The foot of the perpendicular drawn from the | vedclass

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(A) The normal vector $\vec{n_1}$ to the plane $\pi_1$ is the vector joining the point $(1, 1, 1)$ and the foot of the perpendicular $(1, 3, 5)$.$\vec

Vedclass · Accepted Answer

$\frac{\pi}{2}$

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(A) The normal vector $\vec{n_1}$ to the plane $\pi_1$ is the vector joining the point $(1, 1, 1)$ and the foot of the perpendicular $(1, 3, 5)$.$\vec{n_1} = (1-1, 3-1, 5-1) = (0, 2, 4)$.We can simplify this to $\vec{n_1} = (0, 1, 2)$.The equation of plane $\pi_1$ passing through $(1, 3, 5)$ is $0(x-1) + 1(y-3) + 2(z-5) = 0$, which simplifies to $y + 2z - 13 = 0$.For plane $\pi_2$, it passes through points $A(2, 2, -1), B(3, 4, 2), C(3, 3, 0)$.The vectors $\vec{AB} = (1, 2, 3)$ and $\vec{AC} = (1, 1, 1)$ lie on the plane $\pi_2$.The normal vector $\vec{n_2}$ is $\vec{AB} 	imes \vec{AC} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 1 & 2 & 3 \ 1 & 1 & 1 \end{vmatrix} = \hat{i}(2-3) - \hat{j}(1-3) + \hat{k}(1-2) = -\hat{i} + 2\hat{j} - \hat{k}$.So, $\vec{n_2} = (-1, 2, -1)$.The angle $	heta$ between the planes is given by $\cos 	heta = \frac{|\vec{n_1} \cdot \vec{n_2}|}{|\vec{n_1}| |\vec{n_2}|}$.$\vec{n_1} \cdot \vec{n_2} = (0)(-1) + (1)(2) + (2)(-1) = 0 + 2 - 2 = 0$.Since the dot product is $0$, the angle $	heta = \frac{\pi}{2}$.

The foot of the perpendicular drawn from the point $(1, 1, 1)$ to the plane $\pi_1$ is $(1, 3, 5)$. If $(2, 2, -1), (3, 4, 2), (3, 3, 0)$ are three points on the plane $\pi_2$, then the angle between the planes $\pi_1$ and $\pi_2$ is

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