Question

Find the angle between the given vectors to the nearest tenth of a degree.

Answer 1

Answer:

A

Step-by-step explanation:

Given

u = <6, -1>

u = 6i-j

and

v=<7,-4>

v=7i-4j

The formula for angle is:

Let x be the angle

$cos\ x = \frac{u.v}{||u||.||v||}$

where ||u|| is the length and u.v is the dot product or scalar product of both vectors

So,

$||u|| = \sqrt{(6)^2+(-1)^2}\\ = \sqrt{36+1}\\ = \sqrt{37}\\ ||v||=\sqrt{(7)^2+(-4)^2}\\ = \sqrt{49+16}\\ = \sqrt{65}$

$u.v = u_1u_2+v_1v_2\\= (6)(7)+(-1)(-4)\\=42+4\\=46$

$cos\ x=\frac{46}{\sqrt{37}\sqrt{65}} \\= \frac{46}{\sqrt{2405} }\\Can\ also\ be\ written\ as:\\= \frac{46}{\sqrt{2405} } * \frac{\sqrt{2405} }{\sqrt{2405}} \\=\frac{46\sqrt{2405} }{2405}$

The calculated angle will be in radians. To find the angle in degrees:

$x = \frac{180}{\pi} cos^{-1} (\frac{46\sqrt{2405} }{2405})\\x = 20.282\\x= 20.3$

Hence Option A is correct ..