A) \[{{\tan }^{-1}}\frac{b}{ac}\]and in the -ve\[45{}^\circ \]direction
B) \[{{\tan }^{-1}}\frac{bc}{a(c-a)}\]and in the +ve\[{{\tan }^{-1}}\frac{bc}{a}\]direction
C) \[{{y}^{2}}=8x\]and in the -ve\[y=x+2\]direction
D) \[{{x}^{2}}+{{y}^{2}}+{{z}^{2}}-6x-12y-2\text{ }z+20=0,\]and in the +ve\[a=\hat{i}+\hat{j}+\hat{k},b=\hat{i}-\hat{j}+2\hat{k}\]direction
Correct Answer: D
Solution :
As the electric field is given by \[a=\hat{i}+\hat{j}+\hat{k},b=\hat{i}-\hat{j}+2\hat{k}\] Along x-direction, \[c=x\hat{i}+(x-2)\hat{j}-\hat{k}\] \[x\] \[R=(3,3\sqrt{3})\] \[\angle PQR\]at \[\sqrt{3}x+y=0\] and is along positive X-direction.
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