# directional vector of given line

i know that if line equation is given by the following

      r={a1*x+b*y+c1*z+d=0 and a2*x+b2*y+c2*z+d2=0}


then directional vector is given by the cross product of this two plane or it is determinant of the following form

 i   j  k
a1   b1 c1
a2   b2  c2


i was trying to determine directional vecotor of r which is determined by the following form

 r={y=2;3x-(3)^(1/2)*z=0}


but my result did not match to solution in source,there directional vector is determined like this r={x=3^(1/2)*lambda(i can't write lambda directly from my keyboard),y=2;and z=3*(lambda) and U_r={3^(1/2),0,3) please help me(U_r is a directional vector)

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## 1 Answer

It seems you either didn't plug in the correct values or your only issue is with the arithmetic of a determinant. After inspection the coefficients are: $$(a_1,b_1,c_1)=(0,1,0);$$ $$(a_2,b_2,c_2)=(3,0,-\sqrt{3}).$$ The (pseudo)determinant that gives the cross product is therefore $$\begin{vmatrix}\mathbf{i} & \mathbf{j} & \mathbf{k} \\ 0 & 1 & 0 \\ 3 & 0 & -\sqrt{3} \end{vmatrix} = \left[(1)(-\sqrt{3})-0\cdot 0\right]\mathbf{i}-\left[(0)(-\sqrt{3})-3\cdot 0\right]\mathbf{j}+\left[ 0\cdot 0-3\cdot 1 \right]\mathbf{k}$$ $$=(-\sqrt{3},0,-3).$$

The negative sign is superfluous so it can be taken out and we arrive at the solution in the source.

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For some reason \left[\right] ends up looking like the floor function. That might be an inherent issue with the renderer or just with my browser or something, I dunno. –  anon Jul 24 '11 at 5:53
yes it is right my wrong action was that instead of 1 in first column of matrix i have wrote 2 thanks very much –  dato datuashvili Jul 24 '11 at 5:56
Hmm, wouldn't $\det$ be superfluous if you're using $|\cdot|$ already? –  Guess who it is. Jul 24 '11 at 6:26