Let a=i+2j+3k, b=2i+3j-5k and c=3i-j+ k be three vectors.

Question

Let $\overset{	o }{a}=\hat{i}+2\hat{j}+3\hat{k},\overset{	o }{b}=2\hat{i}+3\hat{j}-5\hat{k}$ and $\overset{	o }{c}=3\hat{i}-\hat{j}+\lambda \hat{k}$ be three vectors. Let $\overset{	o }{r}$ be a unit vector along $\vec{b}+\vec{c}$. If $\vec{r}\cdot \vec{a}=3$, then $3\lambda $ is equal to:

Accepted Answer

$\begin{gathered} \vec{a}=\hat{i}+2\hat{j}+3\hat{k} \ \vec{b}=2\hat{i}+3\hat{j}-5\hat{k} \ \vec{c}=3\hat{i}-\hat{j}+\lambda \hat{k} \ \vec{b}+\vec{c}=5\hat{i}+2\hat{j}+(\lambda -5)\hat{k} \end{gathered}$

$\vec{r}$ is a unit vector along $\vec{b}+\vec{c}$

$	herefore \, \vec{r}=\frac{5\hat{i}+2\hat{j}+(\lambda -5)\hat{k}}{\sqrt{25+4+(\lambda -5{)}^{2}}}$

Now, $\vec{r}\cdot \vec{a}=3$

$\frac{1}{\sqrt{29+(\lambda -5{)}^{2}}}[5+4+3(\lambda -5)]=3$

Squaring both sides

$\begin{gathered} \Rightarrow \, \frac{1}{29+(\lambda -5{)}^{2}}[9+3(\lambda -5{)}^{2}]=9 \ \Rightarrow \, [3+(\lambda -5){]}^{2}=29+(\lambda -5{)}^{2} \ \Rightarrow \, 9+(\lambda -5{)}^{2}+6(\lambda -5)=29+(\lambda -5{)}^{2} \ \Rightarrow \, 9+6(\lambda -5)=29 \ \Rightarrow \, \lambda =\frac{20}{6}+5=\frac{25}{3} \ 	herefore \, 3\lambda =3	imes \frac{25}{3}=25 \end{gathered}$

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