The equation of the line passing through (1, | vedclass

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(A) Let the direction ratios of the line be $(l, m, n)$.Since the line is parallel to the planes $x - y + 2z = 5$ and $3x + y + z = 6$,it must be perp

Vedclass · Accepted Answer

$\frac{x - 1}{-3} = \frac{y - 2}{5} = \frac{z - 3}{4}$

Vedclass · Answer

(A) Let the direction ratios of the line be $(l, m, n)$.Since the line is parallel to the planes $x - y + 2z = 5$ and $3x + y + z = 6$,it must be perpendicular to the normals of these planes.The normals are $\vec{n_1} = (1, -1, 2)$ and $\vec{n_2} = (3, 1, 1)$.Thus,$l - m + 2n = 0$ and $3l + m + n = 0$.Using the cross product to find the direction ratios $(l, m, n) = \vec{n_1} 	imes \vec{n_2}$:$l = (-1)(1) - (2)(1) = -3$$m = (2)(3) - (1)(1) = 5$$n = (1)(1) - (-1)(3) = 4$So,the direction ratios are $(-3, 5, 4)$.The line passes through $(1, 2, 3)$,so its equation is $\frac{x - 1}{-3} = \frac{y - 2}{5} = \frac{z - 3}{4}$.

The equation of the line passing through $(1, 2, 3)$ and parallel to the planes $x - y + 2z = 5$ and $3x + y + z = 6$ is:

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