Question

Find the magnitude of u × v and the unit vector parrallel to u × v in the direction of u × v. u = 2i + 2j - k, v = -i + k

Answer 1

Answer:

magnitude = 3

unit vector = $\frac{2i}{3} - \frac{j}{3} - \frac{2k}{3}$

Explanation:

Given vectors:

u = 2i + 2j - k

v = -i + k = -i + 0j + k

(a) u x v is the cross product of u and v, and is given by;

$u X v = \left[\begin{array}{ccc}i&j&k\\2&2&-1\\-1&0&1\end{array}\right]$

u x v = i(2+0) - j(2 - 1) + k(0 - 2)

u x v = 2i - j - 2k

Now the magnitude of u x v is calculated as follows:

| u x v | = $\sqrt{2^2 + (-1)^2 + (-2)^2}$

| u x v | = $\sqrt{4 + 1 + 4}$

| u x v | = $\sqrt{9}$

| u x v | = 3

Therefore, the magnitude of u x v is 3

(b) The unit vector û parallel to u x v in the direction of u x v is given by the ratio of u x v and the magnitude of u x v. i.e

û = $\frac{u X v}{|u X v|}$        

u x v = 2i - j - 2k        [calculated in (a) above]

|u x v| = 3                   [calculated in (a) above]

∴ û = $\frac{2i - j - 2k}{3}$

∴ û = $\frac{2i}{3} - \frac{j}{3} - \frac{2k}{3}$