A symmetrical form of the line of intersectio | vedclass

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(B) The given equations of the planes are $x - ay = b$ and $z - cy = d$. To find the direction ratios $(l, m, n)$ of the line of intersection,we note

Vedclass · Accepted Answer

$\frac{x - b - a}{a} = \frac{y - 1}{1} = \frac{z - d - c}{c}$

Vedclass · Answer

(B) The given equations of the planes are $x - ay = b$ and $z - cy = d$. To find the direction ratios $(l, m, n)$ of the line of intersection,we note that the line is perpendicular to the normals of both planes. The normals are $\vec{n_1} = (1, -a, 0)$ and $\vec{n_2} = (0, -c, 1)$. The direction ratios are given by the cross product $\vec{n_1} 	imes \vec{n_2} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 1 & -a & 0 \ 0 & -c & 1 \end{vmatrix} = \hat{i}(-a) - \hat{j}(1) + \hat{k}(-c) = (-a, -1, -c)$. This is equivalent to the direction ratios $(a, 1, c)$. Now,we find a point on the line. Let $y = 1$. Then $x = a + b$ and $z = c + d$. Thus,the point is $(a + b, 1, c + d)$. The symmetric form of the line is $\frac{x - (a + b)}{a} = \frac{y - 1}{1} = \frac{z - (c + d)}{c}$.

$A$ symmetrical form of the line of intersection of the planes $x = ay + b$ and $z = cy + d$ is

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