Question

The vectors $2\hat{i}+3\hat{j},5\hat{i}+6\hat{j}$ and $8\hat{i}+\lambda \hat{j}$ have their initial points at $(1,1)$. The value of $\lambda$ so that the vectors terminate on one straight line is

• A. 9
• B. 6
• C. 3
• D. 0

Answer 1

The correct option is A 9

Let:

$\overrightarrow{a}=2\overrightarrow{i}+3\overrightarrow{j},\overrightarrow{b}=5\overrightarrow{i}+6\overrightarrow{j},\overrightarrow{c}=8\overrightarrow{i}+\lambda \overrightarrow{j}$

Initial points given (1,1)

$\therefore \overrightarrow{OA}=(2-1)\overrightarrow{i}+(3-1)\overrightarrow{j}=\overrightarrow{i}+2\overrightarrow{j}$

$\overrightarrow{OB}=(5-1)\overrightarrow{i}+(6-1)\overrightarrow{j}=4\overrightarrow{i}+5\overrightarrow{j}$

$\overrightarrow{OC}=(8-1)\overrightarrow{i}+(\lambda -1)\overrightarrow{j}=7\overrightarrow{i}+(\lambda -1)\overrightarrow{j}$

Now,

$\overrightarrow{AB}=\overrightarrow{OB}-\overrightarrow{OA}$

$\Rightarrow \overrightarrow{AB}=(4\overrightarrow{i}+5\overrightarrow{j})-(\overrightarrow{i}+2\overrightarrow{j})$

$\Rightarrow \overrightarrow{AB}=3\overrightarrow{i}+3\overrightarrow{j}$

and

$\overrightarrow{AC}=\overrightarrow{OC}-\overrightarrow{OA}$

$\Rightarrow \overrightarrow{AC}=(7\overrightarrow{i}+(\lambda -1\left)\overrightarrow{j}\right)-(\overrightarrow{i}+2\overrightarrow{j})$

$\Rightarrow \overrightarrow{AC}=6\overrightarrow{i}+(\lambda -3)\overrightarrow{j}$

$As\overrightarrow{AB}and\overrightarrow{AC}terminateonstraightline$

$\therefore \overrightarrow{AB}and\overrightarrow{AC}areparallel$

$\Rightarrow \overrightarrow{AB}=k\cdot \overrightarrow{AC},wherekisscalar.$

$\Rightarrow (3\overrightarrow{i}+3\overrightarrow{j})=k(6\overrightarrow{i}+(\lambda -3\left)\overrightarrow{j}\right)$

Now comparing the coefficients,

$\Rightarrow 3=6kand3=k(\lambda -3)$

$\Rightarrow k=\frac{1}{2}and3=k(\lambda -3)$

Putting the value of k,

$\Rightarrow \lambda =9$