Question

The points with position vectors $20\hat{i}+p\hat{j},5\hat{i}-\hat{j}$ and $10\hat{i}-13\hat{j}$ are collinear. The value of $p$ is

• A. $7$
• B. $-37$
• C. $-7$
• D. $37$

Answer 1

The correct option is B $-37$

$20\hat{i}+p\hat{j},5\hat{i}-\hat{j}$ and $10\hat{i}-13\hat{j}$ are collinear.

$\therefore \left[abc\right]=0\to$ property

$\therefore a.(b\times c)=0$

$\therefore (20\hat{i}+p\hat{j}).\left(\right(5\hat{i}-\hat{j})\times (10\hat{i}-13\hat{j}\left)\right)=0$

$\therefore (20\hat{i}+p\hat{j}).(-65\hat{k}+10\hat{k})=0$

We can't get up from this

Now,

By using slope method

$\frac{-1-p}{5-20}=\frac{-13-(-1)}{10-5}$

$\therefore \frac{-1-p}{-15-3}=\frac{-12}{5}$

$\therefore \frac{p+1}{3}=-12$

$\therefore p=-36-1=-37$