TS EAMCET 2023 (Online) 12th May Evening Shift | Vector Algebra Question 26 | Mathematics

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

$\sin 1$

$\cos 1$

TS EAMCET 2023 (Online) 12th May Morning Shift

MCQ (Single Correct Answer)

-0

If the position vectors of $\mathbf{P}$ and $\mathbf{Q}$ are $\hat{\mathbf{i}}+2 \hat{\mathbf{j}}-7 \hat{\mathbf{k}}$ and $5 \hat{\mathbf{i}}-3 \hat{\mathbf{j}}+4 \hat{\mathbf{k}}$ respectively, then the cosine of the angle between $P Q$ and $Z$-axis is

$\frac{4}{\sqrt{162}}$

$\frac{11}{\sqrt{162}}$

$\frac{5}{\sqrt{162}}$

$\frac{-5}{\sqrt{162}}$

TS EAMCET 2023 (Online) 12th May Morning Shift

MCQ (Single Correct Answer)

-0

$\mathbf{a}, \mathbf{b}, \mathbf{c}$ are three-unit yectors such that $|\mathbf{a}+\mathbf{b}+\mathbf{c}|=1$ and $\mathbf{a}$ is perpendicular to $\mathbf{b}$. If $\mathbf{c}$ makes angles $\alpha, \beta$ with $\mathbf{a}, \mathbf{b}$ respectively, then $\cos \alpha+\cos \beta=$

-1

-2

TS EAMCET 2023 (Online) 12th May Morning Shift

MCQ (Single Correct Answer)

-0

If $\mathbf{a}$ is a vector such that $\mathbf{a} \times \hat{\mathbf{i}}=\hat{\mathbf{j}}+\hat{\mathbf{k}}$ and $\mathbf{a} \cdot \hat{\mathbf{i}}=1$, then equation of the line passing through the point $\hat{\mathbf{i}}+\hat{\mathbf{j}}+\hat{\mathbf{k}}$ and parallel to $\mathbf{a}$ is

$\mathbf{r}=(t+1) \hat{i}+(1-t) \hat{j}+(t+1) \hat{k}$

$r=(t+1) \hat{\mathbf{i}}-(2 t-1) \hat{\mathbf{j}}+t \hat{\mathbf{k}}$

$\mathbf{r}=\hat{\mathbf{i}}+t \hat{\mathbf{j}}-t \hat{\mathbf{k}}$c

$\mathbf{r}=5 t \hat{\mathbf{i}}+7 t \hat{\mathbf{j}}+\hat{\mathbf{k}}$

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