The angle between the line frac{x - 2}{a} = f | vedclass

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(D) The direction ratios of the line are $\vec{v} = (a, b, c)$.The normal vector to the plane $ax + by + cz + 6 = 0$ is $\vec{n} = (a, b, c)$.Since th

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$90$

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(D) The direction ratios of the line are $\vec{v} = (a, b, c)$.The normal vector to the plane $ax + by + cz + 6 = 0$ is $\vec{n} = (a, b, c)$.Since the direction ratios of the line are proportional to the direction ratios of the normal to the plane,the line is perpendicular to the plane.The angle $	heta$ between a line with direction vector $\vec{v}$ and a plane with normal vector $\vec{n}$ is given by $\sin 	heta = \frac{|\vec{v} \cdot \vec{n}|}{|\vec{v}| |\vec{n}|}$.Substituting the values,$\sin 	heta = \frac{|a^2 + b^2 + c^2|}{\sqrt{a^2 + b^2 + c^2} \sqrt{a^2 + b^2 + c^2}} = \frac{a^2 + b^2 + c^2}{a^2 + b^2 + c^2} = 1$.Therefore,$\sin 	heta = 1$,which implies $	heta = 90^o$.

The angle between the line $\frac{x - 2}{a} = \frac{y - 2}{b} = \frac{z - 2}{c}$ and the plane $ax + by + cz + 6 = 0$ is ......... $^o$

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