Question

Find the magnitude of component of 3i-2j k along the vector 12i 3j-4k

Answer 1

Let $\theta$ be the angle between the vectors 3i-2j+k and 12i+3j-4k. Then the magnitude of the component of along 12i+3j-4k is

$|3i-2j+k |\cos \theta=|3i-2j+k | \frac{(3i-2j+k ).(12i+3j-4k )}{|3i-2j+k ||12i+3j-4k|} \\ |3i-2j+k |\cos \theta= \frac{(3i-2j+k ).(12i+3j-4k )}{|12i+3j-4k|} \\ |3i-2j+k |\cos \theta= \frac{36-6-4}{\sqrt{12^2+3^2+4^2}} \\ |3i-2j+k |\cos \theta= \frac{26}{13} \\ |3i-2j+k |\cos \theta=2$