The equation of a line passing through the po | vedclass

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(B) line equally inclined to the axes has direction cosines $(l, m, n)$ such that $|l| = |m| = |n|$.Since $l^2 + m^2 + n^2 = 1$,we have $3l^2 = 1$,whi

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$x + 3 = y - 2 = z + 4$

Vedclass · Answer

(B) line equally inclined to the axes has direction cosines $(l, m, n)$ such that $|l| = |m| = |n|$.Since $l^2 + m^2 + n^2 = 1$,we have $3l^2 = 1$,which implies $l = \pm \frac{1}{\sqrt{3}}$.Thus,the direction ratios $(a, b, c)$ can be taken as $(1, 1, 1)$ or $(1, -1, 1)$ etc.The standard form of the line passing through $(x_1, y_1, z_1)$ with direction ratios $(a, b, c)$ is $\frac{x - x_1}{a} = \frac{y - y_1}{b} = \frac{z - z_1}{c}$.Substituting the point $(-3, 2, -4)$ and direction ratios $(1, 1, 1)$,we get $\frac{x - (-3)}{1} = \frac{y - 2}{1} = \frac{z - (-4)}{1}$.This simplifies to $\frac{x + 3}{1} = \frac{y - 2}{1} = \frac{z + 4}{1}$,which is equivalent to $x + 3 = y - 2 = z + 4$.

The equation of a line passing through the point $(-3, 2, -4)$ and equally inclined to the axes is:

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