The equation of the straight line 3x + 2y - z | vedclass

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(B) To find the symmetrical form of the line of intersection of two planes $P_1: 3x + 2y - z - 4 = 0$ and $P_2: 4x + y - 2z + 3 = 0$,we first find the

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$\frac{x + 2}{3} = \frac{y - 5}{-2} = \frac{z}{5}$

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(B) To find the symmetrical form of the line of intersection of two planes $P_1: 3x + 2y - z - 4 = 0$ and $P_2: 4x + y - 2z + 3 = 0$,we first find the direction vector $\vec{v}$ of the line,which is the cross product of the normals $\vec{n_1} = (3, 2, -1)$ and $\vec{n_2} = (4, 1, -2)$.$\vec{v} = \vec{n_1} 	imes \vec{n_2} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 3 & 2 & -1 \ 4 & 1 & -2 \end{vmatrix} = \hat{i}(-4 + 1) - \hat{j}(-6 + 4) + \hat{k}(3 - 8) = -3\hat{i} + 2\hat{j} - 5\hat{k}$.We can take the direction ratios as $(3, -2, 5)$.Now,find a point on the line by setting $z = 0$ in the plane equations:$3x + 2y = 4$ and $4x + y = -3$.Multiplying the second equation by $2$: $8x + 2y = -6$.Subtracting the first from this: $5x = -10 \implies x = -2$.Substituting $x = -2$ into $4x + y = -3$: $4(-2) + y = -3 \implies y = 5$.So,the point is $(-2, 5, 0)$.The symmetrical form is $\frac{x - x_1}{a} = \frac{y - y_1}{b} = \frac{z - z_1}{c}$,which gives $\frac{x + 2}{3} = \frac{y - 5}{-2} = \frac{z}{5}$.

The equation of the straight line $3x + 2y - z - 4 = 0$ and $4x + y - 2z + 3 = 0$ in the symmetrical form is:

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