Find a vector in the opposite direction of $a = (5, - 12)$ that has magnitude 6.

a. $(30, - 72)$
b. $(- 30, 72)$
c. $LaTeX: \left(\frac{-72}{13},\frac{30}{13}\right)$
d. $LaTeX: \left(\frac{-30}{13},\frac{72}{13}\right)$

The correct answer is D. To see why, view each of the steps below.

Step 1

The magnitude of a is given by $LaTeX: \parallel a \parallel =\sqrt{5^2+(-12)^2}=\sqrt{25+144}=\sqrt{169}=13$

Step 2

A unit vector u in the direction of a can be found evaluating $LaTeX: a\frac{1}{\parallel a \parallel}$

$LaTeX: u=a\frac{1}{\parallel a \parallel}=\left(5,-12\right)\frac{1}{13}=\left(\frac{5}{13},-\frac{12}{13}\right)$

Multiplying u by 6 gives us a vector of magnitude 6 in the direction of a, so we will multiply u by –6 to obtain the desired vector. $LaTeX: -6u=-6\left(\frac{5}{13}.-\frac{12}{13}\right)=\left(\frac{-30}{13},\frac{72}{13}\right)$

All Steps

The magnitude of a is given by $LaTeX: \parallel a \parallel =\sqrt{5^2+(-12)^2}=\sqrt{25+144}=\sqrt{169}=13$

A unit vector u in the direction of a can be found evaluating $LaTeX: a\frac{1}{\parallel a \parallel}$
$LaTeX: u=a\frac{1}{\parallel a \parallel}=\left(5,-12\right)\frac{1}{13}=\left(\frac{5}{13},-\frac{12}{13}\right)$

Step 1

Step 2

Step 3

All Steps