(A) Let $\overrightarrow{OA} = 6b - 2a$ and $\overrightarrow{OP} = a - b$. Let $\overrightarrow{OB} = r$.Since $P$ divides $AB$ in the ratio $1 : 2$,b

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(A) Let $\overrightarrow{OA} = 6b - 2a$ and $\overrightarrow{OP} = a - b$. Let $\overrightarrow{OB} = r$.Since $P$ divides $AB$ in the ratio $1 : 2$,by the section formula,the position vector of $P$ is given by:$\overrightarrow{OP} = \frac{1(\overrightarrow{OB}) + 2(\overrightarrow{OA})}{1 + 2}$Substituting the given values:$a - b = \frac{r + 2(6b - 2a)}{3}$$3(a - b) = r + 12b - 4a$$3a - 3b = r + 12b - 4a$$r = 3a - 3b - 12b + 4a$$r = 7a - 15b$Thus,the position vector of $B$ is $7a - 15b$.

Class 12 Mathematics Vector Algebra Basic , Modulus and Algebra of vectors

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$A$ and $B$ are two points. The position vector of $A$ is $6b - 2a$. $A$ point $P$ divides the line segment $AB$ in the ratio $1 : 2$. If $a - b$ is the position vector of $P$,then the position vector of $B$ is given by:

A
$7a - 15b$
B
$7a + 15b$
C
$15a - 7b$
D
$15a + 7b$

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Basic , Modulus and Algebra of vectors

If the position vectors of the points $A, B, C, D$ are $2i + 3j + 5k, i + 2j + 3k, -5i + 4j - 2k$ and $i + 10j + 10k$ respectively,then:

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If $\bar{a}, \bar{b}, \bar{c}$ are non-zero vectors such that no two of them are parallel,and $\bar{a} + \bar{b}$ is parallel to $\bar{c}$,and $\bar{b} + \bar{c}$ is parallel to $\bar{a}$,then $\bar{a} + \bar{b} + \bar{c} = $

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Let $ABCD$ be a quadrilateral. If $E$ and $F$ are the midpoints of the diagonals $AC$ and $BD$ respectively and $(\overrightarrow{AB}-\overrightarrow{BC})+(\overrightarrow{AD}-\overrightarrow{DC})= k \overrightarrow{FE}$,then $k$ is equal to

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If the position vectors of points $P, Q, R,$ and $S$ are $2\hat{i} + 3\hat{j} + 5\hat{k}$,$\hat{i} + 2\hat{j} + 3\hat{k}$,$-5\hat{i} + 4\hat{j} - 2\hat{k}$,and $\hat{i} + 10\hat{j} + 10\hat{k}$ respectively,then:

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If the vectors $\hat{i}+2 \hat{j}+x \hat{k}$ and $y \hat{i}+6 \hat{j}+4 \hat{k}$ are collinear,then the values of $x$ and $y$ are respectively,

EasyMHT CET 2020

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$A$ and $B$ are two points. The position vector of $A$ is $6b - 2a$. $A$ point $P$ divides the line segment $AB$ in the ratio $1 : 2$. If $a - b$ is the position vector of $P$,then the position vector of $B$ is given by:

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If the position vectors of the points $A, B, C, D$ are $2i + 3j + 5k, i + 2j + 3k, -5i + 4j - 2k$ and $i + 10j + 10k$ respectively,then:

If $\bar{a}, \bar{b}, \bar{c}$ are non-zero vectors such that no two of them are parallel,and $\bar{a} + \bar{b}$ is parallel to $\bar{c}$,and $\bar{b} + \bar{c}$ is parallel to $\bar{a}$,then $\bar{a} + \bar{b} + \bar{c} = $

Let $ABCD$ be a quadrilateral. If $E$ and $F$ are the midpoints of the diagonals $AC$ and $BD$ respectively and $(\overrightarrow{AB}-\overrightarrow{BC})+(\overrightarrow{AD}-\overrightarrow{DC})= k \overrightarrow{FE}$,then $k$ is equal to

If the position vectors of points $P, Q, R,$ and $S$ are $2\hat{i} + 3\hat{j} + 5\hat{k}$,$\hat{i} + 2\hat{j} + 3\hat{k}$,$-5\hat{i} + 4\hat{j} - 2\hat{k}$,and $\hat{i} + 10\hat{j} + 10\hat{k}$ respectively,then:

If the vectors $\hat{i}+2 \hat{j}+x \hat{k}$ and $y \hat{i}+6 \hat{j}+4 \hat{k}$ are collinear,then the values of $x$ and $y$ are respectively,

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