Let the points P, Q and R have position vecto | vedclass

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(B) The plane passes through the origin $O(0, 0, 0)$ and points $Q(1, 3, 4)$ and $R(2, 1, -2)$.The equation of the plane passing through the origin is

Vedclass · Accepted Answer

$3$

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(B) The plane passes through the origin $O(0, 0, 0)$ and points $Q(1, 3, 4)$ and $R(2, 1, -2)$.The equation of the plane passing through the origin is given by the determinant form:$\left| \begin{matrix} x & y & z \ 1 & 3 & 4 \ 2 & 1 & -2 \end{matrix} ight| = 0$Expanding the determinant:$x(3(-2) - 4(1)) - y(1(-2) - 4(2)) + z(1(1) - 3(2)) = 0$$x(-6 - 4) - y(-2 - 8) + z(1 - 6) = 0$$-10x + 10y - 5z = 0$Dividing by $-5$,we get the equation of the plane $OQR$ as:$2x - 2y + z = 0$The distance $d$ of point $P(3, -2, -1)$ from the plane $Ax + By + Cz + D = 0$ is given by $d = \frac{|Ax_1 + By_1 + Cz_1 + D|}{\sqrt{A^2 + B^2 + C^2}}$.Substituting the values:$d = \frac{|2(3) - 2(-2) + 1(-1)|}{\sqrt{2^2 + (-2)^2 + 1^2}}$$d = \frac{|6 + 4 - 1|}{\sqrt{4 + 4 + 1}}$$d = \frac{|9|}{\sqrt{9}} = \frac{9}{3} = 3$Thus,the distance is $3$.

Let the points $P, Q$ and $R$ have position vectors $\overrightarrow{r_1} = 3i - 2j - k, \overrightarrow{r_2} = i + 3j + 4k$ and $\overrightarrow{r_3} = 2i + j - 2k$ respectively relative to an origin $O$. Then the distance of $P$ from the plane $OQR$ is:

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