Let $A$ and $B$ be two points. The position vector of $A$ is $6b - 2a$. Point $P$ divides the line segment $AB$ in the ratio $1 : 2$. If $a - b$ is the position vector of $P$,what is the position vector of $B$?

  • A
    $7a - 15b$
  • B
    $7a + 15b$
  • C
    $22/3$
  • D
    $15a + 7b$

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Let the position vectors of three vertices of a triangle be $4 \overrightarrow{p} + \overrightarrow{q} - 3 \overrightarrow{r}$,$-5 \overrightarrow{p} + \overrightarrow{q} + 2 \overrightarrow{r}$,and $2 \overrightarrow{p} - \overrightarrow{q} + 2 \overrightarrow{r}$. If the position vectors of the orthocenter $(O)$ and the circumcenter $(C)$ of the triangle are $\frac{\overrightarrow{p} + \overrightarrow{q} + \overrightarrow{r}}{4}$ and $\alpha \overrightarrow{p} + \beta \overrightarrow{q} + \gamma \overrightarrow{r}$ respectively,then $\alpha + 2 \beta + 5 \gamma$ is equal to:

Let $a, b, c$ be three non-coplanar vectors such that $r_1 = a - b + c$,$r_2 = b + c - a$,$r_3 = c + a + b$,and $r = 2a - 3b + 4c$. If $r = \lambda_1 r_1 + \lambda_2 r_2 + \lambda_3 r_3$,then:

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The point $B$ divides the arc $AC$ of a quadrant of a circle in the ratio $1 : 2$. If $O$ is the centre and $\overrightarrow{OA} = \mathbf{a}$ and $\overrightarrow{OB} = \mathbf{b}$,then the vector $\overrightarrow{OC}$ is

Let $\vec{w}=\hat{i}+\hat{j}-2 \hat{k}$,and $\vec{u}$ and $\vec{v}$ be two vectors such that $\vec{u} \times \vec{v}=\vec{w}$ and $\vec{v} \times \vec{w}=\vec{u}$. Let $\alpha, \beta, \gamma$ and $t$ be real numbers such that $\vec{u}=\alpha \hat{i}+\beta \hat{j}+\gamma \hat{k}$,$-t \alpha+\beta+\gamma=0$,$\alpha-t \beta+\gamma=0$,and $\alpha+\beta-t \gamma=0$. Match each entry in List-$I$ to the correct entry in List-$II$ and choose the correct option.
List-$I$List-$II$
$(P)$ $|\vec{v}|^2$ is equal to$(1)$ $0$
$(Q)$ If $\alpha=\sqrt{3}$,then $\gamma^2$ is equal to$(2)$ $1$
$(R)$ If $\alpha=\sqrt{3}$,then $(\beta+\gamma)^2$ is equal to$(3)$ $2$
$(S)$ If $\alpha=\sqrt{2}$,then $t+3$ is equal to$(4)$ $3$
$(5)$ $5$

Observe the following lists. Then the correct match for List-$I$ from List-$II$ is:
List-$I$List-$II$
$(A)$ $[\mathbf{a} \mathbf{b} \mathbf{c}]$$1. |\mathbf{a}||\mathbf{b}|\cos(\mathbf{a}, \mathbf{b})$
$(B)$ $(\mathbf{c} \times \mathbf{a}) \times \mathbf{b}$$2. (\mathbf{a} \cdot \mathbf{c})\mathbf{b} - (\mathbf{a} \cdot \mathbf{b})\mathbf{c}$
$(C)$ $\mathbf{a} \times (\mathbf{b} \times \mathbf{c})$$3. \mathbf{a} \cdot (\mathbf{b} \times \mathbf{c})$
$(D)$ $\mathbf{a} \cdot \mathbf{b}$$4. |\mathbf{a}||\mathbf{b}|$
$5. (\mathbf{b} \cdot \mathbf{c})\mathbf{a} - (\mathbf{a} \cdot \mathbf{b})\mathbf{c}$

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