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Raphael Pimenta · Apr 2, 2007, 09:41 AM

Given that a x b is perpendicular to each one of the vectors a and b, determine a unit vector which is perpendicular to each one of the vectors a = 2i - j + k and b = 3i + 4j - k.
Calculate also the sine of the angle, between a and b.

galactus · Apr 3, 2007, 10:37 AM

To find the sine of the angle between the vectors use the cross product.

$sin({\theta})=\frac{||a\times{b}||}{||a|| \;\ ||b||}$

$\frac{||-3i+5j+11k||}{2\sqrt{39}}$

$||-3i+5j+11k||=\sqrt{(-3)^{2}+5^{2}+11^{2}}=\sqrt{155}$

So, we have:

$\frac{\sqrt{155}}{2\sqrt{39}}=\frac{sqrt{6045}}{78 }$

$sin^{-1}(\frac{\sqrt{6045}}{78})=85.40776 \;\ degrees$

Please, check the calculations.

asterisk_man · Apr 3, 2007, 11:33 AM

Look here for how to calculate the cross product if you didn't know: Cross Product -- from Wolfram MathWorld

galactus already gave you the cross product: -3i+5j+11k
Now you just need to find the normal
The normal of a vector is the vector with each term divided by the magnitude of the vector.
For example, the normal of 3i+4j is 3/5i+4/5j since the magnitude of 3i+4j is 5

(note, I didn't double check any of galactus' work so make sure to do that yourself!)