The direction ratios of the line of intersect | vedclass

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(A) The direction ratios of the line of intersection of two planes $a_1x + b_1y + c_1z + d_1 = 0$ and $a_2x + b_2y + c_2z + d_2 = 0$ are given by the

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$3, 1, -2$

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(A) The direction ratios of the line of intersection of two planes $a_1x + b_1y + c_1z + d_1 = 0$ and $a_2x + b_2y + c_2z + d_2 = 0$ are given by the cross product of their normal vectors $\vec{n_1} = (a_1, b_1, c_1)$ and $\vec{n_2} = (a_2, b_2, c_2)$.Given planes are $x - y + z - 5 = 0$ and $x - 3y + 0z - 6 = 0$.The normal vectors are $\vec{n_1} = (1, -1, 1)$ and $\vec{n_2} = (1, -3, 0)$.The direction ratios $(l, m, n)$ are given by $\vec{n_1} 	imes \vec{n_2} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 1 & -1 & 1 \ 1 & -3 & 0 \end{vmatrix}$.$= \hat{i}(0 - (-3)) - \hat{j}(0 - 1) + \hat{k}(-3 - (-1))$$= \hat{i}(3) - \hat{j}(-1) + \hat{k}(-2)$$= 3\hat{i} + 1\hat{j} - 2\hat{k}$.Thus,the direction ratios are $3, 1, -2$.

The direction ratios of the line of intersection of the planes $x-y+z-5=0$ and $x-3y-6=0$ are:

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