The direction cosines of the normal drawn to | vedclass

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Question

(A) Let the points be $A(2,-1,5)$,$B(1,-3,4)$,and $C(5,2,1)$.The equation of the plane passing through these three points is given by the determinant

Vedclass · Accepted Answer

$\frac{11}{\sqrt{179}}, \frac{-7}{\sqrt{179}}, \frac{3}{\sqrt{179}}$

Vedclass · Answer

(A) Let the points be $A(2,-1,5)$,$B(1,-3,4)$,and $C(5,2,1)$.The equation of the plane passing through these three points is given by the determinant form:$\begin{vmatrix} x-x_1 & y-y_1 & z-z_1 \ x_2-x_1 & y_2-y_1 & z_2-z_1 \ x_3-x_1 & y_3-y_1 & z_3-z_1 \end{vmatrix} = 0$Substituting the coordinates:$\begin{vmatrix} x-2 & y+1 & z-5 \ 1-2 & -3-(-1) & 4-5 \ 5-2 & 2-(-1) & 1-5 \end{vmatrix} = 0$$\begin{vmatrix} x-2 & y+1 & z-5 \ -1 & -2 & -1 \ 3 & 3 & -4 \end{vmatrix} = 0$Expanding along the first row:$(x-2)(8 - (-3)) - (y+1)(4 - (-3)) + (z-5)(-3 - (-6)) = 0$$(x-2)(11) - (y+1)(7) + (z-5)(3) = 0$$11x - 22 - 7y - 7 + 3z - 15 = 0$$11x - 7y + 3z - 44 = 0$The normal vector to the plane is $\vec{n} = 11\hat{i} - 7\hat{j} + 3\hat{k}$.The direction ratios are $(11, -7, 3)$.The magnitude of the normal vector is $\sqrt{11^2 + (-7)^2 + 3^2} = \sqrt{121 + 49 + 9} = \sqrt{179}$.The direction cosines are $\frac{11}{\sqrt{179}}, \frac{-7}{\sqrt{179}}, \frac{3}{\sqrt{179}}$.Thus,the correct option is $A$.

The direction cosines of the normal drawn to the plane passing through the points $(2,-1,5)$,$(1,-3,4)$,and $(5,2,1)$ are

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