Question

Find the acute angle lies between the vectors a=3i+4j and b=7i+j

Answer 1

Answer:

The acute angle lies between the vectors a=3i+4j and b=7i+j is 45°

Step-by-step explanation:

The given vectors are:

a = 3i + 4j

b = 7i + j

The acute angle between vectors a and and b is given by the formula:

`\theta=\cos ^(-1)(a.b)/(|a\mleft\Vert b\mright|)`

The scalar product of vectors a and b is:

a.b = (3i + 4j).(7i + j)

a.b = (3x7) + (4x1)

a.b = 21 + 4

a.b = 25

The magnitude of a is:

`\begin{gathered} |a|=\sqrt[]{3^2+4^2} \\ |a|=\sqrt[]{9+16} \\ |a|=\sqrt[]{25} \\ |a|=5 \end{gathered}`

The magnitude of b is:

`\begin{gathered} |b|=\sqrt[]{7^2+1^2} \\ |b|=\sqrt[]{49+1} \\ |b|=\sqrt[]{50} \\ |b|=5\sqrt[]{2} \end{gathered}`

Substituting the values of a.b, |a|, and |b| into the formula for the acute angle.

`\begin{gathered} \theta=\cos ^(-1)(a.b)/(|a\Vert b|) \\ \theta=\cos ^(-1)\frac{25}{5*5\sqrt[]{2}} \\ \theta=\cos ^(-1)\frac{25}{25\sqrt[]{2}} \\ \theta=\cos ^(-1)\frac{1}{\sqrt[]{2}} \\ \theta=45^0 \end{gathered}`

Therefore, the acute angle lies between the vectors a=3i+4j and b=7i+j is 45°