## Related questions with answers

Points P and Q have position vectors p = 2i − j - 3k and q = i + 4j - k. (a) Find the position vector of the midpoint M of [PQ]. (b) Point R lies on the line (PQ) such that QR = QM. Find the coordinates of R if R and M are distinct points.

Solution

Verifieda)

Given that $P$ and $Q$ have position vectors $p=2i-j-3k$ and $q=i+4j-k$

Let $m=xi+yj+zk$ represents the position vector of $M$ the midpoint of $\left[PQ\right]$

$\overrightarrow {PQ}=q-p=i+4j-k-\left(2i-j-3k\right)=-i+5j+2k$

Since $M$ is the midpoint of $\left[PQ\right]$, then

$\overrightarrow {PM}=\frac{1}{2}\overrightarrow {PQ}=-\frac{1}{2}i+\frac{5}{2}j+k$

$\overrightarrow {OM}=\overrightarrow {OP}+\overrightarrow {PM}$

$xi+yj+zk=2i-j-3k-\frac{1}{2}i+\frac{5}{2}j+k=\frac{3}{2}i-\frac{3}{2}j-2k$

the position vector of $M$ the midpoint of $\left[PQ\right]$ is

$m=\frac{3}{2}i-\frac{3}{2}j-2k$

b)

Given that the point $R$ lies on the line $\left(PQ\right)$ such that

$QR=QM$

Let $r=xi+yj+zk$ represent the position vector of the point $R$

$\overrightarrow {OR}=\overrightarrow {OP}+\overrightarrow {PR}=\overrightarrow {OP}+\frac{3}{2}\overrightarrow {PQ}$

$xi+yj+zk=2i-j-3k+\frac{3}{2}\left(-i+5j+2k\right)$

$=2i-j-3k+\left(-\frac{3}{2}i+\frac{15}{2}j+3k\right)=\frac{1}{2}i+\frac{13}{2}j$

Then

$R\left(\frac{1}{2},\frac{13}{2},0\right)$

## Create an account to view solutions

## Create an account to view solutions

## Recommended textbook solutions

#### Mathematics for the IB Diploma Standard Level

ISBN: 9781107613065Ben Woolley, Paul Fannon, Vesna Kadelburg#### Big Ideas Math Geometry: A Common Core Curriculum

1st Edition•ISBN: 9781608408399 (1 more)Boswell, Larson## More related questions

1/4

1/7