Question

If the points with position vectors $10\hat{i}+\lambda \hat{j},3\hat{i}-\hat{j}$ and $4\hat{i}+5\hat{j}$ are collinear then $\lambda$ is

• A. $41$
• B. $-41$
• C. $42$
• D. $-42$

Answer 1

The correct option is A $41$

position vectors : $(10\hat{i}+\lambda \hat{j})a,(3\hat{i}-\hat{j})b,(4\hat{i}+5\hat{j})c$

since they are collinear, so

$\overrightarrow{AB}$ must parallel to $\overrightarrow{BC}$

$\therefore \overrightarrow{AB}=b-a=(3-10)\hat{i}+(-1-\lambda )\hat{j})$

$=(-7\hat{i}+(-1-\lambda \left)\hat{j}\right)$

$\overrightarrow{BC}=\overline{C}-a=(4-3)\hat{i}+(5-(-1\left)\right)\hat{j}$

$=\hat{i}+6\hat{j}$

$\therefore \frac{-7}{1}=\frac{-1-\lambda }{6}$

$\therefore -42=-1-\lambda$

$\lambda =-1+42$

$\lambda =41$ (option A)