{\mathbf{j}}+7 \hat{\mathbf{k}}, \\
\mathbf{c} & =-3 \hat{\mathbf{i}}-2 \hat{\mathbf{j}}-5 \hat{\mathbf{k}}
\end{aligned}
$$
II. $\triangle A B C$ is an isosceles triangle
C. $$
\begin{aligned}
&\text {  }\\
&\begin{aligned}
& a=2 \hat{i}-\hat{j}+\hat{k}, \\
& b=\hat{i}-3 \hat{j}-5 \hat{k}, \\
& c=-3 \hat{i}-4 \hat{j}-4 \hat{k}
\end{aligned}
\end{aligned}
$$
III. $\triangle A B C$ is a right-angled triangle
D. $$
\begin{aligned}
& a=\hat{\mathbf{i}}+\hat{\mathbf{j}}+\hat{\mathbf{k}}, \\
& b=\hat{\mathbf{i}}+2 \hat{\mathbf{j}}+3 \hat{\mathbf{k}}, \\
& c=2 \hat{\mathbf{i}}-\hat{\mathbf{j}}+\hat{\mathbf{k}},
\end{aligned}
$$
IV. $\triangle A B C$ is a right-angled isosceles triangle
V. $$
\triangle A B C \text { is an equilateral triangle }
$$

$$ \text { The correct match is } $$

A
A B C D
I IV III II
B
A B C D
I II III IV
C
A B C D
V I IV II
D
A B C D
V I III II
2
TS EAMCET 2020 (Online) 11th September Morning Shift
MCQ (Single Correct Answer)
+1
-0

Let $\mathbf{a}=\hat{\mathbf{i}}+2 \hat{\mathbf{j}}-2 \hat{\mathbf{k}}$ and $\mathbf{b}=2 \hat{\mathbf{i}}-\hat{\mathbf{j}}-2 \hat{\mathbf{k}}$. If the orthogonal projection vector of $\mathbf{a}$ on $\mathbf{b}$ be $\mathbf{x}$ and the orthogonal projection vector of $\mathbf{b}$ on $\mathbf{a}$ be $\mathbf{y}$, then $|\mathbf{x}-\mathbf{y}|=$

A

$\frac{4}{9} \sqrt{26}$

B

$\frac{8}{9} \sqrt{10}$

C

$\frac{4}{9} \sqrt{10}$

D

$\frac{8}{9} \sqrt{26}$

3
TS EAMCET 2020 (Online) 11th September Morning Shift
MCQ (Single Correct Answer)
+1
-0

Let $\mathbf{p}, \mathbf{q}, \mathbf{r}$ be three non-coplanar vectors and $\left[\begin{array}{lll}\mathbf{p} & \mathbf{q} & \mathbf{r}] \mathbf{a}=\mathbf{q} \times \mathbf{r},\left[\begin{array}{lll}\mathbf{p} & \mathbf{q} & \mathbf{r}] \mathbf{b}=\mathbf{r} \times \mathbf{p},[ \end{array}[\mathbf{p}\right. \\ \mathbf{q} & \mathbf{r}] \mathbf{c}=\mathbf{p} \times \mathbf{q} \text {. If }\end{array}\right. \mathbf{a}, \mathbf{b}, \mathbf{c}$ denote the coterminous edges of a parallelopiped, then its height with the base having a and $\mathbf{c}$ is

A

$|\mathbf{p}|$

B

$\frac{1}{|\mathbf{a}|}$

C

$\frac{1}{|b|}$

D

$\frac{1}{|\mathbf{q}|}$

4
TS EAMCET 2020 (Online) 11th September Morning Shift
MCQ (Single Correct Answer)
+1
-0

If $\mathbf{b}, \mathbf{c}$ are non collinear vectors, $|\mathbf{c}| \neq 0$, $\mathbf{a} \times(\mathbf{b} \times \mathbf{c})+(\mathbf{a} \cdot \mathbf{b}) \mathbf{b}=(4-2 \beta-\sin \alpha) \mathbf{b}+\left(\beta^2-1\right) \mathbf{c}$ and $(\mathbf{c} \cdot \mathbf{c}) \mathbf{a}=\mathbf{c}$, then the scalars $\alpha$ and $\beta$ are

A

$\alpha=\frac{\pi}{2}+\frac{n \pi}{3}, n \in \mathbf{Z} ; \beta=1$

B

$\alpha=\frac{\pi}{2}+2 n \pi, n \in \mathbf{Z} ; \beta=1$

C

$\alpha=\frac{\pi}{2}+(2 n+1) \frac{\pi}{2}, n \in \mathbf{Z}, \beta=2$

D

$\alpha=(2 n+1) \frac{\pi}{2}, n \in \mathbf{Z}, \beta=\frac{3}{2}$

TS EAMCET Subjects

Browse all chapters by subject