Cross Product of Two Vectors: Finding Component Along Direction of C

Midas_Touch
I have two vectors A = 2x + 3y - 4z and B = -6x - 4y + z. The problem asks me to find the component of A X B along the direction of C = x - y + z. So I did put A and B into a matrix, but I didn't get the correct answer, which is -14.4. What am I doing wrong?
 
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First, find the vector D = A X B.
Then find the unit vector in the C direction.
Finally dot D with the unit vector of C.
 
mezarashi said:
First, find the vector D = A X B.
Then find the unit vector in the C direction.
Finally dot D with the unit vector of C.

I got -15 instead of -14.4 (the answer from the back of the book). I am not sure why it's a little off.
 
I can only suggest an algebriac mistake.
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
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