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Question

(B) The direction ratios ($D$.$R$.) of the line segment $AB$ are given by $(-1-4, 2-5, 1-(-10)) = (-5, -3, 11)$.Since the plane is perpendicular to $A

Vedclass · Accepted Answer

$10x + 6y - 22z - 135 = 0$

Vedclass · Answer

(B) The direction ratios ($D$.$R$.) of the line segment $AB$ are given by $(-1-4, 2-5, 1-(-10)) = (-5, -3, 11)$.Since the plane is perpendicular to $AB$,the normal vector to the plane is $\vec{n} = (-5, -3, 11)$.The midpoint $P$ of $AB$ is $\left(\frac{4-1}{2}, \frac{5+2}{2}, \frac{-10+1}{2}\right) = \left(\frac{3}{2}, \frac{7}{2}, -\frac{9}{2}\right)$.The equation of the plane passing through $(x_0, y_0, z_0)$ with normal $(a, b, c)$ is $a(x-x_0) + b(y-y_0) + c(z-z_0) = 0$.Substituting the values: $-5(x - \frac{3}{2}) - 3(y - \frac{7}{2}) + 11(z + \frac{9}{2}) = 0$.Multiplying by $2$: $-5(2x - 3) - 3(2y - 7) + 11(2z + 9) = 0$.$-10x + 15 - 6y + 21 + 22z + 99 = 0$.$-10x - 6y + 22z + 135 = 0$.Multiplying by $-1$: $10x + 6y - 22z - 135 = 0$.

The equation of the plane passing through the midpoint of the line segment joining the points $A(4, 5, -10)$ and $B(-1, 2, 1)$ and perpendicular to $AB$ is:

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