A plane P contains the line of intersection o | vedclass

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(B) The equation of the family of planes passing through the intersection of the two given planes is given by $(x+y+z-6) + \lambda(2x+3y+4z+5) = 0$.Si

Vedclass · Accepted Answer

$620$

Vedclass · Answer

(B) The equation of the family of planes passing through the intersection of the two given planes is given by $(x+y+z-6) + \lambda(2x+3y+4z+5) = 0$.Since the plane $P$ passes through the point $(0, 2, -2)$,we substitute these coordinates into the equation:$(0+2-2-6) + \lambda(2(0)+3(2)+4(-2)+5) = 0$$-6 + \lambda(6-8+5) = 0$$-6 + 3\lambda = 0 \implies \lambda = 2$.Substituting $\lambda = 2$ back into the family equation:$(x+y+z-6) + 2(2x+3y+4z+5) = 0$$x+y+z-6 + 4x+6y+8z+10 = 0$$5x+7y+9z+4 = 0$.The distance $d$ of the point $(12, 12, 18)$ from the plane $5x+7y+9z+4 = 0$ is given by:$d = \frac{|5(12) + 7(12) + 9(18) + 4|}{\sqrt{5^2 + 7^2 + 9^2}}$$d = \frac{|60 + 84 + 162 + 4|}{\sqrt{25 + 49 + 81}}$$d = \frac{310}{\sqrt{155}}$.The square of the distance is $d^2 = \frac{310^2}{155} = \frac{96100}{155} = 620$.

$A$ plane $P$ contains the line of intersection of the planes $\vec{r} \cdot (\hat{i}+\hat{j}+\hat{k}) = 6$ and $\vec{r} \cdot (2\hat{i}+3\hat{j}+4\hat{k}) = -5$. If $P$ passes through the point $(0, 2, -2)$,then the square of the distance of the point $(12, 12, 18)$ from the plane $P$ is

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