Question

a vector is equal to 5 ICAP - 2 J cap - 2 k cap and b vector is equal to ICAP + 3 J cap - 5 k cap find angle between a vector and b vector ​

Answer 1

Answer:

I hope it would help you

thank you

Answer 2

Answer:

$\theta=\cos^{-1}(\frac{12}{5\sqrt{42}})$

Step-by-step explanation:

Given that

$\vec{a}=5i-j-2k,\vec{b}=i+3j-5k$

Angle between two vectors is given by the formula

$\theta=\cos^{-1}(\frac{\vec a\cdot\vec b}{|a||b|})$

Now, we find below values

$\vec a\cdot\vec b=(5i-j-2k)\cdot(i+3j-5k)\\\\=5-3+10\\\\=12$

And the modulus of these vectors are

$|a|=\sqrt{5^2+(-1)^2+(-2)^2}\\\\=\sqrt{25+1+4}\\\\\sqrt{30}$

$|b|=\sqrt{1^2+3^2+(-5)^2}\\\\=\sqrt{1+9+25}\\\\\sqrt{35}$

Therefore, angle between vectors a and b is

$\theta=\cos^{-1}(\frac{12}{\sqrt{30}\sqrt{35}})\\\\\theta=\cos^{-1}(\frac{12}{\sqrt{1050}})\\\\\theta=\cos^{-1}(\frac{12}{5\sqrt{42}})$

Option (3) is correct.