The equation of the plane passing through (1, | vedclass

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(B) Let the equation of the plane passing through $(1, 1, 1)$ be $a(x - 1) + b(y - 1) + c(z - 1) = 0$ $\dots (i)$.Since it passes through $(1, -1, -1)

Vedclass · Accepted Answer

$x + y - z - 1 = 0$

Vedclass · Answer

(B) Let the equation of the plane passing through $(1, 1, 1)$ be $a(x - 1) + b(y - 1) + c(z - 1) = 0$ $\dots (i)$.Since it passes through $(1, -1, -1)$, we have $a(1 - 1) + b(-1 - 1) + c(-1 - 1) = 0$, which simplifies to $-2b - 2c = 0$, or $b + c = 0$ $\dots (ii)$.The plane is perpendicular to $2x - y + z + 5 = 0$, so the normal vectors are perpendicular. Thus, $2a - b + c = 0$ $\dots (iii)$.From $(ii)$, $c = -b$. Substituting into $(iii)$, we get $2a - b - b = 0$, which means $2a = 2b$, or $a = b$.Let $a = 1$, then $b = 1$ and $c = -1$.Substituting these into $(i)$, we get $1(x - 1) + 1(y - 1) - 1(z - 1) = 0$.$x - 1 + y - 1 - z + 1 = 0$, which simplifies to $x + y - z - 1 = 0$.

The equation of the plane passing through $(1, 1, 1)$ and $(1, -1, -1)$ and perpendicular to $2x - y + z + 5 = 0$ is:

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