course Phy 241
To obtain the unit vector of the vector that is perpendicular to A=-2.00i +3.00j+4.00k B=3.00i+1.00j-3.00k, I first used the vector product. The equation for the vector product is as follows Cx=AyBz-AzBy Cy=AzBx-AxBz Cz=AxBy-AyBx. The results for A and B given in the problem are as follows:
Cx=(3)(-3)-(4)(1)=-13
Cy=(4)(3)-(-2)(-3)=6
Cz=(-2)(1)-(3)(3)=-9 This adds up to -11
Now we can use the Pythagorean theorem to get the vector C. The square root of the quantity, (-13)^2 + 6^2 + (-9)^2. This is about 16.9. Is the unit vector then just defined as –i+j-k?
Not quite. -i + j - k has magnitude `sqrt(1^2 + 1^2 + 1^2) = sqrt(3), so it is definitely not a unit vector.
A x B = -13 i + 6 j - 9 k so its magnitude is
|A x B| = sqrt(13^2 + 6^2 + 11^2) = sqrt(326), about 18.1 (a little larger than you 16.9, due to the difference in Cz).
The unit vector would then be
A x B / | A x B | = (-13 i + 6 j - 11 k) / sqrt(326) = -13/sqrt(326) i + 6 /sqrt(326)j - 11 / sqrt(326) k.
Approximated to 10 significant figures this is
- 0.7200034083·i + 0.3323092653·j - 0.6092336531·k.