Question

Find a unit vector that has the same direction as the given vector. 8i ? j + 4k

b) Find a vector that has the same direction as <-2,4,2> but has length 6

c) If v lies in the first quadrant and makes an angle ?/3 with the positive x-axis and |v| = 8, find v in component form.

Answer 1

Answer with Step-by-step explanation:

a.Let given vector

a=8i-j+4k

$\mid a\mid=\sqrt{(8)^2+(-1)^2+(4)^2}=9$

By using the formula

Magnitude of a vector=$\sqrt{x^2+y^2+z^2}$

Where x=Coefficient of i

y=Coefficient of j

z=Coefficient of k

Unit vector=$\hat{a}=\frac{a}{\mid a\mid}$

By using the formula

The unit vector=$\hat{a}=\frac{8i-j+4k}{9}=\frac{8}{9}i-\frac{1}{9}j+\frac{4}{9}k$

b.Let vector b=-2i+4j+2k

$\mid b\mid=\sqrt{(-2)^2+4^2+2^2}=2\sqrt 6$

$\hat{b}=\frac{-2i+4j+2k}{2\sqrt 6}$

Length of vector=6

Therefore, the vector in the direction of <-2,4,2> with length 6 is given by =$6\hat{b}=6\times \frac{-2i+4j+2k}{2\sqrt 6}$

The vector in the direction of <-2,4,2> with length 6 is given by =$-\sqrt 6(-i+2j+k)$

c.$\theta=\frac{\pi}{3}$

$\mid v\mid=8$

Let v=$v_x i+v_y j$

$v_x=\mid v\mid cos\theta$

$v_x=8cos\frac{\pi}{3}=8\times \frac{1}{2}=4$

$cos\frac{\pi}{3}=\frac{1}{2}$

$v_y=\mid v\mid sin\theta$

$v_y=8\times sin\frac{\pi}{3}=8\times \frac{\sqrt 3}{2}=4\sqrt 3$

$sin\frac{\pi}{3}=\frac{\sqrt 3}{2}$

Therefore, the vector v in component form=$4i+4\sqrt 3j$