Recent content by Kaede_N9

Using your method I almost obtained the correct answer. P(-1,0,1); [x y z]^T = [3 -1 4]^T +t[3 -2 0] 3(x-1) - 2(y-0) - 0(z-1) = 3x -3-2y or -3 = 3x -2 and x = 3+3t y = -1-2t z = 4 --- -3 = 3(3 +3t) - 2(-1-2t) -3 = 11 + 13t -14/13 = t --- x = 3 - 42/13 = -3/13 y = -1 + 28/13 = 15/13 z = 4...

Yes, it was a typo. The point should be (3, -1, 4) , and not (3, 1, -4). For the the line, x= 3+ 3t, y= 1- 2t, z= 4 , should 1 or 4, in y and z respectively, be negative? (since the point, at the time you typed your response, could have been (3, -1, 4) or (3, 1, -4). Thank you for your...

Homework Statement In each case find the shortest distance from the point P to the line, and find the point Q on the line closest to P. P(-1,0,1); [x y z]^T = [3 1 -4]^T +t[3 -2 0]Homework Equations Projection equation The Attempt at a Solution Let P_o be (3,1-4) Vector V = Vector P_o to P...

Okay, I re did it and obtained the direction vector AB which is (2,-1,-1). Using the same principle of t=1/4, I obtained (x,y,z) through the scalar equations, to be (3/2,-3/2,7/4). What did I do wrong?

Homework Statement Find all points C on the line through A(1, -1, 2) and B(2, 0, 1) such that vectors llACll= 2 llBCll Homework Equations Not sure. The Attempt at a Solution I found the equation of the line for vector AB: (1,2,-1) +t(2,0,1) Then found the scalar equation...

Original question: For the following pairs of compounds, predict which one will be the strongest acid and justify your choice. Answer: Between HCl and H2SO4, H2SO4 has a lower pKa (~-9) than HCl (~7) and so it is a stronger acid compared to HCl. I wanted to answer this question without...

Homework Statement Which compound is more acidic? (without using pkA value) HCl or H2SO4 Homework Equations Electronegativity. Inductive effect. The Attempt at a Solution Since sulfur is more electronegative compared to cholrine, H2SO4 is more acidic than HCl... So then I...

For the original integrand, we know that it is finite on the interval from 0 to 1 by plugging in 0 and 1? It seems like the original integrand does not have an asymptote like 1/x^2 does.

Homework Statement Find the values of p for which the integral converges and evaluate the integral for those values of p. ∫ 0->1 1/(x^p) dx Homework Equations None. The Attempt at a Solution First thought: Since we must evaluate 0 to 1, 1/0 is undefined so maybe 1/ (0^0) = 1...

1/(x+1) if x = ∞ 1/(∞+1) ≈ 0 0 ≈ 1/(∞+1) ≈ 1/(∞) ≈ 0

∫0->∞ x/(x^3 + 1) dx. Use comparison theorem to determine whether the integral is convergent of divergent. Homework Equations None. The Attempt at a Solution ∫0->∞ x/(x^3 + 1) dx = ∫0->∞ x/(x^3) dx = ∫0->∞ 1/(x^2) dx From my class I learned that ∫1->∞ 1/(x^2) dx , is convergent But...