Let alpha, beta, gamma, delta be real numbers | vedclass

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(C) Let the points be $P(1, 0, -1)$ and $Q(3, 2, -1)$. The mirror image of $P$ with respect to the plane is $Q$. The midpoint $R$ of the line segment

Vedclass · Accepted Answer

$A, B, C$

Vedclass · Answer

(C) Let the points be $P(1, 0, -1)$ and $Q(3, 2, -1)$. The mirror image of $P$ with respect to the plane is $Q$. The midpoint $R$ of the line segment $PQ$ lies on the plane. $R = \left( \frac{1+3}{2}, \frac{0+2}{2}, \frac{-1-1}{2} ight) = (2, 1, -1)$. Since $R$ lies on the plane $\alpha x + \beta y + \gamma z = \delta$,we have: $2\alpha + \beta - \gamma = \delta$  --- $(1)$ The normal vector to the plane is $\vec{n} = (\alpha, \beta, \gamma)$. The vector $\vec{PQ} = (3-1, 2-0, -1-(-1)) = (2, 2, 0)$ is parallel to the normal vector $\vec{n}$. Thus,$\frac{\alpha}{2} = \frac{\beta}{2} = \frac{\gamma}{0} = k$ (where $k 
eq 0$). This gives $\alpha = 2k$,$\beta = 2k$,and $\gamma = 0$. Given $\alpha + \gamma = 1$,we have $2k + 0 = 1$,so $k = \frac{1}{2}$. Therefore,$\alpha = 1$,$\beta = 1$,and $\gamma = 0$. Substituting these into $(1)$: $2(1) + 1(1) - 0 = \delta \implies \delta = 3$. Now,check the options: $(A)$ $\alpha + \beta = 1 + 1 = 2$ (True) $(B)$ $\delta - \gamma = 3 - 0 = 3$ (True) $(C)$ $\delta + \beta = 3 + 1 = 4$ (True) $(D)$ $\alpha + \beta + \gamma = 1 + 1 + 0 = 2 
eq \delta$ (False) Thus,statements $A, B, C$ are true.

Point	Plane
$(-6, 0, 0)$	$2x - 3y + 6z - 2 = 0$

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