Question

if $\hat{r}=3\hat{i}+1\hat{j}-5k,\hat{a}=2\hat{i}-\hat{j}+\hat{k},\widehat{b=\hat{i}+3\hat{j}-2\hat{k}}$ and $\hat{c}=-2\hat{i}+\hat{j}-3\hat{k}$ such that $\hat{r}=l\hat{a}+m\hat{b}+n\hat{n}$ then

• A. $m,\frac{1}{2},n$ are in $A.p$
• B. $l,m,n$ are in $A.p$
• C. $l,m,n$ are in $H.p$
• D. $noneoftheoption$

Answer 1

Answer: (D)

$noneoftheoption$

Given $\overrightarrow{r}=3\hat{i}+\hat{j}-5\hat{k},\overrightarrow{a}=2\hat{i}-\hat{j}+\hat{k},\overrightarrow{b}=\hat{i}+\hat{j}-3\hat{k}$

If $\overrightarrow{r}=l\overrightarrow{a}+m\overrightarrow{b}+n\overrightarrow{c}$

$l(2\hat{i}-\hat{j}+\hat{k})+m(\hat{i}+3\hat{j}-2\hat{k})+n(-2\hat{i}+\hat{j}-3\hat{k})$

Comparing $2l+m-2n=3$ ...... $\left(1\right)$

$-2l+3m+n=1$ ..... $\left(2\right)$

$l-2m-3n=-5$ ...... $\left(3\right)$

$\left(1\right)+2\times \left(2\right),7m=5\Rightarrow m=\frac{5}{7}$

$\left(2\right)+\left(3\right),m-2n=-4\Rightarrow n=\frac{33}{14}$

Substituting in $\left(3\right),l=\frac{7}{2}$

For, $m,\frac{1}{2},n$ to be in $A.P,\frac{m+n}{2}=\frac{\frac{5}{7}+\frac{33}{14}}{2}=\frac{301}{98\times 2}\ne \frac{1}{2}$

$\therefore$ It is not correct

For $l,m,n$ are in $A.P,\frac{l+n}{2}=\frac{\frac{7}{2}+\frac{33}{14}}{2}=\frac{164}{4\times 14}=\frac{41}{\sqrt{14}}\ne m$

It is also not correct

For $l,m,n$ to be in $H.P\frac{\frac{1}{l}+\frac{1}{n}}{2}=\frac{\frac{2}{7}+\frac{14}{33}}{2}=\frac{82}{7\times 33}\ne \frac{1}{m}$

Not correct

For $m,l,n$ to be in $G.Pmn=\frac{5}{7}\left(\frac{33}{14}\right)=\frac{165}{98}\ne l^2$

$\therefore$ None of the option matches