If P,Q,and R are the feet of the perpendicula | vedclass

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(B) The given planes are $P_1: x + 2y + 2z - 2 = 0$ and $P_2: 2x - 2y + z + 8 = 0$. First,we check the angle between the planes by their normals $\vec

Vedclass · Accepted Answer

$1.5$

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(B) The given planes are $P_1: x + 2y + 2z - 2 = 0$ and $P_2: 2x - 2y + z + 8 = 0$. First,we check the angle between the planes by their normals $\vec{n_1} = (1, 2, 2)$ and $\vec{n_2} = (2, -2, 1)$. $\vec{n_1} \cdot \vec{n_2} = (1)(2) + (2)(-2) + (2)(1) = 2 - 4 + 2 = 0$. Since the normals are perpendicular,the planes $P_1$ and $P_2$ are perpendicular. Let $A = (1, 1, 1)$. $P$ is the foot of the perpendicular from $A$ to $P_1$,and $Q$ is the foot of the perpendicular from $A$ to $P_2$. $R$ is the foot of the perpendicular from $A$ to the line of intersection $L = P_1 \cap P_2$. Since $P_1 \perp P_2$,the points $P, Q, R$ form a right-angled triangle at $R$ in the plane containing $A$ and the line $L$. The distance $AP$ is the perpendicular distance from $A$ to $P_1$: $AP = \frac{|1(1) + 2(1) + 2(1) - 2|}{\sqrt{1^2 + 2^2 + 2^2}} = \frac{|3|}{3} = 1$. The distance $AQ$ is the perpendicular distance from $A$ to $P_2$: $AQ = \frac{|2(1) - 2(1) + 1(1) + 8|}{\sqrt{2^2 + (-2)^2 + 1^2}} = \frac{|9|}{3} = 3$. In the rectangle formed by $A, P, R, Q$,the sides are $PR = AQ = 3$ and $QR = AP = 1$. The area of $\Delta PQR = \frac{1}{2} 	imes PR 	imes QR = \frac{1}{2} 	imes 3 	imes 1 = 1.5$.

If $P$,$Q$,and $R$ are the feet of the perpendiculars drawn from the point $A(1, 1, 1)$ to the planes $P_1: x + 2y + 2z = 2$,$P_2: 2x - 2y + z = -8$,and to the line of intersection of $P_1$ and $P_2$ respectively,then the area of $\Delta PQR$ is:

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