Question

If $\overrightarrow{b}=3\hat{i}-4\hat{i}$ and $\overrightarrow{a}$ are collinear and if a makes an obtuse angle with $\hat{i}$ and $\left|\begin{array}{c}\overrightarrow{a}\end{array}\right|=10$ then $\overrightarrow{a}=$

• A. $6\hat{i}+8\hat{j}$
• B. $-6\hat{i}+8\hat{j}$
• C. $-6\hat{i}-8\hat{j}$
• D. $6\hat{i}-8\hat{j}$

Answer 1

Answer: (B)

$-6\hat{i}+8\hat{j}$

If $\overrightarrow{a}$ is collinear of $\overrightarrow{b}$, then
$\overrightarrow{a}=k\overrightarrow{b}$
We have $\left|\overrightarrow{a}\right|=10$
So $\left|\overrightarrow{kb}\right|=10$
$k\left|\overrightarrow{b}\right|=10$
$k\times 5=10$
$k=2$
Now since $\overrightarrow{a}$ make Obtuse angle with $\hat{i}$

$\therefore$ $\overrightarrow{a}.\hat{i}<0$
$k\overrightarrow{b}\cdot \hat{i}<0$
$k<0$
So $\overrightarrow{a}=-6\hat{i}+8\hat{j}$