The direction cosines of the normal to the pl | vedclass

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(B) The equation of the plane is given by $2x + 3y - 6z = 5$.Comparing this with the general equation of a plane $Ax + By + Cz = D$,the direction rati

Vedclass · Accepted Answer

$\frac{2}{7}, \frac{3}{7}, -\frac{6}{7}$

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(B) The equation of the plane is given by $2x + 3y - 6z = 5$.Comparing this with the general equation of a plane $Ax + By + Cz = D$,the direction ratios of the normal to the plane are $(A, B, C) = (2, 3, -6)$.To find the direction cosines $(l, m, n)$,we divide the direction ratios by the magnitude of the normal vector,which is $\sqrt{A^2 + B^2 + C^2}$.Magnitude $= \sqrt{2^2 + 3^2 + (-6)^2} = \sqrt{4 + 9 + 36} = \sqrt{49} = 7$.Therefore,the direction cosines are $\left( \frac{2}{7}, \frac{3}{7}, -\frac{6}{7} \right)$.

The direction cosines of the normal to the plane $2x + 3y - 6z = 5$ are

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