Let Pi be a plane containing the points (0,-5 | vedclass

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(B) The equation of the plane $\Pi$ passing through points $A(0,-5,-1), B(1,-2,5), C(-3,5,0)$ is given by the determinant equation: $\begin{vmatrix} x

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$\frac{14}{\sqrt{682}}$

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(B) The equation of the plane $\Pi$ passing through points $A(0,-5,-1), B(1,-2,5), C(-3,5,0)$ is given by the determinant equation: $\begin{vmatrix} x-0 & y+5 & z+1 \ 1-0 & -2+5 & 5+1 \ -3-0 & 5+5 & 0+1 \end{vmatrix} = 0$ $\begin{vmatrix} x & y+5 & z+1 \ 1 & 3 & 6 \ -3 & 10 & 1 \end{vmatrix} = 0$ Expanding along the first row: $x(3-60) - (y+5)(1+18) + (z+1)(10+9) = 0$ $-57x - 19(y+5) + 19(z+1) = 0$ Dividing by $-19$: $3x + y + 5 - z - 1 = 0 \Rightarrow 3x + y - z + 4 = 0$ The normal vector to the plane is $\vec{n} = 3\hat{i} + \hat{j} - \hat{k}$. The unit normal vector is $\hat{n} = \frac{3\hat{i} + \hat{j} - \hat{k}}{\sqrt{3^2 + 1^2 + (-1)^2}} = \frac{3\hat{i} + \hat{j} - \hat{k}}{\sqrt{11}}$. The line $L$ is parallel to the vector $\vec{v} = \hat{i} + 5\hat{j} - 6\hat{k}$. The unit vector along the line $L$ is $\hat{u} = \frac{\hat{i} + 5\hat{j} - 6\hat{k}}{\sqrt{1^2 + 5^2 + (-6)^2}} = \frac{\hat{i} + 5\hat{j} - 6\hat{k}}{\sqrt{62}}$. The length of the projection of $\hat{n}$ on line $L$ is $|\hat{n} \cdot \hat{u}|$: $|\hat{n} \cdot \hat{u}| = \left| \left( \frac{3\hat{i} + \hat{j} - \hat{k}}{\sqrt{11}} ight) \cdot \left( \frac{\hat{i} + 5\hat{j} - 6\hat{k}}{\sqrt{62}} ight) ight|$ $= \left| \frac{3(1) + 1(5) + (-1)(-6)}{\sqrt{11} \cdot \sqrt{62}} ight| = \left| \frac{3 + 5 + 6}{\sqrt{682}} ight| = \frac{14}{\sqrt{682}}$.

Let $\Pi$ be a plane containing the points $(0,-5,-1), (1,-2,5), (-3,5,0)$ and $L$ be a line passing through the point $(0,-5,-1)$ and parallel to the vector $\hat{i}+5\hat{j}-6\hat{k}$. Then the length of the projection of the unit normal vector to the plane $\Pi$ on the line $L$ is

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