1
TS EAMCET 2020 (Online) 11th September Evening Shift
MCQ (Single Correct Answer)
+1
-0
$\mathbf{x}, \mathbf{y}, \mathbf{z}$ are three vectors each of magnitude $\sqrt{2}$ and each making an angle $60^{\circ}$ with one another. If $\mathbf{a}=\mathbf{x} \times(\mathbf{y} \times \mathbf{z}), \mathbf{b}=\mathbf{y} \times(\mathbf{z} \times \mathbf{x}), \mathbf{c}=\mathbf{x} \times \mathbf{y}$, then $\mathbf{x}=$
A

$\frac{1}{2}[(\mathrm{a}+\mathrm{b}) \times \mathrm{c}-(\mathrm{a}+\mathrm{b})]$

B

$\frac{1}{2}[c+a-b]$

C

$\frac{1}{2}[(\mathbf{a}+\mathbf{b}) \times \mathbf{c}+(\mathbf{a}+\mathbf{b})]$

D

$\frac{1}{2}[(\mathbf{a} \times \mathbf{b}) \times \mathbf{c}-\mathbf{a}+\mathbf{b}]$

2
TS EAMCET 2020 (Online) 11th September Evening Shift
MCQ (Single Correct Answer)
+1
-0

Let $\mathbf{a}=2 \hat{\mathbf{i}}-2 \hat{\mathbf{j}}+\hat{\mathbf{k}}, \mathbf{b}=-\hat{\mathbf{j}}+\hat{\mathbf{k}}$. If $\mathbf{c}$ is a vector such that $\mathbf{a} \cdot \mathbf{c}=|\mathbf{c}|,|\mathbf{c}-\mathbf{a}|=2 \sqrt{2}$ and the angle between $\mathbf{a} \times \mathbf{b}$ and $\mathbf{c}$ is $\frac{\pi}{3}$, then $|(\mathbf{a} \times \mathbf{b}) \times \mathbf{c}|=$

A

$3 \sqrt{3}$

B

$\frac{3}{2}$

C

$\frac{3 \sqrt{3}}{2}$

D

0

3
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
4
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

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

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