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

$\mathbf{1}, \mathbf{m}, \mathbf{n}$ are three unit vectors in a right handed system and $L$ is a line through the points $A, B, C$ whose position vectors are $p \mathbf{1}+7 \mathbf{m}-6 \mathbf{n}, 2 \mathbf{1}+5 \mathbf{m}-4 \mathbf{n}$ and $1+4 \mathbf{m}-3 \mathbf{n}$ respectively. If the equation of the plane containing $L$ and the points ( $-p, p, p+1$ ) is $a x+b y+c z=1$, then $p(a+b+c)=$

A

0

B

$\frac{-40}{19}$

C

$\frac{40}{19}$

D

-6

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

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

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