Three Dimensional Geometry TG EAPCET Previous Year Chapters wise Questions

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To find the unit vector normal to the plane $\pi$, we first calculate the cross product of the vectors $\hat{i}+3\hat{k}$ and $2\hat{i}+\hat{j}-\hat{k}$. The cross product is calculated as follows:$n=\begin{vmatrix}\hat{i}&\hat{j}&\hat{k}\\1&0&3\\2&1&-1\end{vmatrix}=(-3)\hat{i}+7\hat{j}+\hat{k}$Now, we find the unit vector $n_{\;\text{unit}\;}$ by dividing this normal vector by its magnitude. The magnitude of $-3\hat{i}+7\hat{j}+\hat{k}$ is:$\sqrt{(-3)^2+7^2+1^2}=\sqrt{9+49+1}=\sqrt{59}$Thus, the unit vector is:$n_{\text{unit}}=\frac{1}{\sqrt{59}}(-3\hat{i}+7\hat{j}+\hat{k})$The equation of the plane is given by:$-3x+7y+z+d=0$Since the plane passes through the point $(-3,7,1)$, we substitute these coordinates into the plane's equation:$-3(-3)+7(7)+1+d=09+49+1+d=0 d=-59$Therefore, the equation of the plane is:$-3x+7y+z-59=0$ The perpendicular distance $p$ from the origin to the plane can be calculated using the formula for the distance from a point to a plane:$p=\frac{\left|0(-3)+0(7)+0(1)-59\right|}{\sqrt{(-3)^2+7^2+1^2}}=\frac{59}{\sqrt{59}}=\sqrt{59}$Given $p$, we calculate:$\sqrt{p^2+5}=\sqrt{(\sqrt{59})^2+5}=\sqrt{59+5}=\sqrt{64}=8$