Vector perpendicular to the plane

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For what value of a is the vector v=(-2,1,a) perpendicular to the plane z=6x-3y+4?

i looked the book and my teacher's solution they are different so i just want to make sure i did this right...


Vector v is perpendicular to the plane if it is parallel to the plane’s normal vector. A vector which is normal (= perpendicular) to the plane is n=(6,-3, 4) Hence, v parallel n if their coordinates are proportional, i.e. −2/6=1/3=a/4, therefore a=4/3



can someone correct me if i am wrong?PLZ
 
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You got one component of the normal vector wrong.
 
how? z=6x-3y+4 don't u just take the coefficient and that is your normal vector?
 
Do you see a 4z in your equation?
 
aw crap so is it -1 then?
 
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since i have two vectors why can't i do Vector1 . Vector2=0?
since the dot product says a.b=0 is perpendicular to each other?
 
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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