Let v =LeftAngleBracket negative 10, 15 RightAngleBracket. What is the approximate direction angle of One-halfv? 34° 56° 124° 304°

Question

asked Jun 6, 2021 215k views

2 Answers

Momin IqbalAhmed · Answer 1 · 2021-06-09T21:31:54+0000

Answer:

The approximate angle of
$\vec u = (1)/(2)\cdot \vec v$ , where
$\vec v = \langle 10, 15\rangle$ , is 56º.

Explanation:

Let
$\vec v = \langle 10, 15\rangle$ and
$\vec u = (1)/(2)\cdot \vec v$ , then the resulting vector is described below:

$\vec u = (1)/(2)\cdot \langle 10,15\rangle$

$\vec u = \left\langle 5,(15)/(2) \right\rangle$

From Trigonometry, we find that direction angle of this vector is defined by the following inverse trigonometric expression:

$\theta = \tan^(-1)\left((u_(y))/(u_(x)) \right)$ (1)

Where:

u_(x) - x-Component of
$\vec u$ , dimensionless.

u_(y) - y-Component of
$\vec u$ , dimensionless.

If we know that
u_(x) = 5 and
u_(y) = (15)/(2) , then the direction angle of the vector is:

$\theta = \tan^(-1)\left(((15)/(2))/(5) \right)$

$\theta =\tan^(-1) (15)/(10)$

$\theta = \tan^(-1) (3)/(2)$

$\theta \approx 56.310^(\circ)$

The approximate angle of
$\vec u = (1)/(2)\cdot \vec v$ , where
$\vec v = \langle 10, 15\rangle$ , is 56º.