The equation of the plane,passing through the | vedclass

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(C) Let the point be $(x_1, y_1, z_1) = (-1, 2, -3)$.The direction ratios of the two lines are $(a_1, b_1, c_1) = (3, 2, -4)$ and $(a_2, b_2, c_2) = (

Vedclass · Accepted Answer

$8x + 14y + 13z + 19 = 0$

Vedclass · Answer

(C) Let the point be $(x_1, y_1, z_1) = (-1, 2, -3)$.The direction ratios of the two lines are $(a_1, b_1, c_1) = (3, 2, -4)$ and $(a_2, b_2, c_2) = (2, -3, 2)$.The equation of the plane passing through $(x_1, y_1, z_1)$ and parallel to vectors $\vec{b_1} = (a_1, b_1, c_1)$ and $\vec{b_2} = (a_2, b_2, c_2)$ is given by the determinant:$\begin{vmatrix} x - x_1 & y - y_1 & z - z_1 \ a_1 & b_1 & c_1 \ a_2 & b_2 & c_2 \end{vmatrix} = 0$Substituting the values:$\begin{vmatrix} x + 1 & y - 2 & z + 3 \ 3 & 2 & -4 \ 2 & -3 & 2 \end{vmatrix} = 0$Expanding along the first row:$(x + 1)(2(2) - (-4)(-3)) - (y - 2)(3(2) - (-4)(2)) + (z + 3)(3(-3) - 2(2)) = 0$$(x + 1)(4 - 12) - (y - 2)(6 + 8) + (z + 3)(-9 - 4) = 0$$-8(x + 1) - 14(y - 2) - 13(z + 3) = 0$$-8x - 8 - 14y + 28 - 13z - 39 = 0$$-8x - 14y - 13z - 19 = 0$Multiplying by $-1$,we get $8x + 14y + 13z + 19 = 0$.

The equation of the plane,passing through the point $(-1, 2, -3)$ and parallel to the lines $\frac{x-1}{3} = \frac{y-2}{2} = \frac{z}{-4}$ and $\frac{x}{2} = \frac{y-1}{-3} = \frac{z-2}{2}$,is

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