TS EAMCET 2020 (Online) 11th September Evening Shift | Vector Algebra Question 110 | Mathematics

$\mathbf{r} \cdot \frac{\mathbf{c} \times \mathbf{a}}{|\mathbf{c} \times \mathbf{a}|}=\left|\frac{\mathbf{a} \times \mathbf{b}}{\mathbf{a} \times \mathbf{c}}\right|$

$\mathbf{r} \cdot \frac{\mathbf{a} \times \mathbf{b}}{|\mathbf{a} \times \mathbf{b}|}=\frac{[\mathbf{a} \mathbf{b c}]}{|\mathbf{b} \times \mathbf{c}|}$

$\mathbf{r} \cdot \frac{\mathbf{b} \times \mathbf{c}}{|\mathbf{b} \times \mathbf{c}|}=\frac{[\mathbf{a} \mathbf{b c}]}{|\mathbf{b} \times \mathbf{c}|}$

$\mathbf{r} \cdot[\mathbf{a} \mathbf{b c}] \mathbf{a}=\frac{|\mathbf{b} \times \mathbf{c}|}{|\mathbf{a} \times \mathbf{c}|}$

TS EAMCET 2020 (Online) 11th September Evening Shift

MCQ (Single Correct Answer)

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$\mathbf{x}, \mathbf{y}, \mathbf{z}$ 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 $\mathbf{x}=$

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

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

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

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

TS EAMCET 2020 (Online) 11th September Evening Shift

MCQ (Single Correct Answer)

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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 $|(\mathbf{a} \times \mathbf{b}) \times \mathbf{c}|=$

$3 \sqrt{3}$

$\frac{3}{2}$

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

TS EAMCET 2020 (Online) 11th September Morning Shift

MCQ (Single Correct Answer)

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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	B	C	D
I	IV	III	II

A	B	C	D
I	II	III	IV

A	B	C	D
V	I	IV	II

A	B	C	D
V	I	III	II

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