Subjective Type

Find the position vector of a point $$R$$ which divides the line joining two points $$P$$ and $$Q$$ whose position vectors are $$\hat {i}+2\hat {j}-\hat {k}$$ and $$-\hat {i}+\hat {j}+\hat {k}$$ respectively, in the ratio $$2:1$$, (i) internally (ii) externally

Solution

The position vector of point $$R$$ dividing the line segment joining two points $$P$$ and $$Q$$ in the ratio $$m:n$$ is given by:
i. Internally:
$$\dfrac {m\vec {Q}+n\vec {P}}{m+n}$$
ii. Externally:
$$\dfrac {m\vec {Q}-n\vec {P}}{m-n}$$
Position vectors of $$P$$ and $$Q$$ are given as:
$$\vec {OP}=\hat {i}+2\hat {j}-\hat {k}$$ and $$\vec {OQ}=-\hat {i}+\hat {j}+\hat {k}$$
(i) The position vector of point $$R$$ which divides the line joining two points $$P$$ and $$Q$$ internally in the ratio $$2:1$$ is given by,
$$\vec {OR}=\dfrac {2(-\hat {i}+\hat {j}+\hat {k})+1(\hat {i}+2\hat {j}-\hat {k})}{2+1}=\dfrac {-2\hat {i}+2\hat {j}+\hat {k}+(\hat {i}+2\hat {j}-\hat {k})}{3}$$
$$=\dfrac {-\hat {i}+4\hat {j}+\hat {k}}{3}=-\dfrac {1}{3}\hat {i}+\dfrac {4}{3}\hat {j}+\dfrac {1}{3}\hat {k}$$
(ii) The position vector of point $$R$$ which divides the line joining two points $$P$$ and $$Q$$ externally in the ratio $$2:1$$ is given by,
$$\vec {OR}=\dfrac {2(-\hat {i}+\hat {j}+\hat {k})-1(\hat {i}+2\hat {j}-\hat {k})}{2-1}=(-2\hat {i}+2\hat {j}+2\hat {k})-(\hat {i}+2\hat {j}-\hat {k})$$
$$=-3\hat {i}+3\hat {k}$$


SIMILAR QUESTIONS

Vector Algebra

Let $$\mathrm{A}\mathrm{B}\mathrm{C}$$ be a triangle and let $$\mathrm{S}$$ be its circumcentre and $$\mathrm{O}$$ be its orthocentre. The $$\overline{\mathrm{S}\mathrm{A}}+\overline{\mathrm{S}\mathrm{B}}+\overline{\mathrm{S}\mathrm{C}}= $$

Vector Algebra

If $$A(\overline{a})$$ , $$B(\overline{b})$$ and $$C(\overline{c})$$ be the vertices of a triangle $$ABC$$ whose circumcentre is the origin then orthocentre is given by

Vector Algebra

Taking $$O$$' as origin and the position vectors of $$A, B$$ are $$\vec i+3\vec{j}-2\vec k, 3\vec{i}+\vec{j}-2\vec{k}$$. The vector $$\overrightarrow{OC}$$ is bisecting the angle $$AOB$$ and if $$C$$ is a point on line $$\overrightarrow{AB}$$ then $$C$$ is

Vector Algebra

$$ABCD$$ is a quadrilateral, $$E$$ is the point of intersection of the line joining the midpoints of the opposite sides. If $$O$$ is any point and $$\vec{OA} + \vec{OB} + \vec{OC} + \vec{OD} = \vec{x OE},$$ then $$x$$ is equal to

Vector Algebra

Let $$ABC$$ be a triangle whose circumcentre is at P. If the position vectors of $$A, B, C$$ and P are $$\vec {a}, \vec {b}, \vec {c}$$ and $$\dfrac {\vec {a} + \vec {b} + \vec {c}}{4}$$ respectively, then the position vector of the orthocentre of this triangle, is:

Vector Algebra

Find the position vector of the mid point of the vector joining the points $$P (2, 3, 4)$$ and $$Q (4, 1, 2)$$.

Vector Algebra

In a quadrilateral $$PQRS,\ \vec{PQ}=\vec{a}, \vec{QR}=\vec{b}, \vec{SP}=\vec{a} - \vec{b}.\ M$$ is the mid-point of $$QR$$ and $$X$$ is a point on $$SM$$ such that $$\vec{SX}=\dfrac{4}{5}\vec{SM}$$, then $$\vec{PX}$$ is

Vector Algebra

$$ABCD$$ is a quadrilateral, $$E$$ is the point of intersection of the line joining the midpoints of the opposite sides. If $$O$$ is any point and $$\vec{OA} + \vec{OB} + \vec{OC} + \vec{OD} = \vec{x OE},$$ then $$x$$ is equal to

Vector Algebra

In triangle $$ABC$$, $$\angle A = 30^o$$, $$H$$ is the orthocentre and $$D$$ is the midpoint of $$BC$$. Segment $$HD$$ is produced to $$T$$ such that $$HD = DT$$. The length $$AT$$ is equal to

Vector Algebra

If $$\overrightarrow{b}$$ is a vector whose initial point divides the join of $$5\widehat{i}$$ and $$5\widehat{j}$$ in the ratio $$k : 1$$ and whose terminal point is the origin and $$|\vec b| \leq \sqrt{37}$$, then $$k$$ lies in the interval

Contact Details