The definition of similar triangles says that 2 triangles are similar if they have the same shape but different size. There are two criteria to check for this:
1) If all angles in one triangle are equal to the angles in another one, then the 2 are equal.
2) If the sides have the same proportions, then the 2 triangles are similar.
1) We have that all the angles of the 2 triangles have an equal angle in the other triangle. In specific, Q is matched to B, P to A and R to C. Hence, since corresponding angles are congruent, the two triangles are similar.
2) Here we are given information about the sides of the triangles, so we will check the second criterion. We form the ratio of the largest sides of each trangle and the shortest sides. 30/5=6. For the shortest sides, 18/3=6. Finally for the middle sides, 24/4=6. Hence, we have that the triangles are similar since the ratios are equal. (it doesn't matter whether we take the bigger or the smaller side as a numerator, as long as we are consistent).
To find W⊥, you can use the Gram-Schmidt process using the usual inner-product and the given 5 independent set of vectors.
<span>Define projection of v on u as </span>
<span>p(u,v)=u*(u.v)/(u.u) </span>
<span>we need to proceed and determine u1...u5 as: </span>
<span>u1=w1 </span>
<span>u2=w2-p(u1,w2) </span>
<span>u3=w3-p(u1,w3)-p(u2,w3) </span>
<span>u4=w4-p(u1,w4)-p(u2,w4)-p(u3,w4) </span>
<span>u5=w5-p(u4,w5)-p(u2,w5)-p(u3,w5)-p(u4,w5) </span>
<span>so that u1...u5 will be the new basis of an orthogonal set of inner space. </span>
<span>However, the given set of vectors is not independent, since </span>
<span>w1+w2=w3, </span>
<span>therefore an orthogonal basis cannot be found. </span>
