Question

The position vector of a point $P$ is $r=xi+yj+zk,$ where $x,y,z\in N$ and $u=i+j+k.$ If $r\cdot u=10$, then the possible positions of $P$ are

• A. $72$
• B. $36$
• C. $60$
• D. $108$

Answer 1

The correct option is B $36$

$\overrightarrow{r}=x\hat{i}+y\hat{j}+z\hat{k}$

$\overrightarrow{u}=\hat{i}+\hat{j}+\hat{k}$

$\overrightarrow{r}\cdot \overrightarrow{u}=10$

$(x\hat{i}+y\hat{j}+z\hat{k})\cdot (\hat{i}+\hat{j}+\hat{k})=10$

$x+y+z=10$

HERE x,y,z are natural number So x,y,z cannot be negative and zero as well

so atleast x,y,z should be 1

so values of x,y,z vary from 2 to 8

So combination of values becomes $(\frac{7}{3})+1$

possible values$=\frac{7!}{4!3!}+1$

possible values$=\frac{7\times 6\times 5}{3\times 2\times 1}+1$

possible values$=35+1$

possible values$=36$