If the plane passing through the points hat{i | vedclass

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(B) The equation of the plane passing through the points $(1, 1, 1)$,$(2, 0, -1)$ and the origin $(0, 0, 0)$ is given by the determinant equation: $\b

Vedclass · Accepted Answer

$\frac{1}{11}(11\hat{i}+9\hat{j}+8\hat{k})$

Vedclass · Answer

(B) The equation of the plane passing through the points $(1, 1, 1)$,$(2, 0, -1)$ and the origin $(0, 0, 0)$ is given by the determinant equation: $\begin{vmatrix} x & y & z \ 1 & 1 & 1 \ 2 & 0 & -1 \end{vmatrix} = 0$ Expanding the determinant,we get: $x(-1 - 0) - y(-1 - 2) + z(0 - 2) = 0$ $-x + 3y - 2z = 0$ or $x - 3y + 2z = 0$ (Equation $i$). The equation of the line passing through the points $(1, 3, -2)$ and $(1, -1, 3)$ is: $\frac{x-1}{1-1} = \frac{y-3}{-1-3} = \frac{z-(-2)}{3-(-2)} = r$ $\frac{x-1}{0} = \frac{y-3}{-4} = \frac{z+2}{5} = r$ Thus,any point on the line is $(1, -4r+3, 5r-2)$. Substituting this point into the plane equation $x - 3y + 2z = 0$: $1 - 3(-4r+3) + 2(5r-2) = 0$ $1 + 12r - 9 + 10r - 4 = 0$ $22r - 12 = 0 \Rightarrow r = \frac{12}{22} = \frac{6}{11}$. Substituting $r = \frac{6}{11}$ back into the point coordinates: $x = 1$,$y = -4(\frac{6}{11}) + 3 = \frac{-24+33}{11} = \frac{9}{11}$,$z = 5(\frac{6}{11}) - 2 = \frac{30-22}{11} = \frac{8}{11}$. The point $A$ is $(1, \frac{9}{11}, \frac{8}{11})$,which in vector form is $\frac{1}{11}(11\hat{i} + 9\hat{j} + 8\hat{k})$. Thus,option $B$ is correct.

If the plane passing through the points $\hat{i}+\hat{j}+\hat{k}$,$2\hat{i}-\hat{k}$ and the origin meets the line passing through the points $\hat{i}+3\hat{j}-2\hat{k}$ and $\hat{i}-\hat{j}+3\hat{k}$ at the point $A$,then $A=$

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