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skelet666 [1.2K]
2 years ago
9

Solve the equation e/3 = 7 e= (answer)

Mathematics
1 answer:
Bond [772]2 years ago
5 0

Answer:

21

Step-by-step explanation:

e/3 = 7 you can rewrite as e/3 = 7/1 and cross multiply e*1 = 3*7 so e= 21

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Use the equation a = IaIâ
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Answer:

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

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

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Step-by-step explanation:

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

The magnitude of <u>a</u> 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

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\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 <u>a</u> is

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

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

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

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