Question

The position vectors of two points $A$ and $B$ are respectively $\hat{i}+\hat{j}-2\hat{k}$ and $\hat{i}-3\hat{j}+\hat{k}$. The position vector of point $C$ on $AB$ such that $AC=15$ unit can be

• A. $\hat{i}-11\hat{j}+7\hat{k}$
• B. $-11\hat{j}+7\hat{k}$
• C. $2\hat{i}-11\hat{j}+7\hat{k}$
• D. $\hat{i}+\hat{j}+17\hat{k}$

Answer 1

Answer: (A)

$\hat{i}-11\hat{j}+7\hat{k}$

Any point on $AB$ , will have the position vector $\hat{i}+\hat{j}-2\hat{k}+t(-4\hat{j}+3\hat{k})$
Let $C$ point be$=(1,1-4t,-2+3t)$
Length $AC=\sqrt{16t^2+(3t{)}^2}=15$
$t=3$
$c=(1,-11,7)$