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Mathematics Question on Vector Algebra

Let, a→=i^+2j^+k^,c→=i^+j^+k^.\overrightarrow{a}=\widehat{i}+2\widehat{j}+\widehat{k}, \overrightarrow{c}=\widehat{i}+\widehat{j}+\widehat{k}. A vector coplanar to a→\overrightarrow{a} and c→\overrightarrow{c} of magnitude 13,\frac{1}{\sqrt{3}}, then the vector is

A

4i^−j^+4k^4\widehat{i}-\widehat{j}+4\widehat{k}

B

4i^−j^−4k^4\widehat{i}-\widehat{j}-4\widehat{k}

C

2i^+j^+k^2\widehat{i}+\widehat{j}+\widehat{k}

D

None of these

Answer

4i^−j^+4k^4\widehat{i}-\widehat{j}+4\widehat{k}

Explanation

Solution

Let vector r→\overrightarrow{r} be coplanar to a→andb→.\overrightarrow{a} and \overrightarrow{b}.
\therefore \hspace25mm \overrightarrow{r}=\overrightarrow{a}+\overrightarrow{b}
\Rightarrow \hspace25mm \overrightarrow{r}=(\widehat{i}+2\widehat{j}+\widehat{k})+t(\widehat{i}-\widehat{j}+\widehat{k})
=i^(1+t)+j^(2−t)+k^(1+t)=\widehat{i}(1+t)+\widehat{j}(2-t)+\widehat{k}(1+t)
The projection of \overrightarrow{r} on \overrightarrow{c}=\frac{1}{\sqrt{3}} \hspace15mm [given]
\Rightarrow \hspace25mm \frac{\overrightarrow{r}.\overrightarrow{c}}{| \overrightarrow{c} |}=\frac{1}{\sqrt{3}}
⇒∣1.(1+t)+1.(2−t)−1.(1+t)∣3=13\Rightarrow \frac{| 1.(1+t)+1.(2-t)-1.(1+t) |}{\sqrt{3}}=\frac{1}{\sqrt{3}}
⇒(2−t)=±1⇒t=1or3\Rightarrow (2-t)=\pm 1\Rightarrow t=1 or 3
when, t =1, we have r→=2i^+j^+2k^\overrightarrow{r} =2\widehat{i}+\widehat{j}+2\widehat{k}
when, t =3, we have r→=4i^−j^+4k^\overrightarrow{r} =4\widehat{i}-\widehat{j}+4\widehat{k}