If the image of the point A(1, 1, 1) with res | vedclass

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(B) The formula for the image $B(\alpha, \beta, \gamma)$ of a point $A(x_1, y_1, z_1)$ with respect to the plane $ax + by + cz + d = 0$ is given by: $

Vedclass · Accepted Answer

$-\frac{28}{9}$

Vedclass · Answer

(B) The formula for the image $B(\alpha, \beta, \gamma)$ of a point $A(x_1, y_1, z_1)$ with respect to the plane $ax + by + cz + d = 0$ is given by: $\frac{\alpha - x_1}{a} = \frac{\beta - y_1}{b} = \frac{\gamma - z_1}{c} = -2 \frac{ax_1 + by_1 + cz_1 + d}{a^2 + b^2 + c^2}$ Here,$(x_1, y_1, z_1) = (1, 1, 1)$ and the plane is $4x + 2y + 4z + 1 = 0$. Calculate the value of the expression: $ax_1 + by_1 + cz_1 + d = 4(1) + 2(1) + 4(1) + 1 = 4 + 2 + 4 + 1 = 11$ $a^2 + b^2 + c^2 = 4^2 + 2^2 + 4^2 = 16 + 4 + 16 = 36$ Substitute these values into the formula: $\frac{\alpha - 1}{4} = \frac{\beta - 1}{2} = \frac{\gamma - 1}{4} = -2 	imes \frac{11}{36} = -\frac{11}{18}$ Now,solve for $\alpha, \beta, \gamma$: $\alpha - 1 = 4 	imes (-\frac{11}{18}) = -\frac{22}{9} \implies \alpha = 1 - \frac{22}{9} = -\frac{13}{9}$ $\beta - 1 = 2 	imes (-\frac{11}{18}) = -\frac{11}{9} \implies \beta = 1 - \frac{11}{9} = -\frac{2}{9}$ $\gamma - 1 = 4 	imes (-\frac{11}{18}) = -\frac{22}{9} \implies \gamma = 1 - \frac{22}{9} = -\frac{13}{9}$ Finally,calculate the sum $\alpha + \beta + \gamma$: $\alpha + \beta + \gamma = -\frac{13}{9} - \frac{2}{9} - \frac{13}{9} = -\frac{28}{9}$

If the image of the point $A(1, 1, 1)$ with respect to the plane $4x + 2y + 4z + 1 = 0$ is $B(\alpha, \beta, \gamma)$,then $\alpha + \beta + \gamma =$

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