Please anyone help me to explain how to find the standard matrix for a rotation of $\pi/2$ radians about the axis determined by $v = (1,1,1)$ using this formula: $$ \begin{bmatrix} a^2(1-\cos \theta)+\cos \theta & ab(1-\cos\theta)-c\sin\theta & ac(1-\cos\theta)+b\sin\theta \\ ab(1-\cos\theta)+c\sin\theta & b^2(1-\cos\theta)+\cos\theta & bc(1-\cos\theta)-a\sin\theta \\ ac(1-\cos\theta)-b\sin\theta & bc(1-\cos\theta)+a\sin\theta & c^2(1-\cos\theta)+\cos\theta \\ \end{bmatrix} $$ Note: This formula requires that the vector defining the axis of rotation have length $1$

  • 1
    $\begingroup$ What is giving you trouble? Scaling the vector to be unit length? Evaluating trig functions at $\pi/2$? Using a formula? $\endgroup$ – rschwieb Apr 15 '17 at 13:47
  • $\begingroup$ because the length of vector should be 1 but in this case the length of vector isn't 1 $\endgroup$ – Steve Apr 15 '17 at 15:56
  • $\begingroup$ then just normalize it. Normalizing it does not change the axis of rotation... $\endgroup$ – rschwieb Apr 15 '17 at 19:35
  • $\begingroup$ Normalizing the vector gives you $u=[ 1/\sqrt{3}, 1/\sqrt{3}, 1/\sqrt{3}]$ $\endgroup$ – Widawensen Apr 19 '17 at 11:30

As you mention above, you need to normalize the vector first. $$v=(1,1,1) \implies v_{norm} = \frac{v}{\|v\|} = \frac{(1,1,1)}{\sqrt{1^2 + 1^2 + 1^2}} = \left(\frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}}\right)$$

Then, take $\theta=\pi/2$ and $(a,b,c) = (1/\sqrt{3},1/\sqrt{3},1/\sqrt{3})$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.