Question

Use the equation a = IaIâ

Answer 1

Answer:

a) $\:\:=\sqrt{14}\cdot \frac{\:\:}{\sqrt{14} }$

b)$\:\:=\sqrt{29} \cdot \frac{\:\:}{\sqrt{29} }$

c) $\:\:=7\cdot \frac{\:\:}{7}$

Step-by-step explanation:

a) Let a=<2,1,-3>

The magnitude of a is $|a|=\sqrt{2^2+1^2+(-3)^2}$

$|a|=\sqrt{4+1+9}=\sqrt{14}$

The unit vector in the direction of a is

$\hat{a}=\frac{\:\:}{\sqrt{14} }$

Using the relation $a=|a|\hat{a}$, we have

$\:\:=\sqrt{14}\cdot \frac{\:\:}{\sqrt{14} }$

b) Let a=2i - 3j + 4k

$|a|=\sqrt{2^2+(-3)^2+4^2}$

$|a|=\sqrt{4+9+16}=\sqrt{29}$

$\hat{a}=\frac{\:\:}{\sqrt{29} }$

Using the relation $a=|a|\hat{a}$, we have

$\:\:=\sqrt{29} \cdot \frac{\:\:}{\sqrt{29} }$

c) Let us first find the sum of <1, 2, -3> and <2, 4, 1> to get:

<1+2, 2+4, -3+1>=<3, 6, -2>

Let a=<3, 6, -2>

The magnitude is

$|a|=\sqrt{3^2+6^2+(-2)^2}$

$|a|=\sqrt{9+36+4}=\sqrt{49}=7$

The unit vector in the direction of a is

$\hat{a}=\frac{\:\:}{7}$

Using the relation $a=|a|\hat{a}$, we have

$\:\:=7\cdot \frac{\:\:}{7}$